Ideal gas law $pV = nRT$

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

1. Exercise 12-1Core

   The volume of the inside of a fridge is $0.250\,\text{m}^3$. When the door has been closed for a long time, the temperature of the air in the fridge is $278\,\text{K}$ and the pressure is $1.01 \cdot 10^5\,\text{Pa}$.

   How many moles of air are in the fridge? Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$.

   molCheck
   Answer

   $10.9\,\text{mol}$

   Show the solution

   Given

   air in a fridge, $V = 0.250\,\text{m}^3$, $T = 278\,\text{K}$, $p = 1.01 \cdot 10^5\,\text{Pa}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   amount of substance $n$ in moles

   Principle

   The air in the fridge is in a fixed container, and its pressure, volume and temperature are all known at the same time. The equation of state of an ideal gas links these three to the amount of substance, and air behaves like an ideal gas at ordinary pressure and temperature.

   Formula

   $pV = nRT$, so $n = \dfrac{pV}{RT}$

   Conversions

   No conversions are needed, because the volume is in cubic metres, the pressure in pascals and the temperature in kelvins. Then $\text{Pa}\cdot\text{m}^3 = \text{J}$, and the joules cancel.

   Substitution

   $n$$= \dfrac{1.01 \cdot 10^5\,\text{Pa} \cdot 0.250\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 278\,\text{K}}$

   $= 10.93\,\text{mol}$

   Answer

   $10.9\,\text{mol}$

   Three significant figures. All the data ($0.250\,\text{m}^3$, $278\,\text{K}$, $1.01 \cdot 10^5\,\text{Pa}$ and $R = 8.31$) are given with three digits.

   Reasonableness

   At room temperature and normal pressure one litre of air contains about $0.041\,\text{mol}$, so $250\,\textsf{l}$ contains about $250 \cdot 0.041 \approx 10\,\text{mol}$. In the cold, slightly more air fits, because $293\,\text{K} / 278\,\text{K} \approx 1.05$. So the result $10.9\,\text{mol}$ is believable, and the mass of the air is about $10.9\,\text{mol} \cdot 29\,\text{g/mol} \approx 0.32\,\text{kg}$.

2. Exercise 12-2Core

   The inside volume of a basketball is $0.00710\,\text{m}^3$, and it contains $0.460\,\text{mol}$ of air. The temperature of the sports hall is $293\,\text{K}$.

   What is the pressure of the air inside the ball? Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Give the answer in kilopascals.

   kPaCheck
   Answer

   $158\,\text{kPa}$

   Show the solution

   Given

   air in a basketball, $V = 0.00710\,\text{m}^3$, $n = 0.460\,\text{mol}$, $T = 293\,\text{K}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   pressure $p$ in kilopascals

   Principle

   The ball contains a certain amount of air in a certain volume and at a certain temperature. The air behaves as an ideal gas, so the equation of state gives the pressure directly from the amount of substance, the temperature and the volume.

   Formula

   $pV = nRT$, so $p = \dfrac{nRT}{V}$

   Conversions

   The data are already in base units, so the substitution gives the pressure in pascals, because $\tfrac{\text{J}}{\text{m}^3} = \text{Pa}$. Finally $1\,\text{kPa} = 1000\,\text{Pa}$.

   Substitution

   $p$$= \dfrac{0.460\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 293\,\text{K}}{0.00710\,\text{m}^3}$

   $= 157\,750\,\text{Pa}$

   $= 157.75\,\text{kPa}$

   Answer

   $158\,\text{kPa}$

   Three significant figures. All the data ($0.460\,\text{mol}$, $0.00710\,\text{m}^3$, $293\,\text{K}$ and $R = 8.31$) are given with three digits.

   Reasonableness

   Atmospheric pressure is about $101\,\text{kPa}$, so the gauge pressure of the ball is about $158\,\text{kPa} - 101\,\text{kPa} \approx 57\,\text{kPa}$. This is a typical gauge pressure for a pumped-up basketball, so the ball is hard but not overinflated. The pressure is clearly greater than atmospheric pressure, as it should be for a fully pumped-up ball.

3. Exercise 12-3

   An automatic life jacket inflates when $0.400\,\text{mol}$ of carbon dioxide is released from the cartridge of the jacket. The temperature of the gas in the jacket is $283\,\text{K}$ and the pressure is $1.05 \cdot 10^5\,\text{Pa}$.

   What is the volume of the gas in the jacket? Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Give the answer in cubic metres.

   m³Check
   Answer

   $0.00896\,\text{m}^3$

   Show the solution

   Given

   carbon dioxide in a life jacket, $n = 0.400\,\text{mol}$, $T = 283\,\text{K}$, $p = 1.05 \cdot 10^5\,\text{Pa}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   volume $V$ in cubic metres

   Principle

   When the gas has been released into the jacket, its pressure is the pressure in the jacket and its temperature is known. The amount of gas is known, so the equation of state gives the volume. The type of gas does not affect the calculation, because the equation of state has no molar mass.

