Changes of state of a gas and Boyle's law

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 11-1Core

   A disposable syringe contains $30.0\,\text{ml}$ of air at a pressure of $101\,\text{kPa}$. The tip of the syringe is closed with a finger, and the plunger is pulled out slowly until the volume of the air is $45.0\,\text{ml}$. The temperature of the air stays the same, and no air enters or leaves the syringe.

   What is the air pressure in the syringe now?

   kPaCheck
   Answer

   $p_2 \approx 67.3\,\text{kPa}$

   Show the solution

   Given

   $V_1 = 30.0\,\text{ml}$, $p_1 = 101\,\text{kPa}$, $V_2 = 45.0\,\text{ml}$, temperature constant, no air enters or leaves the syringe

   Find

   the air pressure $p_2$ in kilopascals

   Principle

   The temperature and the amount of air stay the same, so this is an isothermal change and Boyle's law holds: the pressure is inversely proportional to the volume, that is, the product $pV$ stays constant.

   Formula

   $p_1 V_1 = p_2 V_2$, so $p_2 = \dfrac{p_1 V_1}{V_2}$

   Conversions

   No units need to be converted: both volumes are in millilitres, so $\text{ml}$ cancels, and the pressure comes out in the same unit as $p_1$.

   Substitution

   $p_2$$= \dfrac{101\,\text{kPa} \cdot 30.0\,\text{ml}}{45.0\,\text{ml}}$

   $= 67.333\,\text{kPa}$

   Answer

   $p_2 \approx 67.3\,\text{kPa}$

   Three significant figures. All three initial values ($101\,\text{kPa}$, $30.0\,\text{ml}$ and $45.0\,\text{ml}$) have three figures, so nothing limits the result to fewer than three.

   Reasonableness

   The volume becomes 1.5 times larger, so the pressure must decrease by the same factor: $101\,\text{kPa} / 1.5 \approx 67\,\text{kPa}$. The pressure is below atmospheric pressure, so the outside air pushes the plunger inwards, which you can feel when pulling the plunger. The product $pV$ before and after is $3030\,\text{kPa}\cdot\text{ml}$ in both cases.

2. Exercise 11-2Core

   The pressure tank of a summer cottage's water system has a sealed air space at the top. When the tank has little water, the volume of the air is $24.0\,\textsf{l}$ and the total pressure is $160.0\,\text{kPa}$. The pump pumps more water into the tank, so the air is compressed and its total pressure rises to $280.0\,\text{kPa}$. The temperature of the air stays the same.

   What is the volume of the air now?

   lCheck
   Answer

   $V_2 \approx 13.7\,\textsf{l}$

   Show the solution

   Given

   $V_1 = 24.0\,\textsf{l}$, $p_1 = 160.0\,\text{kPa}$, $p_2 = 280.0\,\text{kPa}$ (total pressures), temperature constant, the air in the tank is sealed

   Find

   the air volume $V_2$ in litres

   Principle

   The air at the top of the tank is sealed and its temperature stays the same, so the volume and pressure of the air follow Boyle's law. The rising water compresses the air and raises its pressure.

   Formula

   $p_1 V_1 = p_2 V_2$, so $V_2 = \dfrac{p_1 V_1}{p_2}$

   Conversions

   No units need to be converted: the kilopascals cancel in the quotient, and the volume comes out directly in litres.

   Substitution

   $V_2$$= \dfrac{160.0\,\text{kPa} \cdot 24.0\,\textsf{l}}{280.0\,\text{kPa}}$

   $= 13.714\,\textsf{l}$

   Answer

   $V_2 \approx 13.7\,\textsf{l}$

   Three significant figures. The limiting value is the volume $24.0\,\textsf{l}$ (three figures); the pressures $160.0\,\text{kPa}$ and $280.0\,\text{kPa}$ are given with four figures, so they do not limit the result.

   Reasonableness

   The pressure increases, so the volume must decrease. The pressure ratio is $280/160 = 1.75$, and $24.0\,\textsf{l} / 1.75 \approx 13.7\,\textsf{l}$. The product $pV$ is $3840\,\text{kPa}\cdot\textsf{l}$ at the start and at the end. The compressed air acts like a spring that pushes water to the taps even when the pump is not running.

3. Exercise 11-3

   The graph shows example data: the pressure $p$ of the air in a disposable syringe as a function of the volume $V$, when the tip of the syringe is closed and the plunger is moved slowly, so the temperature of the air stays the same.

   What is the air pressure when its volume is $25\,\text{ml}$? Read the values you need from the graph.

   kPaCheck
   Answer

   $p_2 = 80\,\text{kPa}$

   Show the solution

   Given

   the measured point of the graph at a grid crossing, $V_1 = 20\,\text{ml}$, $p_1 = 100\,\text{kPa}$; $V_2 = 25\,\text{ml}$; the temperature and the amount of air stay the same

   Find

   the air pressure $p_2$ in kilopascals

   Principle

   The amount of air and the temperature stay the same, so the air follows Boyle's law: the product $pV$ is constant. So the pressure is not read from a line segment drawn between the points, because the $(V, p)$ graph is a curve and not a straight line.

   Reading the graph

   Choose a point that lies exactly at a grid crossing: $(20\,\text{ml}$; $100\,\text{kPa})$. The point $(40\,\text{ml}$; $50\,\text{kPa})$ also gives the same product $pV = 2000\,\text{kPa}\cdot\text{ml}$, which supports that the temperature stayed the same.

   Formula

   $p_1 V_1 = p_2 V_2$, so $p_2 = \dfrac{p_1 V_1}{V_2}$

   Conversions

   Not needed: the millilitres cancel in the quotient, and the pressure comes out in kilopascals.

   Substitution

   $p_2$$= \dfrac{100\,\text{kPa} \cdot 20\,\text{ml}}{25\,\text{ml}}$

   $= 80\,\text{kPa}$

   Answer

   $p_2 = 80\,\text{kPa}$

   Two significant figures like the volume $25\,\text{ml}$ and the volume $20\,\text{ml}$ read from the graph.

   Reasonableness

   The volume $25\,\text{ml}$ lies between the measured $20\,\text{ml}$ and $30\,\text{ml}$, so the pressure must lie between their pressures $100\,\text{kPa}$ and $67\,\text{kPa}$, and $80\,\text{kPa}$ does. A value read from a line segment drawn between the points would be about $83\,\text{kPa}$, which is too large, because the curve runs below the segment.

4. Exercise 11-4

   A balloon at a birthday party contains $3.0\,\textsf{l}$ of air in a living room at $22\,{}^\circ\text{C}$. The balloon is taken to the balcony in a frost of $-18\,{}^\circ\text{C}$, where its air cools to the temperature of the outside air. The air pressure in the balloon stays the same.

   What is the volume of the air in the balloon on the balcony?

   lCheck
   Answer

   $V_2 \approx 2.6\,\textsf{l}$

   Show the solution

   Given

   $V_1 = 3.0\,\textsf{l}$, $\theta_1 = 22\,{}^\circ\text{C}$, $\theta_2 = -18\,{}^\circ\text{C}$, pressure constant

   Find

   the air volume $V_2$ in litres

   Principle

   The pressure and the amount of air stay the same, so the change is isobaric and the volume is directly proportional to the temperature in kelvins: $V/T$ stays constant. The proportionality holds only for the absolute temperature.

   Formula

   $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$, so $V_2 = \dfrac{V_1 T_2}{T_1}$

   Conversions

   The temperatures in kelvins:

   $T_1$$= (22 + 273.15)\,\text{K}$

   $= 295.15\,\text{K}$

   $T_2$$= (-18 + 273.15)\,\text{K}$

   $= 255.15\,\text{K}$

   Substitution

   $V_2$$= \dfrac{3.0\,\textsf{l} \cdot 255.15\,\text{K}}{295.15\,\text{K}}$

   $= 2.5934\,\textsf{l}$

   Answer

   $V_2 \approx 2.6\,\textsf{l}$

   Two significant figures. The limiting value is the volume $3.0\,\textsf{l}$; the temperatures in kelvins, $295\,\text{K}$ and $255\,\text{K}$ (from whole degrees), have three figures and do not limit the result.