   Formula

   $pV = nRT$, so $V = \dfrac{nRT}{p}$

   Conversions

   No conversions are needed: the amount of substance is in moles, the temperature in kelvins and the pressure in pascals, so the volume comes out in cubic metres.

   Substitution

   $V$$= \dfrac{0.400\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 283\,\text{K}}{1.05 \cdot 10^5\,\text{Pa}}$

   $= 0.008959\,\text{m}^3$

   Answer

   $0.00896\,\text{m}^3$

   Three significant figures. All the data ($0.400\,\text{mol}$, $283\,\text{K}$, $1.05 \cdot 10^5\,\text{Pa}$ and $R = 8.31$) are given with three digits.

   Reasonableness

   The result is about $9.0\,\textsf{l}$. At room temperature a mole of gas takes up about $24\,\textsf{l}$, so $0.400\,\text{mol}$ takes up about $0.400 \cdot 24\,\textsf{l} \approx 9.6\,\textsf{l}$, and at a slightly lower temperature the volume is a little smaller. The mass of the carbon dioxide is only $0.400\,\text{mol} \cdot 44\,\text{g/mol} \approx 18\,\text{g}$, which is the size of a cartridge.

4. Exercise 12-4

   On the jetty of a cottage there is a dark plastic barrel with its lid closed. The volume of the barrel is $0.200\,\text{m}^3$, and it contains $8.40\,\text{mol}$ of air. On a sunny afternoon the pressure of the air in the barrel is $1.13 \cdot 10^5\,\text{Pa}$.

   What is the temperature of the air in the barrel in kelvins? Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$.

   KCheck
   Answer

   $324\,\text{K}$

   Show the solution

   Given

   air in a barrel, $V = 0.200\,\text{m}^3$, $n = 8.40\,\text{mol}$, $p = 1.13 \cdot 10^5\,\text{Pa}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   temperature $T$ in kelvins

   Principle

   The barrel is closed, so the amount of air and the volume are known, and the pressure has been measured. The equation of state gives the temperature when the air behaves as an ideal gas.

   Formula

   $pV = nRT$, so $T = \dfrac{pV}{nR}$

   Conversions

   No conversions are needed: the pressure is in pascals, the volume in cubic metres and the amount of substance in moles, so the result comes out directly in kelvins.

   Substitution

   $T$$= \dfrac{1.13 \cdot 10^5\,\text{Pa} \cdot 0.200\,\text{m}^3}{8.40\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}}$

   $= 323.8\,\text{K}$

   Answer

   $324\,\text{K}$

   Three significant figures. All the data ($0.200\,\text{m}^3$, $8.40\,\text{mol}$, $1.13 \cdot 10^5\,\text{Pa}$ and $R = 8.31$) are given with three digits.

   Reasonableness

   In degrees Celsius the result is $323.8 - 273.15 \approx 51\,{}^\circ\text{C}$, which is a possible temperature for a dark barrel in the summer sun. With the same amount of substance and volume, a pressure of $101\,\text{kPa}$ would correspond to a temperature of $16\,{}^\circ\text{C}$, so the pressure has risen by about $12\,\%$ as the temperature rose. The pressure grows with the temperature, as it should.

5. Exercise 12-5

   The gas cylinder of a home carbonation machine contains $425\,\text{g}$ of carbon dioxide. The molar mass of carbon dioxide is $44\,\text{g/mol}$.

   How many moles of carbon dioxide are in the cylinder?

   molCheck
   Answer

   $9.7\,\text{mol}$

   Show the solution

   Given

   carbon dioxide, $m = 425\,\text{g}$, $M = 44\,\text{g/mol}$

   Find

   amount of substance $n$ in moles

   Principle

   The molar mass tells the mass of one mole, so the total mass of the gas divided by the molar mass gives the number of moles. The equation of state is not needed, because pressure and temperature do not affect the amount of substance.

   Formula

   $M = \dfrac{m}{n}$, so $n = \dfrac{m}{M}$

   Conversions

   No conversions are needed, because the mass and the molar mass are both in grams. The grams cancel, and the mole is left.

   Substitution

   $n$$= \dfrac{425\,\text{g}}{44\,\text{g/mol}}$

   $= 9.659\,\text{mol}$

   Answer

   $9.7\,\text{mol}$

   Two significant figures. The limiting datum is the molar mass $44\,\text{g/mol}$ (two digits); the mass $425\,\text{g}$ has three digits.

   Reasonableness

   If the molar mass is rounded to $40\,\text{g/mol}$, you get about $425 / 40 \approx 10.6\,\text{mol}$, so the result $9.7\,\text{mol}$ is of the same order of magnitude and a little smaller, because the real molar mass is greater. Checking backwards: $9.659\,\text{mol} \cdot 44\,\text{g/mol} = 425\,\text{g}$. At room temperature the gas would take up about $9.7 \cdot 24\,\textsf{l} \approx 230\,\textsf{l}$.

6. Exercise 12-6

   What does the amount of substance $n$ describe in the equation of state of an ideal gas $pV = nRT$?

   Choose an option.

   • The density of the gas, because in a dense gas there are more particles in the same space.
   • The number of particles in moles, because one mole is always the same number of particles.
   • The speed of the particles, because faster particles hit the wall more often.
   • The mass of the gas, because a heavier gas contains more matter and causes a higher pressure.

   Answer
   I have done my solution.
   Answer

   The number of particles in moles, because one mole is always the same number of particles.

   Show the solution

   Find

   what the amount of substance $n$ describes in the equation of state $pV = nRT$.

   Principle

   The mole is a unit of the number of particles, and one mole always contains $6.022 \cdot 10^{23}$ particles, whatever the substance. The equation of state links the pressure, volume and temperature of a gas to its number of particles.

   Model answer

   The amount of substance gives the number of particles of the gas in moles, $n = N/N_\text{A}$. It is not the mass of the gas: the mass is $m = nM$, and the molar mass $M$ does not appear in the equation of state. Nor is it the density or the speed of the particles, because the volume $V$ and the temperature $T$ are quantities of their own in the equation.

7. Exercise 12-7

   A balloon is filled with helium in a room where the temperature is $21\,{}^\circ\text{C}$ and the air pressure is $101\,\text{kPa}$. The volume of the balloon is then $2.50\,\textsf{l}$.

   The balloon is taken outside, where the temperature is $-9\,{}^\circ\text{C}$ and the air pressure is $98\,\text{kPa}$. The pressure of the gas in the balloon is always the same as the pressure of the surrounding air, and no gas escapes.

   What is the volume of the balloon outside, in litres?

   lCheck
   Answer

   $2.3\,\textsf{l}$

   Show the solution

   Given

   helium, $V_1 = 2.50\,\textsf{l}$, $\theta_1 = 21\,{}^\circ\text{C}$, $p_1 = 101\,\text{kPa}$; $\theta_2 = -9\,{}^\circ\text{C}$, $p_2 = 98\,\text{kPa}$

   Find

   volume $V_2$ in litres

   Principle

   No gas escapes, so the amount of substance $n$ stays the same. In both states the equation of state $pV = nRT$ gives the same number $nR = pV/T$, so $p_1V_1/T_1 = p_2V_2/T_2$.

   Formula

   $\dfrac{p_1V_1}{T_1}$$= \dfrac{p_2V_2}{T_2}$

   so $V_2 = \dfrac{p_1 V_1 T_2}{p_2 T_1}$

   Conversions

   $T_1$$= (21 + 273.15)\,\text{K}$

   $= 294.15\,\text{K}$

   $T_2$$= (-9 + 273.15)\,\text{K}$

   $= 264.15\,\text{K}$

   The units of pressure and volume cancel, so they do not need to be converted.

   Substitution

   $V_2$$= \dfrac{101\,\text{kPa} \cdot 2.50\,\textsf{l} \cdot 264.15\,\text{K}}{98\,\text{kPa} \cdot 294.15\,\text{K}}$

   $= 2.314\,\textsf{l}$

   Answer

   $2.3\,\textsf{l}$

   Two significant figures, because the pressure $98\,\text{kPa}$ is given with two digits (intermediate result $2.314\,\textsf{l}$).

   Reasonableness

   The temperature falls by the ratio $264.15 : 294.15 \approx 0.90$, which decreases the volume by a tenth. The pressure falls by the ratio $98 : 101 \approx 0.97$, which increases the volume by three per cent. The ratio of the volumes is $0.90 : 0.97 \approx 0.93$ and $0.93 \cdot 2.50\,\textsf{l} \approx 2.3\,\textsf{l}$, which agrees with the result.

8. Exercise 12-8

   A student calculates the amount of substance of a gas when the pressure is $150\,\text{kPa}$, the volume is $0.0020\,\text{m}^3$ and the temperature is $300\,\text{K}$. The student writes

   $n = \dfrac{150 \cdot 0.0020}{8.31 \cdot 300} = 1.2 \cdot 10^{-4}\,\text{mol}$

   The result seems small to the student, because two litres of gas surely contains more than a thousandth of a mole. What is the error in the calculation?

   Choose an option.

   • The volume has been substituted in cubic metres, although the equation of state requires it in litres.
   • The temperature has been substituted in degrees Celsius instead of kelvins, so the denominator is too small.
   • The molar mass of the gas is missing from the calculation, so the amount of substance cannot be calculated from the equation of state.
   • The pressure has been substituted in kilopascals instead of pascals, so the numerator is too small.

   Answer
   I have done my solution.
   Answer

   The pressure has been substituted in kilopascals instead of pascals, so the numerator is too small.

   Show the solution

   Find

   what the error in the student's calculation is.

   Principle

   The equation of state works correctly only when the quantities are in base units: pressure in pascals, volume in cubic metres and temperature in kelvins. Otherwise a unit factor is left in the result.

   Model answer

   The student substituted the pressure $150\,\text{kPa}$ as the number $150$ instead of in pascals. The correct pressure is $150\,000\,\text{Pa}$, which gives $n = \dfrac{150\,000 \cdot 0.0020}{8.31 \cdot 300} \approx 0.12\,\text{mol}$. The student's result is thus a thousand times too small, as the student's own estimate hinted. The temperature and the volume were substituted in the correct units, and no molar mass is needed.