   Reasonableness

   The temperature drops by $40\,\text{K}$, which is about $14\,\%$ of the original $295\,\text{K}$, so the volume decreases by about $14\,\%$, and $3.0\,\textsf{l} \cdot 0.86 \approx 2.6\,\textsf{l}$. The balloon shrinks, which shows as wrinkled rubber. Calculated in degrees Celsius,

   $V_2$$= 3.0\,\textsf{l} \cdot \dfrac{-18}{22}$

   would be negative, which is impossible. So the temperatures must be used in kelvins.

5. Exercise 11-5

   The lid of a chest freezer is opened, and the air space of the freezer fills with room air at $22\,{}^\circ\text{C}$. The lid is closed tight, so the air pressure in the freezer is $101\,\text{kPa}$. The air cools to $-18\,{}^\circ\text{C}$, and its volume stays the same.

   What is the air pressure in the freezer now? (This is why the lid is hard to open again right after closing it.)

   kPaCheck
   Answer

   $p_2 \approx 87.3\,\text{kPa}$

   Show the solution

   Given

   $p_1 = 101\,\text{kPa}$, $\theta_1 = 22\,{}^\circ\text{C}$, $\theta_2 = -18\,{}^\circ\text{C}$, volume constant

   Find

   the air pressure $p_2$ in kilopascals

   Principle

   The lid is tight and the volume of the freezer does not change, so the amount of air and the volume stay the same. The change is isochoric, and the pressure is directly proportional to the temperature in kelvins: $p/T$ stays constant.

   Formula

   $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$, so $p_2 = \dfrac{p_1 T_2}{T_1}$

   Conversions

   The temperatures in kelvins:

   $T_1$$= (22 + 273.15)\,\text{K}$

   $= 295.15\,\text{K}$

   $T_2$$= (-18 + 273.15)\,\text{K}$

   $= 255.15\,\text{K}$

   Substitution

   $p_2$$= \dfrac{101\,\text{kPa} \cdot 255.15\,\text{K}}{295.15\,\text{K}}$

   $= 87.312\,\text{kPa}$

   Answer

   $p_2 \approx 87.3\,\text{kPa}$

   Three significant figures. The pressure $101\,\text{kPa}$ and the temperatures in kelvins, $295\,\text{K}$ and $255\,\text{K}$ (from whole degrees $22\,{}^\circ\text{C}$ and $-18\,{}^\circ\text{C}$), all have three figures.

   Reasonableness

   The absolute temperature decreases by about $14\,\%$, so the pressure decreases by the same amount: $101\,\text{kPa} \cdot 0.86 \approx 87\,\text{kPa}$. The pressure drops by about $14\,\text{kPa}$, so the outside air pushes on the lid harder than the inside air. A completely tight lid of half a square metre would experience an extra force of about $14\,\text{kPa} \cdot 0.5\,\text{m}^2 \approx 7\,\text{kN}$. In practice air leaks past the seals, and the lid opens again after a moment.

6. Exercise 11-6

   A sealed amount of air is in a disposable syringe, and the plunger is moved slowly so that the temperature of the air stays the same.

   What does the graph look like in which the air pressure $p$ is plotted as a function of the volume $V$?

   Choose an option.

   • A falling curve, because the product $pV$ is constant and the pressure decreases as the volume increases.
   • A falling straight line, because the pressure $p$ decreases by the same amount for every added millilitre.
   • A horizontal line, because the temperature $T$ is constant and therefore the pressure is constant too.
   • A rising straight line, because when the volume $V$ increases there are more particles in the air and the pressure increases.

   Answer
   I have done my solution.
   Answer

   A falling curve, because the product $pV$ is constant and the pressure decreases as the volume increases.

   Show the solution

   Find

   what the $(V, p)$ graph of an isothermal change looks like.

   Principle

   The amount of sealed air stays the same and its temperature stays the same, so Boyle's law holds: $pV$ is constant. Then $p = \text{constant} \cdot \dfrac{1}{V}$, that is, the pressure is inversely proportional to the volume.

   Model answer

   The graph is a falling curve, because the product $pV$ is constant: when the volume doubles, the pressure halves, and when the volume halves, the pressure doubles. The curve approaches the axes but never meets them. A falling straight line would be wrong, because the pressure would not decrease by the same amount for every added millilitre. A rising graph is wrong, because the amount of air is the same all the time, and when the volume increases the particles hit the walls less often. A horizontal graph is wrong, because a constant temperature does not mean a constant pressure: when the volume changes, the pressure changes.

   Reading the graph

   The shape of the graph is deduced from the product $pV = \text{constant}$: choose the volumes $V$, $2V$ and $4V$, and the corresponding pressures are $p$, $\dfrac{p}{2}$ and $\dfrac{p}{4}$. The points do not lie on one straight line but form a falling curve.

7. Exercise 11-7

   The cylinder of an air compressor contains $1.50\,\textsf{l}$ of air at a pressure of $101\,\text{kPa}$. The plunger slowly compresses the air into a smaller space until the pressure is $240\,\text{kPa}$. The temperature of the air stays the same, and no air escapes from the cylinder.

   What is the volume of the air now?

   lCheck
   Answer

   $V_2 \approx 0.63\,\textsf{l}$

   Show the solution

   Given

   $V_1 = 1.50\,\textsf{l}$, $p_1 = 101\,\text{kPa}$, $p_2 = 240\,\text{kPa}$, temperature constant, no air escapes

   Find

   the air volume $V_2$ in litres

   Principle

   The amount and the temperature of the air stay the same, so Boyle's law holds: the pressure is inversely proportional to the volume and the product $pV$ stays constant.

   Formula

   $p_1 V_1 = p_2 V_2$, so $V_2 = \dfrac{p_1 V_1}{p_2}$

   Conversions

   No units need to be converted: the kilopascals cancel in the quotient, and the volume comes out directly in litres.

   Substitution

   $V_2$$= \dfrac{101\,\text{kPa} \cdot 1.50\,\textsf{l}}{240\,\text{kPa}}$

   $= 0.6312\,\textsf{l}$

   Answer

   $V_2 \approx 0.63\,\textsf{l}$

   Two significant figures like the pressure $240\,\text{kPa}$.

   Reasonableness

   The pressure increases, so the volume must decrease. The pressure ratio is $240 : 101 \approx 2.4$, and $1.50\,\textsf{l} : 2.4 \approx 0.63\,\textsf{l}$. The product $pV$ is $101\,\text{kPa} \cdot 1.50\,\textsf{l} = 151.5\,\text{kPa}\cdot\textsf{l}$ at the start and $240\,\text{kPa} \cdot 0.6312\,\textsf{l} = 151.5\,\text{kPa}\cdot\textsf{l}$ at the end.

8. Exercise 11-8

   A student measures the pressure $p$ of a sealed amount of air at different volumes $V$ at a constant temperature and plots the points in a $(1/V,\,p)$ graph. Within the measurement uncertainty, the points lie on a straight line through the origin.

   What does this show?

   Choose an option.

   • The temperature rose during the measurements, because the points $(V,\,p)$ are not on a curve.
   • The pressure is directly proportional to the volume, because $1/V$ increases with the volume.
   • The pressure is inversely proportional to the volume, so the product $pV$ is constant.
   • The pressure is inversely proportional to the square of the volume, so the product $pV^2$ is constant.

   Answer
   I have done my solution.
   Answer

   The pressure is inversely proportional to the volume, so the product $pV$ is constant.

   Show the solution

   Find

   what a straight line through the origin in a $(1/V,\,p)$ graph shows.

   Principle

   A straight line through the origin has the form $p = k \cdot \dfrac{1}{V}$, and its slope is $k = pV$. If the points lie, within the measurement uncertainty, on such a line, the product $pV$ is constant.