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

1. Exercise 12-9Core

   The pressure gauge of the oxygen cylinder of an ambulance is broken. The volume of the cylinder is $10.0\,\textsf{l}$, and weighing shows that the cylinder contains $2.40\,\text{kg}$ of oxygen. The temperature in the garage is $15\,{}^\circ\text{C}$.

   What is the pressure of the oxygen in the cylinder in bars ($1\,\text{bar}$ = $100\,\text{kPa}$)? The molar mass of oxygen is $32.0\,\text{g/mol}$. Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$ and assume that oxygen obeys the equation of state of an ideal gas.

   barCheck
   Answer

   $180\,\text{bar}$

   Show the solution

   Given

   oxygen, $V = 10.0\,\textsf{l}$, $m = 2.40\,\text{kg}$, $\theta = 15\,{}^\circ\text{C}$, $M = 32.0\,\text{g/mol}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   pressure $p$ in bars, where $1\,\text{bar} = 100\,\text{kPa}$

   Principle

   The oxygen in the cylinder is in a fixed volume at a known temperature, and its amount of substance can be found by weighing from the mass and the molar mass. We assume that oxygen is an ideal gas, so the equation of state gives the pressure.

   Formula

   $pV = nRT$ and $n = \dfrac{m}{M}$, so $p = \dfrac{nRT}{V} = \dfrac{mRT}{MV}$

   Conversions

   $V$$= 10.0\,\textsf{l}$

   $= 0.0100\,\text{m}^3$

   $M$$= 32.0\,\text{g/mol}$

   $= 0.0320\,\text{kg/mol}$

   $T$$= (15 + 273.15)\,\text{K}$

   $= 288.15\,\text{K}$

   Finally $1\,\text{bar} = 10^5\,\text{Pa}$.

   Substitution

   $n$$= \dfrac{2.40\,\text{kg}}{0.0320\,\text{kg/mol}}$

   $= 75.0\,\text{mol}$

   $p$$= \dfrac{75.0\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 288.15\,\text{K}}{0.0100\,\text{m}^3}$

   $= 1.796 \cdot 10^7\,\text{Pa}$

   $= 179.6\,\text{bar}$

   Answer

   $180\,\text{bar}$

   Three significant figures. All the data ($10.0\,\textsf{l}$, $2.40\,\text{kg}$, $32.0\,\text{g/mol}$ and $R = 8.31$) are given with three digits, and the $288\,\text{K}$ obtained from the whole degrees $15\,{}^\circ\text{C}$ also has three digits.

   Reasonableness

   The pressure of a full oxygen cylinder is typically about $200\,\text{bar}$, so an almost full cylinder is believable. Order of magnitude: a mole of gas takes up about $24\,\textsf{l}$ at normal pressure, so $75\,\text{mol}$ would take up about $1800\,\textsf{l}$, and compressed into $10\,\textsf{l}$ the pressure is about $1800 / 10 \approx 180$ times atmospheric pressure. At such a high pressure a real gas differs from an ideal gas by a few percent, but the problem asked for the ideal gas value.

2. Exercise 12-10Core

   In a chemistry storeroom there is a gas cylinder whose label is worn illegible.

   A $1.00\,\textsf{l}$ glass flask emptied with a vacuum pump is filled with the gas from the cylinder, and the mass of the flask increases by $1.15\,\text{g}$. The pressure of the gas in the flask is $101\,\text{kPa}$ and the temperature is $22\,{}^\circ\text{C}$.

   Calculate the molar mass of the gas in grams per mole. Which gas is it, when the options are helium ($4.0\,\text{g/mol}$), nitrogen ($28\,\text{g/mol}$), oxygen ($32\,\text{g/mol}$), argon ($40\,\text{g/mol}$) and carbon dioxide ($44\,\text{g/mol}$)? Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$.

   g/molCheck
   Answer

   $27.9\,\text{g/mol}$

   Show the solution

   Given

   unknown gas, $V = 1.00\,\textsf{l}$, increase in mass $m = 1.15\,\text{g}$, $p = 101\,\text{kPa}$, $\theta = 22\,{}^\circ\text{C}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   molar mass $M$ and the type of gas

   Principle

   The flask was a vacuum, so the increase in mass is the mass of the gas. The equation of state gives the amount of substance from the pressure, the volume and the temperature, and the molar mass is the mass divided by the amount of substance. The equation of state does not contain the type of gas, so the type is identified only from the molar mass.

   Formula

   $pV = nRT$ and $n = \dfrac{m}{M}$, so $\dfrac{pV}{RT} = \dfrac{m}{M}$ and $M = \dfrac{mRT}{pV}$

   Conversions

   $V$$= 1.00\,\textsf{l}$

   $= 0.00100\,\text{m}^3$

   $p$$= 101\,\text{kPa}$

   $= 101\,000\,\text{Pa}$

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

   $= 295.15\,\text{K}$

   The mass is left in grams, so the result is in $\text{g/mol}$.

   Substitution

   $M$$= \dfrac{1.15\,\text{g} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 295.15\,\text{K}}{101\,000\,\text{Pa} \cdot 0.00100\,\text{m}^3}$

   $= 27.93\,\tfrac{\text{g}}{\text{mol}}$

   Answer

   $27.9\,\text{g/mol}$

   Three significant figures. All the data ($1.00\,\textsf{l}$, $1.15\,\text{g}$, $101\,\text{kPa}$ and $R = 8.31$) are given with three digits, and $295\,\text{K}$ (from the whole degrees $22\,{}^\circ\text{C}$) also has three digits. The gas is nitrogen ($28\,\text{g/mol}$).

   Reasonableness

   The amount of substance is

   $\dfrac{101\,000\,\text{Pa} \cdot 0.00100\,\text{m}^3}{8.31 \cdot 295.15}$

   $= 0.0412\,\text{mol}$

   which fits the fact that a litre of gas at room temperature contains about $0.04\,\text{mol}$. With the same amount of substance the mass would have been $0.16\,\text{g}$ for helium, $1.32\,\text{g}$ for oxygen, $1.65\,\text{g}$ for argon and $1.81\,\text{g}$ for carbon dioxide, so only nitrogen ($1.15\,\text{g}$) fits.

3. Exercise 12-11

   At the end of the intake stroke, the cylinder of a tractor's diesel engine contains $0.500\,\textsf{l}$ of air. The pressure of the air is $101\,\text{kPa}$ and the temperature is $30\,{}^\circ\text{C}$.

   At the end of the compression stroke the volume of the air is $0.0300\,\textsf{l}$ and the pressure is $4.60\,\text{MPa}$. The valves are closed, so no air can escape from the cylinder.

   What is the temperature of the air at the end of compression in degrees Celsius? Diesel fuel ignites by itself at a temperature of about $250\,{}^\circ\text{C}$. Does the fuel injected into the cylinder ignite?

   °CCheck
   Answer

   $555\,{}^\circ\text{C}$

   Show the solution

   Given

   air in a cylinder, initially $V_1 = 0.500\,\textsf{l}$, $p_1 = 101\,\text{kPa}$, $\theta_1 = 30\,{}^\circ\text{C}$; finally $V_2 = 0.0300\,\textsf{l}$, $p_2 = 4.60\,\text{MPa}$

   Find

   final temperature $\theta_2$ in degrees Celsius, and whether the fuel ignites ($250\,{}^\circ\text{C}$)

   Principle

   The valves are closed, so the amount of air stays the same and $nR$ is constant. Then the equation of state gives the combined gas law between the initial and final state.

   Formula

   $\dfrac{p_1 V_1}{T_1}$$= \dfrac{p_2 V_2}{T_2}$

   so

   $T_2$$= T_1 \cdot \dfrac{p_2 V_2}{p_1 V_1}$

   $\theta_2 = T_2 - 273.15\,\text{K}$

   Conversions

   $T_1$$= (30 + 273.15)\,\text{K}$

   $= 303.15\,\text{K}$

   The pressures and volumes appear as ratios, so they cancel when they are in the same units: $4.60\,\text{MPa} = 4600\,\text{kPa}$ and the litres stay as litres.

   Substitution

   $T_2$$= 303.15\,\text{K} \cdot \dfrac{4600\,\text{kPa} \cdot 0.0300\,\textsf{l}}{101\,\text{kPa} \cdot 0.500\,\textsf{l}}$

   $= 828.4\,\text{K}$

   $\theta_2$$= 828.4 - 273.15$

   $= 555.3\,{}^\circ\text{C}$

   Answer

   $555\,{}^\circ\text{C}$

   Three significant figures. All the data ($0.500\,\textsf{l}$, $101\,\text{kPa}$, $0.0300\,\textsf{l}$, $4.60\,\text{MPa}$ and $303\,\text{K}$ from the whole degrees $30\,{}^\circ\text{C}$) have three digits, and subtracting $273.15\,\text{K}$ from the temperature $828\,\text{K}$ keeps the units. The fuel ignites, because $555\,{}^\circ\text{C}$ is clearly above $250\,{}^\circ\text{C}$.

   Reasonableness

   The volume is about $17$ times smaller and the pressure about $46$ times greater ($0.500/0.0300 \approx 16.7$ and $4600/101 \approx 45.5$), so the product $pV$ and with it the temperature in kelvins grow by a factor of about $45.5 / 16.7 \approx 2.73$: $303\,\text{K} \cdot 2.73 \approx 828\,\text{K}$. The result agrees, and the temperature rises when a gas is compressed, so a diesel engine does not need a spark plug.

4. Exercise 12-12

   Two identical, closed and rigid containers, each with a volume of $2.00\,\textsf{l}$, contain gas samples A and B.

   When the containers are heated, their pressure is measured at different temperatures. The graph shows the measurement results.