   Model answer

   The line shows that the pressure is inversely proportional to the volume and that the product $pV$ is constant, that is, Boyle's law holds within the measurement uncertainty. The claim of direct proportionality to the volume is wrong, because when the volume increases the quantity $1/V$ decreases. The claim of a rising temperature is wrong, because the points were plotted in a $(1/V,\,p)$ graph and not in a $(V,\,p)$ graph; a rising temperature would take the points off the line. The claim about the square is wrong, because then the points would lie on a curve and not on a line.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 11-9Core

   The cylinder of a firefighter's breathing apparatus has a volume of $6.8\,\textsf{l}$, and in a fully filled cylinder the air pressure is $300.0\,\text{bar}$ ($1\,\text{bar}$ = $100\,\text{kPa}$).

   How many cubic metres would this air take up if it were released to an atmospheric pressure of $101\,\text{kPa}$? The temperature stays the same. Assume that Boyle's law holds even at such a high pressure.

   m³Check
   Answer

   $V_2 \approx 2.0\,\text{m}^3$

   Show the solution

   Given

   $V_1 = 6.8\,\textsf{l}$, $p_1 = 300.0\,\text{bar}$, $p_2 = 101\,\text{kPa}$, $1\,\text{bar} = 100\,\text{kPa}$, temperature constant

   Find

   the air volume $V_2$ at outside air pressure in cubic metres

   Principle

   The amount of air and the temperature stay the same when the air is released, so Boyle's law holds (the exercise tells us to assume it holds even at high pressure). The air expands until its pressure is the same as the atmospheric pressure.

   Formula

   $p_1 V_1 = p_2 V_2$, so $V_2 = \dfrac{p_1 V_1}{p_2}$

   Conversions

   Pressures to the same unit:

   $p_1$$= 300.0\,\text{bar}$

   $= 300.0 \cdot 100\,\text{kPa}$

   $= 30\,000\,\text{kPa}$

   (the factor $100\,\text{kPa/bar}$ is a definition and does not limit). The result comes out in litres, and $1\,\text{m}^3 = 1000\,\textsf{l}$.

   Substitution

   $V_2$$= \dfrac{30\,000\,\text{kPa} \cdot 6.8\,\textsf{l}}{101\,\text{kPa}}$

   $= 2019.8\,\textsf{l}$

   $= 2.0198\,\text{m}^3$

   Answer

   $V_2 \approx 2.0\,\text{m}^3$

   Two significant figures. The limiting value is the volume $6.8\,\textsf{l}$ (two figures); the pressure $300.0\,\text{bar}$ has four figures and $101\,\text{kPa}$ has three, so neither limits the result.

   Reasonableness

   The pressure is about 300 times the atmospheric pressure, so the volume is about $300 \cdot 6.8\,\textsf{l} \approx 2000\,\textsf{l}$. If a firefighter breathes $50\,\textsf{l}$ per minute, the air lasts about $40\,\text{min}$, which matches real equipment. At such a high pressure air no longer follows Boyle's law exactly: the product $pV$ is about $10\,\%$ larger than the law predicts, so the real volume is about $1.8\,\text{m}^3$.

2. Exercise 11-10Core

   An empty glass bottle is closed with a cork at $20\,{}^\circ\text{C}$, so the air pressure in the bottle is $101\,\text{kPa}$. The bottle is forgotten on the sauna bench. The cork pops out when the air pressure in the bottle is $125\,\text{kPa}$. The volume of the bottle stays the same.

   At what temperature does the cork pop out? Give the answer in degrees Celsius.

   °CCheck
   Answer

   $\theta_2 \approx 90\,{}^\circ\text{C}$

   Show the solution

   Given

   $\theta_1 = 20\,{}^\circ\text{C}$, $p_1 = 101\,\text{kPa}$, $p_2 = 125\,\text{kPa}$, volume constant

   Find

   the temperature $\theta_2$ in degrees Celsius when the cork pops out

   Principle

   The bottle is sealed and its volume stays the same, so the amount of air and the volume are constant. The change is isochoric, and the pressure is directly proportional to the temperature in kelvins: $p/T$ stays constant.

   Formula

   $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$, so $T_2 = \dfrac{T_1 p_2}{p_1}$ and $\theta_2 = T_2 - 273.15\,\text{K}$ in degrees Celsius

   Conversions

   $T_1$$= (20 + 273.15)\,\text{K}$

   $= 293.15\,\text{K}$

   At the end the kelvins are converted back to degrees Celsius by subtracting $273.15$. The pressures do not need to be converted because they are in the same unit.

   Substitution

   $T_2$$= \dfrac{293.15\,\text{K} \cdot 125\,\text{kPa}}{101\,\text{kPa}}$

   $= 362.81\,\text{K}$

   so

   $\theta_2$$= (362.81 - 273.15)\,{}^\circ\text{C}$

   $= 89.66\,{}^\circ\text{C}$

   Answer

   $\theta_2 \approx 90\,{}^\circ\text{C}$

   In whole degrees Celsius. The temperature in kelvins, $T_2 = 362.8\,\text{K} \approx 363\,\text{K}$, has three figures, because the pressures $101\,\text{kPa}$ and $125\,\text{kPa}$ and $T_1 = 293\,\text{K}$ (from the whole degree $20\,{}^\circ\text{C}$) have three figures. When $273.15\,\text{K}$ is subtracted, the precision of the temperature is in units, so the result is given in whole degrees: $90\,{}^\circ\text{C}$.

   Reasonableness

   The pressure increases by a factor of $125/101 \approx 1.24$, so the temperature in kelvins must increase by $24\,\%$: $293\,\text{K} \cdot 1.24 \approx 363\,\text{K}$. The sauna benches are often at 80–$90\,{}^\circ\text{C}$, so the cork can really pop out. If the calculation were done in degrees Celsius, the result would be

   $20\,{}^\circ\text{C} \cdot \dfrac{125}{101}$$\approx 25\,{}^\circ\text{C}$

   and the cork would pop out already in a room. That is not true.

3. Exercise 11-11

   In a physics lesson the tip of a disposable syringe was connected to a pressure gauge, and the plunger of the syringe was moved slowly so that the temperature of the air stayed the same.

   The measured volumes $V$ (ml) were 10.0; 12.5; 15.0; 20.0; 25.0 and 30.0, and the corresponding pressures $p$ (kPa) were 203; 159; 134; 99; 81 and 66.

   Calculate a column $1/V$ in the table, plot $p$ as a function of $1/V$ and fit a straight line through the origin to the points. The slope of the line is $pV$. Give it in joules.

   JCheck
   Answer

   $pV \approx 2.0\,\text{J}$

   Show the solution

   Given

   the measurements in the table $V = 10.0$; $12.5$; $15.0$; $20.0$; $25.0$; $30.0\,\text{ml}$ and $p = 203$; $159$; $134$; $99$; $81$; $66\,\text{kPa}$, temperature constant

   Find

   the slope $pV$ of the straight line fitted to the $(1/V,\,p)$ graph, in joules

   Principle

   The temperature and the amount of air stay the same, so by Boyle's law $pV$ is constant. Then $p = pV \cdot \dfrac{1}{V}$ is a proportionality, and the $(1/V,\,p)$ graph is a straight line through the origin whose slope is $pV$.

   Reading the graph

   Calculate the column $1/V$ (see Conversions) and plot the points $(1/V,\,p)$. The slope of a line through the origin is $p/(1/V) = pV$, so each point gives its own value: $pV = 2030$; $1988$; $2010$; $1980$; $2025$; $1980\,\text{kPa}\cdot\text{ml}$. The line is placed so that the points are spread evenly on both sides of it, and its slope is approximately the mean of these values,

   $\dfrac{12\,013}{6}\,\text{kPa}\cdot\text{ml}$$\approx 2000\,\text{kPa}\cdot\text{ml}$

   Check from the graph: the line passes through the point $(0.100\,\tfrac{1}{\text{ml}}$; $200\,\text{kPa})$, and the measured point $(0.100$; $203)$ is slightly above it. The slope is read from the line and not from a single measured point, because every measurement has an error.