   How many moles more gas is there in container A than in container B? Read the values you need from the graph. Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$ and assume that the gases obey the equation of state of an ideal gas.

   molCheck
   Answer

   $n_\text{A} - n_\text{B} \approx 0.024\,\text{mol}$

   Show the solution

   Given

   the volume of each container $V = 2.00\,\textsf{l}$; the measurement points of the graph at grid crossings, $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   the difference of the amounts of substance $n_\text{A} - n_\text{B}$

   Principle

   The containers are closed and rigid, so the amount of gas and the volume stay the same and the lines of the graph follow the equation of state. When the pressure of both series is read at the same temperature, the equation of state gives both amounts of substance, and the difference is obtained by subtracting.

   Reading the graph

   Both series are read at the same temperature $T = 400\,\text{K}$, which is a vertical grid line: gas A $p_\text{A} = 100\,\text{kPa}$ and gas B $p_\text{B} = 60\,\text{kPa}$. The pressures of the points at $T = 200\,\text{K}$ are half of these, so the lines pass through the origin, as $p = nRT/V$ requires.

   Formula

   $pV = nRT$, so

   $n$$= \dfrac{pV}{RT}$

   $n_\text{A} - n_\text{B}$$= \dfrac{(p_\text{A} - p_\text{B})\,V}{RT}$

   Conversions

   $p_\text{A} - p_\text{B}$$= (100 - 60)\,\text{kPa}$

   $= 40\,\text{kPa}$

   $= 40\,000\,\text{Pa}$

   $V$$= 2.00\,\textsf{l}$

   $= 0.00200\,\text{m}^3$

   The temperature is already in kelvins.

   Substitution

   $n_\text{A} - n_\text{B}$$= \dfrac{40\,000\,\text{Pa} \cdot 0.00200\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 400\,\text{K}}$

   $= \dfrac{80\,\text{J}}{3324\,\tfrac{\text{J}}{\text{mol}}}$

   $= 0.0241\,\text{mol}$

   Answer

   $n_\text{A} - n_\text{B} \approx 0.024\,\text{mol}$

   Two significant figures, because the volume and the pressure readings from the graph are known to two digits. Separately $n_\text{A} \approx 0.060\,\text{mol}$ and $n_\text{B} \approx 0.036\,\text{mol}$. The points are read from grid crossings, so the accepted range is narrow: $0.0233$–$0.0247\,\text{mol}$.

   Reasonableness

   The slope of line A is greater, so there is more gas in A and the difference is positive. The ratio of the pressures $100 : 60$ is also the ratio of the amounts of substance: $0.060 / 0.036 \approx 1.7$.

5. Exercise 12-13

   A motorist suspects that the tyre of a passenger car is leaking. The motorist measures the total pressure of the air in the tyre before driving and immediately after a long drive (example data set). The volume of the tyre is assumed to stay the same.

   a) Calculate how high the pressure in the tyre would be after the drive if no air had leaked.

   b) What percentage of the original amount of air in the tyre has leaked? Justify with the equation of state why measuring the pressure alone is not enough to show a leak.

   The answer box takes the result of part b).

   Temperature and total pressure of the air in the tyre (example data set)

   Moment	$\theta\,({}^\circ\text{C})$	$p\,(\text{kPa})$
   Before the drive	5.0	300
   After the drive	26.0	305

   %Check
   Answer

   $5.5\,\%$

   Show the solution

   Given

   tyre, $\theta_1 = 5.0\,{}^\circ\text{C}$, $p_1 = 300\,\text{kPa}$ before the drive; $\theta_2 = 26.0\,{}^\circ\text{C}$, $p_2 = 305\,\text{kPa}$ after the drive; volume $V$ constant

   Find

   the leaked share $\dfrac{n_1 - n_2}{n_1}$ as a percentage

   Principle

   The volume is constant, so the equation of state $pV = nRT$ gives the amount of substance $n = \dfrac{pV}{RT}$ in both states. The temperature rises during the drive, so the pressure would rise even in a tyre that does not leak, and the leak shows only when the amounts of substance are compared.

   Formula

   $n_1 = \dfrac{p_1V}{RT_1}$ and $n_2 = \dfrac{p_2V}{RT_2}$, so

   $\dfrac{n_2}{n_1}$$= \dfrac{p_2 T_1}{p_1 T_2}$

   and the leaked share is $1 - \dfrac{n_2}{n_1}$

   Conversions

   $T_1$$= (5.0 + 273.15)\,\text{K}$

   $= 278.15\,\text{K}$

   $T_2$$= (26.0 + 273.15)\,\text{K}$

   $= 299.15\,\text{K}$

   The pressures are in the same unit, so they cancel.

   Substitution

   a) Without a leak the amount of substance and the volume do not change, so

   $p_2'$$= p_1 \dfrac{T_2}{T_1}$

   $= 300\,\text{kPa} \cdot \dfrac{299.15\,\text{K}}{278.15\,\text{K}}$

   $= 322.6\,\text{kPa}$

   b)

   $\dfrac{n_2}{n_1}$$= \dfrac{305\,\text{kPa} \cdot 278.15\,\text{K}}{300\,\text{kPa} \cdot 299.15\,\text{K}}$

   $= 0.9453$

   so the leaked share is $1 - 0.9453 = 0.0547$

   Answer

   $5.5\,\%$

   Two significant figures: the share is the difference $1 - 0.945$, so the third digit of the pressures decides the second digit. In a tyre that does not leak the pressure after the drive would be $323\,\text{kPa}$, but the measured pressure is $305\,\text{kPa}$. The rise in pressure $300 \to 305\,\text{kPa}$ does not alone show a leak, because the rise in temperature would raise the pressure even without a leak.

   Reasonableness

   The measured pressure is $305 : 322.6 \approx 0.945$ of the pressure of a tyre that does not leak, so about $94.5\,\%$ of the air is left and the leaked share is about $5.5\,\%$. The temperature rose by about $7.5\,\%$ ($278 \to 299\,\text{K}$), but the pressure by only $1.7\,\%$, so some air must have escaped.

6. Exercise 12-14

   Students study how the volume of air depends on temperature. The air is in a syringe that is heated and whose piston moves freely. The pressure stays at atmospheric pressure, $101\,\text{kPa}$, and the amount of air stays constant. The measurement results are in the table (example data set). Draw the $(T, V)$ graph and fit a line to the points. The points are usually not exactly on a line because of measurement uncertainty.

   a) Determine the slope $\Delta V / \Delta T$ of the line from two points chosen on the line, not from two measurement points.

   b) Calculate from the slope the amount of substance $n$ of the air in the syringe in millimoles. Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Justify briefly why the line should pass through the origin when the temperature is in kelvins.

   The answer box takes the result of part b).

   Volume of air at constant pressure (example data set)

   $T\,(\text{K})$	$V\,(\text{cm}^3)$
   280	115.0
   300	124.1
   320	131.2
   340	140.5
   360	147.6

   mmolCheck
   Answer

   $5.0\,\text{mmol}$

   Show the solution

   Given

   $p = 101\,\text{kPa}$ constant, amount of air constant; $(T, V)$ points in the table; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   amount of substance $n$ in millimoles

   Principle

   The pressure and the amount of substance are constant, so the equation of state gives $V = \dfrac{nR}{p}\,T$. The volume is directly proportional to the temperature in kelvins, and the slope of the line in the $(T, V)$ graph is $\dfrac{nR}{p}$.

   Reading the graph

   Plot the points and draw a line that passes between them. The least-squares line is $V = 0.408\,\tfrac{\text{cm}^3}{\text{K}} \cdot T + 1.1\,\text{cm}^3$, and your own line can be compared with it. Read two points far apart from the line: the values of the line are $(280\,\text{K}$; $115.4\,\text{cm}^3)$ and $(360\,\text{K}$; $148.0\,\text{cm}^3)$, not the table values $115.0\,\text{cm}^3$ and $147.6\,\text{cm}^3$. The slope is

   $k$$= \dfrac{(148.0 - 115.4)\,\text{cm}^3}{(360 - 280)\,\text{K}}$

   $= 0.4075\,\tfrac{\text{cm}^3}{\text{K}}$

   The constant term of the line, $1.1\,\text{cm}^3$, is small compared with the volumes $115$–$148\,\text{cm}^3$, so the line passes almost through the origin.

   Formula

   $k = \dfrac{nR}{p}$, so $n = \dfrac{kp}{R}$

   Conversions

   $k$$= 0.4075\,\tfrac{\text{cm}^3}{\text{K}}$

   $= 0.4075 \cdot 10^{-6}\,\tfrac{\text{m}^3}{\text{K}}$

   $p$$= 101\,\text{kPa}$

   $= 101\,000\,\text{Pa}$

   Finally $1\,\text{mol} = 1000\,\text{mmol}$.

   Substitution

   $n$$= \dfrac{0.4075 \cdot 10^{-6}\,\tfrac{\text{m}^3}{\text{K}} \cdot 101\,000\,\text{Pa}}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}}$

   $= 4.953 \cdot 10^{-3}\,\text{mol}$

   Answer

   $5.0\,\text{mmol}$

   Two significant figures, because the measurement uncertainty of the fitted line is about $\pm 0.1\,\text{mmol}$ (intermediate result $4.95\,\text{mmol}$). The line must pass through the origin, because $V = nRT/p$ is directly proportional to the temperature when $T$ is in kelvins: at $T = 0$ the volume would be zero.

   Reasonableness

   A volume of $131\,\text{cm}^3$ at $320\,\text{K}$ and a pressure of $101\,\text{kPa}$ give

   $n$$\approx \dfrac{101\,000 \cdot 131 \cdot 10^{-6}}{8.31 \cdot 320}$

   $\approx 5.0 \cdot 10^{-3}\,\text{mol}$

   which agrees with the result. At room temperature $0.13\,\textsf{l}$ of air contains about $0.13 \cdot 0.041\,\text{mol} \approx 5\,\text{mmol}$.