   Formula

   $k$$= \dfrac{\Delta p}{\Delta(1/V)}$

   $= \dfrac{p}{1/V}$

   $= pV$

   so $pV = \dfrac{p}{1/V}$

   Conversions

   $1/V = 0.100$; $0.0800$; $0.0667$; $0.0500$; $0.0400$; $0.0333\,\tfrac{1}{\text{ml}}$. Unit:

   $1\,\text{kPa} \cdot 1\,\text{ml}$$= 1000\,\text{Pa} \cdot 10^{-6}\,\text{m}^3$

   $= 0.001\,\text{J}$

   Substitution

   A point read from the line:

   $pV$$= \dfrac{200\,\text{kPa}}{0.100\,\tfrac{1}{\text{ml}}}$

   $= 2000\,\text{kPa}\cdot\text{ml}$

   $= 2000 \cdot 0.001\,\text{J}$

   $= 2.00\,\text{J}$

   Answer

   $pV \approx 2.0\,\text{J}$

   Two significant figures. The limiting values are the pressures $99\,\text{kPa}$, $81\,\text{kPa}$ and $66\,\text{kPa}$ in the table, which are given with two figures; the volumes are given with three figures, and the three-figure pressures $203$, $159$ and $134\,\text{kPa}$ do not limit the result.

   Reasonableness

   The values of the product $pV$ are $1980$–$2030\,\text{kPa}\cdot\text{ml}$, so they differ from the mean by less than $2\,\%$. So the slope of the line is almost the same as $pV$ for every point, and Boyle's law holds within the accuracy of the measurements. If the law did not hold, the points would not lie on a straight line.

4. Exercise 11-12

   The envelope of a hot-air balloon has a volume of $2800\,\text{m}^3$, and there is an opening at the bottom of the envelope. The envelope is full of air at $10\,{}^\circ\text{C}$, which is heated with a burner to $100\,{}^\circ\text{C}$. The pressure inside the envelope stays equal to the outside air pressure.

   All the air that was originally in the envelope warms to $100\,{}^\circ\text{C}$.

   How many cubic metres of this $100\,{}^\circ\text{C}$ air do not fit in the envelope but have been pushed out through the opening?

   m³Check
   Answer

   $\Delta V \approx 890\,\text{m}^3$

   Show the solution

   Given

   $V_1 = 2800\,\text{m}^3$, $\theta_1 = 10\,{}^\circ\text{C}$, $\theta_2 = 100\,{}^\circ\text{C}$, pressure constant

   Find

   the volume $V_2$ of the original air at $100\,{}^\circ\text{C}$ and the volume $\Delta V$ pushed out of the envelope, in cubic metres

   Principle

   The bottom of the envelope is open, so the pressure stays equal to the outside air pressure. When the same amount of air is heated at constant pressure, the volume is directly proportional to the temperature in kelvins. The volume of the envelope is constant, $V_1$, so the extra air is pushed out through the opening.

   Formula

   $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$, so $V_2 = \dfrac{V_1 T_2}{T_1}$ and the volume pushed out is $\Delta V = V_2 - V_1$

   Conversions

   The temperatures in kelvins:

   $T_1$$= (10 + 273.15)\,\text{K}$

   $= 283.15\,\text{K}$

   $T_2$$= (100 + 273.15)\,\text{K}$

   $= 373.15\,\text{K}$

   Substitution

   $V_2$$= \dfrac{2800\,\text{m}^3 \cdot 373.15\,\text{K}}{283.15\,\text{K}}$

   $= 3690.0\,\text{m}^3$

   so

   $\Delta V$$= 3690.0\,\text{m}^3 - 2800\,\text{m}^3$

   $= 890.0\,\text{m}^3$

   Answer

   $\Delta V \approx 890\,\text{m}^3$

   Two significant figures. The limiting value is the temperature difference $\Delta T = 100\,{}^\circ\text{C} - 10\,{}^\circ\text{C} = 90\,\text{K}$, which as a difference of whole degrees has two figures; the volume $2800\,\text{m}^3$ has at least two figures and does not limit the result. When $\Delta V = V_1 (T_2 - T_1)/T_1$ is calculated directly from the difference, the temperature $T_1 = 283\,\text{K}$ has three figures and the result is $890\,\text{m}^3$.

   Reasonableness

   The temperature in kelvins increases by a factor of $373/283 \approx 1.32$, so the volume increases by about $32\,\%$: $2800\,\text{m}^3 \cdot 0.32 \approx 890\,\text{m}^3$. About a quarter of the warm air ($890/3690 \approx 24\,\%$) does not fit in the envelope. So less air is left in the envelope than when it was cold, and the balloon becomes lighter. A ten-fold volume calculated in degrees Celsius would be absurd.

5. Exercise 11-13

   Two sealed amounts of air are heated in a water bath. In vessel A the pressure $p$ of the air is measured and in vessel B the volume $V$ of the air. The air in both is at the temperature $\theta$ of the water bath (table, example data).

   a) For every row of vessel A calculate the ratios $p/\theta$ and $p/T$, and for every row of vessel B the ratio $V/T$, where $T$ is the temperature in kelvins. Using your calculations, deduce which vessel shows an isochoric and which an isobaric change of state, and justify why the temperature must be in kelvins.

   b) Predict the pressure of vessel A when the water bath boils and its temperature is $100\,{}^\circ\text{C}$.

   Write the result of part b) in the answer box.

   Air pressure in vessel A and volume in vessel B (example data)

   $\theta\,({}^\circ\text{C})$	$p_\text{A}\,(\text{kPa})$	$V_\text{B}\,(\text{ml})$
   20.0	99.8	14.7
   40.0	106.4	15.7
   60.0	113.4	16.6
   80.0	119.8	17.6

   kPaCheck
   Answer

   $p \approx 127\,\text{kPa}$

   Show the solution

   Given

   from the table the temperatures $\theta = 20.0$; $40.0$; $60.0$; $80.0\,{}^\circ\text{C}$, the pressures of vessel A $p_\text{A} = 99.8$; $106.4$; $113.4$; $119.8\,\text{kPa}$ and the volumes of vessel B $V_\text{B} = 14.7$; $15.7$; $16.6$; $17.6\,\text{ml}$

   Find

   a) the changes of state and the comparison of the ratios; b) the pressure $p$ of vessel A at $100\,{}^\circ\text{C}$ in kilopascals

   Principle

   If the pressure of vessel A is directly proportional to the temperature in kelvins, $p/T$ is constant, and the change is isochoric (the volume and the amount of air stay the same). In vessel B the volume changes: if $V/T$ is constant, the pressure is constant and the change is isobaric. The proportionality holds only for the temperature in kelvins.

   Reading the graph

   The task does not require plotting: the points $(T,\,p)$ and $(T,\,V)$ would lie on straight lines through the origin if the ratios are constant, so the ratios are calculated directly from the rows of the table and compared. The temperatures in kelvins are $T = 293.15$; $313.15$; $333.15$; $353.15\,\text{K}$.

   Model answer a)

   In vessel A the ratio $p/T$ is $0.3404$; $0.3398$; $0.3404$; $0.3392\,\tfrac{\text{kPa}}{\text{K}}$, so it is constant to within about $0.3\,\%$. In vessel B the ratio $V/T$ is $0.0501$; $0.0501$; $0.0498$; $0.0498\,\tfrac{\text{ml}}{\text{K}}$, so it is also constant within the measurement uncertainty. So vessel A shows an isochoric change and vessel B an isobaric change. The small variation is due to the random variation of the measurements.
   In degrees Celsius the ratio of vessel A is $p/\theta = 4.99$; $2.66$; $1.89$; $1.50\,\tfrac{\text{kPa}}{{}^\circ\text{C}}$. These decrease sharply, so the pressure is not proportional to the temperature in degrees Celsius but to the temperature in kelvins.

   Formula

   $\dfrac{p}{T} = k$, so $p = kT$

   Conversions

   $T = \theta + 273.15\,\text{K}$, so

   $T_{100}$$= (100 + 273.15)\,\text{K}$

   $= 373.15\,\text{K}$

   The coefficient $k$ is the mean of the four ratios of vessel A.

   Substitution

   The sum of the ratios is $1.3598\,\tfrac{\text{kPa}}{\text{K}}$, so $k = 0.33995\,\tfrac{\text{kPa}}{\text{K}}$.

   $$\begin{aligned} p &= kT \\ &= 0.33995\,\tfrac{\text{kPa}}{\text{K}} \cdot 373.15\,\text{K} \\ &= 126.9\,\text{kPa} \end{aligned}$$

   Answer

   $p \approx 127\,\text{kPa}$

   Three significant figures like the pressures in the table.

   Reasonableness

   The temperature in kelvins increases from $80\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$ in the ratio $373.15 : 353.15 \approx 1.057$, and $119.8\,\text{kPa} \cdot 1.057 \approx 126.6\,\text{kPa}$. The same calculation from the initial state:

   $99.8\,\text{kPa} \cdot \dfrac{373.15}{293.15}$

   $\approx 127.0\,\text{kPa}$

   The result is greater than the last measured value $119.8\,\text{kPa}$, as it should be when the temperature rises.

6. Exercise 11-14

   A student verifies Boyle's law. She connects a disposable syringe to a pressure gauge with a thin hose and moves the plunger slowly so that the temperature stays the same. She reads the volume $V$ of the air from the scale of the syringe and the pressure $p$ from the gauge (table, example data). The student calculates the product $pV$ for every row and notices that it is not constant, although the temperature stayed the same. However, the hose and the gauge also contain air, whose volume $V_\text{h}$ is the same in all the measurements and which the student did not take into account.

   a) Calculate the product $pV$ for every row and explain why it increases with the volume. Is the error random or systematic? Justify your answer.

   b) The total volume of the sealed air is $V + V_\text{h}$, and Boyle's law holds for it. Determine the volume $V_\text{h}$ of the air in the hose and the gauge using the first and the last row of the table.

   Write the result of part b) in the answer box.

   Air pressure as a function of the volume of the syringe (example data)

   $V\,(\text{ml})$	$p\,(\text{kPa})$
   10.0	200
   20.0	109
   30.0	75.0
   40.0	57.1

   mlCheck
   Answer

   $V_\text{h} \approx 2.0\,\text{ml}$

   Show the solution

   Given

   from the table $V = 10.0$; $20.0$; $30.0$; $40.0\,\text{ml}$ and $p = 200$; $109$; $75.0$; $57.1\,\text{kPa}$, temperature constant; the volume $V_\text{h}$ of the air in the hose and the gauge is constant but unknown

   Find

   a) the behaviour of the product $pV$ and the type of error; b) the volume $V_\text{h}$ in millilitres

   Principle

   The sealed air in the syringe and the hose has a total volume $V + V_\text{h}$. The temperature and the amount of air stay the same, so Boyle's law holds for all the air: $p\,(V + V_\text{h})$ is constant. The student calculated the product using only the volume of the syringe.

   Reading the graph

   From the table the product $pV$ is calculated for every row, and for part b) the first row $(10.0\,\text{ml}$; $200\,\text{kPa})$ and the last row $(40.0\,\text{ml}$; $57.1\,\text{kPa})$ are chosen.

   Model answer a)

   The products are $pV = 2000$; $2180$; $2250$; $2284\,\text{kPa}\cdot\text{ml}$. They increase with the volume, because there is in reality always $V_\text{h}$ more air than is read from the scale. When $V$ is small, this extra volume is a large part of the total volume and the product $pV$ stays smaller than the true value. When $V$ is large, the share is smaller and the product is closer to the true value. The error is systematic: it is in the same direction in every measurement and the volume is the same, so it does not vary randomly. Repeating the measurements would not remove it; it is corrected by taking the volume of the hose into account.

   Formula

   $p_1 (V_1 + V_\text{h}) = p_4 (V_4 + V_\text{h})$, so

   $V_\text{h}$$= \dfrac{p_4 V_4 - p_1 V_1}{p_1 - p_4}$

   Conversions

   Not needed: the kilopascals and millilitres are the same on both sides of the equation, and the result comes out in millilitres.

   Substitution

   The products are

   $p_4 V_4$$= 57.1\,\text{kPa} \cdot 40.0\,\text{ml}$

   $= 2284\,\text{kPa}\cdot\text{ml}$

   $p_1 V_1$$= 200\,\text{kPa} \cdot 10.0\,\text{ml}$

   $= 2000\,\text{kPa}\cdot\text{ml}$

   The pressure difference is $p_1 - p_4 = 142.9\,\text{kPa}$.

   $$\begin{aligned} V_\text{h} &= \dfrac{284\,\text{kPa}\cdot\text{ml}}{142.9\,\text{kPa}} \\ &= 1.99\,\text{ml} \end{aligned}$$

   Answer

   $V_\text{h} \approx 2.0\,\text{ml}$

   Two significant figures. The difference of the products $284\,\text{kPa}\cdot\text{ml}$ is small compared with the products, so the result is sensitive to the readings and cannot be given more precisely.

   Reasonableness

   The products $p\,(V + V_\text{h})$ with $V_\text{h} = 2.0\,\text{ml}$ are $200 \cdot 12.0 = 2400$; $109 \cdot 22.0 = 2398$; $75.0 \cdot 32.0 = 2400$ and $57.1 \cdot 42.0 = 2398\,\text{kPa}\cdot\text{ml}$. They are constant to within $0.1\,\%$, so the corrected Boyle's law fits all four rows. A hose volume of two millilitres is reasonable: a thin hose and the inside of the gauge fit in a small space.

7. Exercise 11-15

   To determine absolute zero, a tight rigid bottle containing air is heated in a water bath, and the air pressure $p$ is measured as a function of the temperature $\theta$ (table, example data). The volume of the bottle and the amount of air stay the same.

   a) Plot the points in a $(\theta,\,p)$ graph, fit a straight line to them and extend the line (extrapolate) to the pressure $p = 0$. What temperature in degrees Celsius do you get for absolute zero?

   b) Compare the result with the known value $-273.15\,{}^\circ\text{C}$ and explain why a real gas still does not reach this point, although the extension of the line passes through it.

   Write the result of part a) in the answer box.

   Air pressure in a rigid bottle (example data)

   $\theta\,({}^\circ\text{C})$	$p\,(\text{kPa})$
   0.0	100.1
   20.0	107.3
   40.0	114.9
   60.0	121.8
   80.0	129.1

   °CCheck
   Answer

   $\theta_0 \approx -276\,{}^\circ\text{C}$

   Show the solution

   Given

   from the table $\theta = 0.0$; $20.0$; $40.0$; $60.0$; $80.0\,{}^\circ\text{C}$ and $p = 100.1$; $107.3$; $114.9$; $121.8$; $129.1\,\text{kPa}$, volume and amount of air constant

   Find

   a) the absolute zero $\theta_0$ in degrees Celsius from the extension of the line; b) the comparison with the known value and the behaviour of the gas

   Principle

   The volume and the amount of air are constant, so the pressure is directly proportional to the temperature in kelvins, $p = kT$. So a line fitted to the $(\theta,\,p)$ points reaches the pressure $p = 0$ at the temperature $T = 0\,\text{K}$, which is absolute zero.

   Reading the graph

   A line is fitted to the points so that the points lie evenly on both sides of it. Two points far apart are read from the line. The fitted line passes less than $0.04\,\text{kPa}$ from the measured points $(0.0\,{}^\circ\text{C}$; $100.1\,\text{kPa})$ and $(80.0\,{}^\circ\text{C}$; $129.1\,\text{kPa})$, so they are suitable as points of the line. The extension of the line to the left of the graph meets the $\theta$ axis where $p = 0$.

   Model answer b)

   The result differs from the known value $-273.15\,{}^\circ\text{C}$ by about $3\,\text{K}$, that is, about $1\,\%$, which is within the measurement uncertainty when the points are less than $0.3\,\text{kPa}$ from the line. The extrapolation reaches about $276\,\text{K}$ below the measured range, although the measurement interval is only $80\,\text{K}$, so even a small error in the slope changes the result a lot. The extension is a prediction of the model: a real gas condenses into a liquid at a much higher temperature, and then the pressure is no longer proportional to the temperature in kelvins and the extension of the line does not hold.

   Formula

   The slope of the line

   $k$$= \dfrac{p_2 - p_1}{\theta_2 - \theta_1}$

   and the line $p = p_1 + k(\theta - \theta_1)$. From the condition $p = 0$ we get $\theta_0 = \theta_1 - \dfrac{p_1}{k}$.

   Conversions

   Not needed: a temperature difference is the same in degrees Celsius and in kelvins, and the result is wanted in degrees Celsius.

   Substitution

   The slope

   $$\begin{aligned} k &= \dfrac{129.1\,\text{kPa} - 100.1\,\text{kPa}}{80.0\,{}^\circ\text{C} - 0.0\,{}^\circ\text{C}} \\ &= 0.3625\,\tfrac{\text{kPa}}{\text{K}} \end{aligned}$$

   and the absolute zero

   $$\begin{aligned} \theta_0 &= 0.0\,{}^\circ\text{C} - \dfrac{100.1\,\text{kPa}}{0.3625\,\tfrac{\text{kPa}}{\text{K}}} \\ &= -276.1\,{}^\circ\text{C} \end{aligned}$$

   Answer

   $\theta_0 \approx -276\,{}^\circ\text{C}$

   Three significant figures like the pressures. Reading from a graph and extrapolating are estimates, so $\pm 3\,\%$ is accepted.

   Reasonableness

   When $\theta = -276\,{}^\circ\text{C}$, the pressure is

   $100.1\,\text{kPa} + 0.3625\,\tfrac{\text{kPa}}{\text{K}} \cdot (-276.1\,\text{K})$

   $\approx 0\,\text{kPa}$

   The result is close to the known value $-273.15\,{}^\circ\text{C}$. If the result were, for example, $-100\,{}^\circ\text{C}$ or $-500\,{}^\circ\text{C}$, the slope would have been read wrongly.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 11-16

   A diver is in a lake at a depth of $20.0\,\text{m}$ and blows an air bubble whose volume is $2.0\,\text{cm}^3$. The bubble rises to the surface.

   The atmospheric pressure is $101\,\text{kPa}$ and the density of water is $1000\,\text{kg/m}^3$. Assume that the temperature of the water is the same all the way and that the air of the bubble does not dissolve in water.

   What is the volume of the bubble at the surface? Using the result, justify why a person diving with compressed air must not hold their breath when rising to the surface.

   cm³Check
   Answer

   $V_2 \approx 5.9\,\text{cm}^3$

   Show the solution

   Given

   $h = 20.0\,\text{m}$, $V_1 = 2.0\,\text{cm}^3$, $p_0 = 101\,\text{kPa}$, $\rho = 1000\,\tfrac{\text{kg}}{\text{m}^3}$, $g = 9.81\,\tfrac{\text{m}}{\text{s}^2}$, temperature constant, air does not dissolve in water

   Find

   the volume $V_2$ of the bubble at the surface in cubic centimetres

   Principle

   Deep down, the bubble is subject to the hydrostatic pressure of the water in addition to the atmospheric pressure. A rising bubble is in balance with the surrounding water, so its pressure is always the pressure of its surroundings. The temperature and the amount of air stay the same, so Boyle's law holds.

   Formula

   $p_1 = p_0 + \rho g h$ and $p_1 V_1 = p_0 V_2$, so

   $V_2$$= \dfrac{p_1 V_1}{p_0}$

   $= \dfrac{(p_0 + \rho g h) V_1}{p_0}$

   Conversions

   The hydrostatic pressure in pascals:

   $\rho g h$$= 196\,200\,\text{Pa}$

   $= 196.2\,\text{kPa}$

   because $1\,\text{kPa} = 1000\,\text{Pa}$. The volumes are in the same unit, so they do not need to be converted.

   Substitution

   $p_1$$= 101\,\text{kPa} + 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 20.0\,\text{m}$

   $= 101\,\text{kPa} + 196.2\,\text{kPa}$

   $= 297.2\,\text{kPa}$

   $V_2$$= \dfrac{297.2\,\text{kPa} \cdot 2.0\,\text{cm}^3}{101\,\text{kPa}}$

   $= 5.885\,\text{cm}^3$

   Answer

   $V_2 \approx 5.9\,\text{cm}^3$

   Two significant figures. The limiting value is the volume of the bubble, $2.0\,\text{cm}^3$ (two figures); the depth $20.0\,\text{m}$ and the atmospheric pressure $101\,\text{kPa}$ have three figures, and the density of water, $1000\,\tfrac{\text{kg}}{\text{m}^3}$, is a known table value that does not limit the result.

   Reasonableness

   Every ten metres of water adds about one atmosphere of pressure, so at $20\,\text{m}$ the pressure is about $3 p_0$ and the volume at the surface is about three times as large: $3 \cdot 2.0\,\text{cm}^3 \approx 6\,\text{cm}^3$. The same applies to the lungs: if a person diving with compressed air held their breath while rising, the air in the lungs would try to expand to almost three times its volume. The lungs cannot stretch that much, and lung tissue can be damaged by far smaller expansion. You must never hold your breath when rising: breathe normally throughout the ascent, so that the expanding air can leave the lungs.

2. Exercise 11-17

   The tyres of a car are filled in a garage at $20\,{}^\circ\text{C}$, and the tyre pressure gauge shows $250.0\,\text{kPa}$. The gauge shows gauge pressure, that is, how much greater the air pressure in the tyre is than the atmospheric pressure. The atmospheric pressure is $101\,\text{kPa}$.

   At night the car stands in the yard, and the air in the tyre cools to $-20\,{}^\circ\text{C}$. The volume of the tyre stays practically the same, and no air escapes.

   What does the gauge show on a frosty morning? Justify why the reading of the gauge cannot be substituted directly into the formula $p_1/T_1 = p_2/T_2$.

   kPaCheck
   Answer

   $\Delta p_2 \approx 202\,\text{kPa}$

   Show the solution

   Given

   gauge pressure $\Delta p_1 = 250.0\,\text{kPa}$, $p_0 = 101\,\text{kPa}$, $\theta_1 = 20\,{}^\circ\text{C}$, $\theta_2 = -20\,{}^\circ\text{C}$, volume constant, no air escapes

   Find

   the reading of the gauge, that is, the gauge pressure $\Delta p_2$ on a frosty morning, in kilopascals

   Principle

   The volume of the tyre and the amount of air stay the same, so the pressure is directly proportional to the temperature in kelvins. The proportionality applies to the total pressure, not to the gauge pressure of the meter: the gauge pressure is the total pressure minus the atmospheric pressure, and there is still air in the tyre even when the gauge shows zero.

   Formula

   $p = \Delta p + p_0$ and $\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$, so $p_2 = \dfrac{p_1 T_2}{T_1}$ and $\Delta p_2 = p_2 - p_0$

   Conversions

   The total pressure

   $p_1$$= 250.0\,\text{kPa} + 101\,\text{kPa}$

   $= 351\,\text{kPa}$

   (the precision of the sum is in units, because $101\,\text{kPa}$ is given to the precision of units). The temperatures in kelvins:

   $T_1$$= (20 + 273.15)\,\text{K}$

   $= 293.15\,\text{K}$

   $T_2$$= (-20 + 273.15)\,\text{K}$

   $= 253.15\,\text{K}$

   Substitution

   $p_2$$= \dfrac{351\,\text{kPa} \cdot 253.15\,\text{K}}{293.15\,\text{K}}$

   $= 303.11\,\text{kPa}$

   so

   $\Delta p_2$$= 303.11\,\text{kPa} - 101\,\text{kPa}$

   $= 202.11\,\text{kPa}$

   Answer

   $\Delta p_2 \approx 202\,\text{kPa}$

   Three significant figures. The atmospheric pressure $101\,\text{kPa}$ is given to the precision of units, so both the total pressure $351\,\text{kPa}$ and $p_2 = 303\,\text{kPa}$ are known to the precision of units (three figures); also $T_1 = 293\,\text{K}$ and $T_2 = 253\,\text{K}$ have three figures. The difference $303 - 101$ keeps the units, so $202\,\text{kPa}$ has three figures.

   Reasonableness

   The total pressure decreases by about $14\,\%$ ($253/293 \approx 0.86$): $351\,\text{kPa} \cdot 0.86 \approx 303\,\text{kPa}$. If the reading were substituted directly into the formula, the result would be $250\,\text{kPa} \cdot 0.86 \approx 216\,\text{kPa}$, which is wrong, because the gauge pressure is not proportional to the temperature. The reading drops from $250\,\text{kPa}$ to $202\,\text{kPa}$, almost a fifth, so it is worth checking the tyres in frosty weather.

3. Exercise 11-18

   The cylinder of a bicycle pump is $40.0\,\text{cm}$ long and equally wide along its whole length. When the plunger is pulled up, the cylinder is full of outside air at a pressure of $101\,\text{kPa}$. The gauge pressure of the tyre is $300.0\,\text{kPa}$.

   The valve of the pump lets air into the tyre only when the air pressure in the cylinder is equal to the air pressure in the tyre.

   How far must the plunger be pressed before air starts to flow into the tyre? Assume that the temperature stays the same and the volume of the hose is negligible. Justify why pumping a hard tyre feels inefficient.

   cmCheck
   Answer

   $s \approx 29.9\,\text{cm}$

   Show the solution

   Given

   cylinder length $L_1 = 40.0\,\text{cm}$, initial pressure $p_1 = p_0 = 101\,\text{kPa}$, gauge pressure of the tyre $\Delta p = 300.0\,\text{kPa}$, cross-sectional area constant, temperature constant

   Find

   the distance $s$ the piston moves before the valve opens, in centimetres

   Principle

   The valve is closed, so the air in the cylinder is sealed and its temperature stays the same: Boyle's law holds. The valve opens only when the pressure in the cylinder equals the total pressure of the tyre, $p_0 + \Delta p$. The cross-sectional area of the cylinder is constant, so the volume is proportional to the length of the air column.

   Formula

   $p_2 = p_0 + \Delta p$ and $p_1 A L_1 = p_2 A L_2$, so $L_2 = \dfrac{p_1 L_1}{p_2}$ and the distance $s = L_1 - L_2$

   Conversions

   The total pressure of the tyre is obtained by adding the atmospheric pressure to the gauge pressure:

   $p_2$$= 300.0\,\text{kPa} + 101\,\text{kPa}$

   $= 401\,\text{kPa}$

   No other conversions are needed, because $A$ and the kilopascals cancel and the lengths are in centimetres.

   Substitution

   $L_2$$= \dfrac{101\,\text{kPa} \cdot 40.0\,\text{cm}}{401\,\text{kPa}}$

   $= 10.075\,\text{cm}$

   so

   $s$$= 40.0\,\text{cm} - 10.075\,\text{cm}$

   $= 29.925\,\text{cm}$

   Answer

   $s \approx 29.9\,\text{cm}$

   Three significant figures. In the difference $40.0\,\text{cm} - 10.1\,\text{cm}$ the decimals are determined by the length $40.0\,\text{cm}$ (tenths of a centimetre), so the result $29.9\,\text{cm}$ has three figures; the pressures $101\,\text{kPa}$ and $401\,\text{kPa}$ also have three figures.

   Reasonableness

   The pressure ratio is $401/101 \approx 4$, so the air column is compressed to about a quarter: $40.0\,\text{cm} / 4 \approx 10\,\text{cm}$. About three quarters of the stroke is used to compress the air, and only during the last $10\,\text{cm}$ does air flow into the tyre. The harder the tyre, the smaller the part of the stroke that fills it, and that is why pumping feels inefficient.

4. Exercise 11-19

   A student studies air sealed in a disposable syringe. At first there is $40.0\,\text{ml}$ of air at a pressure of $101\,\text{kPa}$ and a temperature of $20.0\,{}^\circ\text{C}$. She compresses the air in two ways. In slow compression the temperature stays the same. In fast compression the plunger is pushed down in a fraction of a second and the pressure is read as soon as the plunger has stopped: the air has had time to give only part of the energy to the surroundings (table, example data).

   a) Calculate the product $pV$ for every row in both series. Deduce in which compression Boyle's law holds.

   b) Explain why the pressure in fast compression is greater than Boyle's law predicts. What happens to the pressure when the plunger is held still and the air cools back to the temperature $20.0\,{}^\circ\text{C}$?

   c) Think of the fast compression as two steps: first an isothermal compression to the volume $10.0\,\text{ml}$ and then the warming of the air at constant volume to the pressure in the table. Use this to calculate the temperature of the air in degrees Celsius right after the fast compression, when the volume is $10.0\,\text{ml}$.

   Write the result of part c) in the answer box.

   Air pressure in slow and fast compression (example data)

   $V\,(\text{ml})$	$p_\text{slow}\,(\text{kPa})$	$p_\text{fast}\,(\text{kPa})$
   40.0	101	101
   30.0	135	141
   20.0	203	226
   10.0	403	481

   °CCheck
   Answer

   $\theta_2 \approx 76\,{}^\circ\text{C}$

   Show the solution

   Given

   initial state $V_1 = 40.0\,\text{ml}$, $p_1 = 101\,\text{kPa}$, $\theta_1 = 20.0\,{}^\circ\text{C}$; from the table the pressures in slow and fast compression at the volumes $40.0$; $30.0$; $20.0$ and $10.0\,\text{ml}$; final volume of the fast compression $V_2 = 10.0\,\text{ml}$ and pressure $p_2 = 481\,\text{kPa}$

   Find

   a) in which compression Boyle's law holds; b) why the pressure is greater in the fast compression and what happens as the air cools; c) the temperature $\theta_2$ of the air in degrees Celsius right after the fast compression

   Principle

   Boyle's law holds when the temperature of the sealed gas stays constant. In fast compression the plunger does work on the gas, but the energy does not have time to transfer completely to the surroundings, so part of the work stays in the gas and the internal energy and the temperature of the gas increase. The amount of air is the same, so the temperature is obtained from two changes of state: first an isothermal compression (Boyle's law) and then warming at constant volume ($p/T$ constant).

   Reading the graph

   From the row $V = 10.0\,\text{ml}$ of the table the pressure of the fast compression, $481\,\text{kPa}$, is read. The comparison pressure for the isothermal compression is calculated from the initial state, because it contains no reading error.

   Model answer a)

   In slow compression the products are $pV = 4040$; $4050$; $4060$ and $4030\,\text{kPa}\cdot\text{ml}$, so they are constant to within about $1\,\%$ and Boyle's law holds. In fast compression the products are $4040$; $4230$; $4520$ and $4810\,\text{kPa}\cdot\text{ml}$, and they increase by as much as about $19\,\%$ from the initial value, so Boyle's law does not hold.

   Model answer b)

   In fast compression the temperature of the gas rises, because part of the work stays as energy of the gas: the mean kinetic energy of the particles increases. The particles of warmer gas hit the walls more often and harder, so the pressure is greater than at a constant temperature. Boyle's law requires a constant temperature, so it does not describe this situation. When the plunger is held still, the volume is constant and the air gives energy to the surroundings until the temperature is again $20.0\,{}^\circ\text{C}$. The pressure then drops back to the value given by Boyle's law, that is, about $404\,\text{kPa}$.

   Formula

   Step 1 (isothermal): $p_1 V_1 = p' V_2$, so $p' = \dfrac{p_1 V_1}{V_2}$. Step 2 (isochoric): $\dfrac{p'}{T_1} = \dfrac{p_2}{T_2}$, so $T_2 = \dfrac{T_1 p_2}{p'}$ and $\theta_2 = T_2 - 273.15\,\text{K}$ in degrees Celsius.

   Conversions

   $T_1$$= (20.0 + 273.15)\,\text{K}$

   $= 293.15\,\text{K}$

   Finally the kelvins are converted back to degrees Celsius by subtracting $273.15$. The pressures and volumes do not need to be converted, because they cancel.

   Substitution

   The isothermal compression gives the pressure

   $$\begin{aligned} p' &= \dfrac{101\,\text{kPa} \cdot 40.0\,\text{ml}}{10.0\,\text{ml}} \\ &= 404\,\text{kPa} \end{aligned}$$

   and the warming at constant volume gives the temperature

   $$\begin{aligned} T_2 &= \dfrac{293.15\,\text{K} \cdot 481\,\text{kPa}}{404\,\text{kPa}} \\ &= 349.0\,\text{K} \end{aligned}$$

   so

   $\theta_2$$= (349.0 - 273.15)\,{}^\circ\text{C}$

   $= 75.9\,{}^\circ\text{C}$

   Answer

   $\theta_2 \approx 76\,{}^\circ\text{C}$

   In whole degrees Celsius. The pressure $481\,\text{kPa}$ has three figures, and the initial temperature $20.0\,{}^\circ\text{C}$ limits the result to whole degrees.

   Reasonableness

   The temperature rises by about $56\,\text{K}$ when the compression is fast and the volume decreases to a quarter. A bicycle pump gets warm when pumping, so the order of magnitude is reasonable. Check: when the gas cools back to $293.15\,\text{K}$ at constant volume, the pressure is

   $481\,\text{kPa} \cdot \dfrac{293.15}{349.0}$

   $\approx 404\,\text{kPa}$

   the same as the pressure given by Boyle's law. The ratio of the pressures $481 : 404 \approx 1.19$ is the same as the ratio of the temperatures in kelvins $349.0 : 293.15 \approx 1.19$.

5. Exercise 11-20

   A freediver fills her lungs at the surface and dives into a lake holding her breath. The volume $V$ of the air in the lungs was measured at different depths $h$ (table, example data). The atmospheric pressure is $101\,\text{kPa}$, the density of water is $1000\,\text{kg/m}^3$ and $g = 9.81\,\text{m/s}^2$. Assume that the temperature and the amount of the air stay the same and that the pressure of the air in the lungs is all the time the same as the pressure of the surrounding water.

   a) Calculate the total pressure $p$ of the surroundings at each depth and the product $pV$. Using your calculations, deduce whether the air in the lungs obeys Boyle's law.

   b) Assume as a model that the lungs cannot be compressed to less than $1.5\,\textsf{l}$. At what depth would the volume of the air in the lungs reach this limit?

   c) What is the ratio of the pressure at the depth of part b) to the pressure at the surface, and what is the ratio of the volumes of the lungs at the surface and at this depth? Compare the ratios.

   Write the result of part b) in the answer box.

   Volume of air in a freediver's lungs (example data)

   $h\,(\text{m})$	$V\,(\textsf{l})$
   0.0	6.0
   5.0	4.0
   10.0	3.0
   20.0	2.0

   mCheck
   Answer

   $h \approx 30\,\text{m}$

   Show the solution

   Given

   from the table the depths $h = 0.0$; $5.0$; $10.0$; $20.0\,\text{m}$ and the volumes $V = 6.0$; $4.0$; $3.0$; $2.0\,\textsf{l}$; $p_0 = 101\,\text{kPa}$, $\rho = 1000\,\tfrac{\text{kg}}{\text{m}^3}$, $g = 9.81\,\tfrac{\text{m}}{\text{s}^2}$, limit $V_\text{limit} = 1.5\,\textsf{l}$; temperature and amount of air constant

   Find

   a) whether the air in the lungs obeys Boyle's law; b) the depth $h$ in metres at which $V = 1.5\,\textsf{l}$; c) the increase of pressure compared with the surface

   Principle

   At the depth $h$ the lungs are subject to the hydrostatic pressure of the water in addition to the atmospheric pressure, $p = p_0 + \rho g h$. The pressure in the lungs is the same as the pressure of the surroundings. The temperature and the amount of the air stay the same, so Boyle's law holds if the product $pV$ is constant.

   Reading the graph

   For each depth in the table the pressure is calculated with the formula $p = p_0 + \rho g h$ and multiplied by the volume. The mean of the products $pV$ gives the constant of Boyle's law, and it is used to calculate the pressure at which $V = 1.5\,\textsf{l}$.

   Model answer a)

   The pressures are $101$; $150$; $199$ and $297\,\text{kPa}$, and the products $pV = 606$; $600$; $597$ and $594\,\text{kPa}\cdot\textsf{l}$. The products decrease steadily, but differ from each other by less than $2\,\%$, which is within the measurement uncertainty, so the air in the lungs obeys Boyle's law.

   Formula

   $p = p_0 + \rho g h$ and $p V = \text{constant}$. At the limit

   $p_\text{limit}$$= \dfrac{pV}{V_\text{limit}}$

   and the depth is

   $h$$= \dfrac{p_\text{limit} - p_0}{\rho g}$

   Conversions

   $\rho g$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}$

   $= 9810\,\tfrac{\text{Pa}}{\text{m}}$

   $= 9.81\,\tfrac{\text{kPa}}{\text{m}}$

   The products $pV$ are calculated in the unit $\text{kPa}\cdot\textsf{l}$, so $V_\text{limit}$ stays in litres and the result comes out in kilopascals.

   Substitution

   The sum of the products is $2397\,\text{kPa}\cdot\textsf{l}$, so $pV = 599.25\,\text{kPa}\cdot\textsf{l}$.

   $$\begin{aligned} p_\text{limit} &= \dfrac{599.25\,\text{kPa}\cdot\textsf{l}}{1.5\,\textsf{l}} \\ &= 399.5\,\text{kPa} \end{aligned}$$

   $$\begin{aligned} h &= \dfrac{399.5\,\text{kPa} - 101\,\text{kPa}}{9.81\,\tfrac{\text{kPa}}{\text{m}}} \\ &= 30.43\,\text{m} \end{aligned}$$

   Answer

   $h \approx 30\,\text{m}$

   Two significant figures like the limit volume $1.5\,\textsf{l}$. If the product corresponding to the surface, $101\,\text{kPa} \cdot 6.0\,\textsf{l} = 606\,\text{kPa}\cdot\textsf{l}$, is used, the result is $31\,\text{m}$, which is of the same size.

   Model answer c)

   The ratio of the pressures is $399.5 : 101 \approx 4.0$, that is, the pressure is four times that at the surface. A water column of thirty metres adds about $3 \cdot 101\,\text{kPa}$ of pressure, and the atmospheric pressure is added on top of that. The ratio of the volumes is $6.0\,\textsf{l} : 1.5\,\textsf{l} = 4.0$, that is, the volume is a quarter. The ratios are inverse, as Boyle's law requires.

   Reasonableness

   The row $h = 20.0\,\text{m}$ of the table gives the pressure $297\,\text{kPa}$, that is, about three times the atmospheric pressure, and the volume $2.0\,\textsf{l}$, that is, a third. Every additional $10\,\text{m}$ adds about $98\,\text{kPa}$ of pressure, that is, about one atmospheric pressure, so at the depth $30\,\text{m}$ the pressure is about $4$ times and the volume a quarter, $6.0\,\textsf{l} : 4 = 1.5\,\textsf{l}$. The depth $30\,\text{m}$ is below the deepest row of the table, $20\,\text{m}$, so this is an extrapolation, whose reliability rests on the assumption of a constant temperature and a constant amount of air.