7. Exercise 12-15

   The table shows the pressure of two equal amounts of gas as a function of volume (example data set). Series A and B were measured at different temperatures, and the volumes in the measurements are different.

   Which series was measured at the higher temperature?

   Pressure and volume of the same amount of gas at two temperatures (example data set)

   $V_\text{A}\,(\textsf{l})$	$p_\text{A}\,(\text{kPa})$	$V_\text{B}\,(\textsf{l})$	$p_\text{B}\,(\text{kPa})$
   1.0	244	1.5	203
   2.0	122	2.5	122
   3.0	81	3.5	87
   4.0	61	4.5	68

   Choose an option.

   • Series A, because its pressure $p$ is lower, and a low pressure means faster-moving particles.
   • Series A, because its pressure falls more slowly as the volume increases, so the gas is warmer.
   • Series B, because its product $pV$ is greater, and the temperature of the same amount of gas is proportional to it.
   • It cannot be deduced, because the volumes of the series are different and the pressure decreases in both.

   Answer
   I have done my solution.
   Answer

   Series B, because its product $pV$ is greater, and the temperature of the same amount of gas is proportional to it.

   Show the solution

   Given

   pressures and volumes of the same amount of gas: series A with volumes $1.0$–$4.0\,\textsf{l}$ and series B with volumes $1.5$–$4.5\,\textsf{l}$

   Find

   which series was measured at the higher temperature.

   Principle

   The amount of gas is the same, so the equation of state $pV = nRT$ implies that the product $pV$ is directly proportional to the temperature. On an isotherm the product $pV$ is constant, and at a higher temperature the product is greater. The volumes of the series are different, so the pressures cannot be compared directly.

   Model answer

   In series A the product $pV$ is about $244\,\text{kPa} \cdot \textsf{l}$ on every row, for example $2.0\,\textsf{l} \cdot 122\,\text{kPa}$. In series B it is about $305\,\text{kPa} \cdot \textsf{l}$, for example $2.5\,\textsf{l} \cdot 122\,\text{kPa}$. The product is greater in series B, so series B was measured at the higher temperature. The ratio of the temperatures is $305 : 244 \approx 1.25$. The same pressure $122\,\text{kPa}$ is at a greater volume in series B, and in the $(V, p)$ coordinate system the isotherm of B lies higher. Both series obey Boyle's law, so the decrease of the pressure as the volume increases does not tell the temperature.

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

1. Exercise 12-16

   At rest, a person breathes into the lungs with one inhalation air that takes up a volume of $0.55\,\textsf{l}$ at a temperature of $37\,{}^\circ\text{C}$. The pressure of the air in the lungs is $101\,\text{kPa}$.

   How many molecules are in one breath? The Avogadro constant is $6.022 \cdot 10^{23}\,\tfrac{1}{\text{mol}}$ and $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Give the answer in the form $a \cdot 10^{22}$. Justify why the calculation does not need information about which gases the air consists of.

   · 10²²Check
   Answer

   $1.3 \cdot 10^{22}$

   Show the solution

   Given

   air in the lungs, $V = 0.55\,\textsf{l}$, $\theta = 37\,{}^\circ\text{C}$, $p = 101\,\text{kPa}$; $N_\text{A} = 6.022 \cdot 10^{23}\,\tfrac{1}{\text{mol}}$, $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   number of molecules $N$

   Principle

   The equation of state gives the amount of substance, and the Avogadro constant converts it into the number of molecules. Air is a mixture of gases, but the equation of state contains only the total amount of substance of all the particles and no molar mass. So you do not need to know which gases the air consists of.

   Formula

   $pV = nRT$ and $N = n N_\text{A}$, so $n = \dfrac{pV}{RT}$ and $N = \dfrac{pV N_\text{A}}{RT}$

   Conversions

   $V$$= 0.55\,\textsf{l}$

   $= 0.00055\,\text{m}^3$

   $p$$= 101\,\text{kPa}$

   $= 101\,000\,\text{Pa}$

   $T$$= (37 + 273.15)\,\text{K}$

   $= 310.15\,\text{K}$

   Substitution

   The numerator is

   $101\,000\,\text{Pa} \cdot 0.00055\,\text{m}^3 \cdot 6.022 \cdot 10^{23}\,\tfrac{1}{\text{mol}}$

   $= 3.345 \cdot 10^{25}\,\tfrac{\text{J}}{\text{mol}}$

   and the denominator is

   $8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 310.15\,\text{K}$

   $= 2577\,\tfrac{\text{J}}{\text{mol}}$

   so

   $$N = \dfrac{3.345 \cdot 10^{25}\,\tfrac{\text{J}}{\text{mol}}}{2577\,\tfrac{\text{J}}{\text{mol}}} = 1.298 \cdot 10^{22}$$

   Answer

   $1.3 \cdot 10^{22}$

   Molecules, two significant figures. The limiting datum is the volume $0.55\,\textsf{l}$ (two digits); the pressure $101\,\text{kPa}$, the temperature $310\,\text{K}$ (from the whole degrees $37\,{}^\circ\text{C}$) and the constants have three or more digits.

   Reasonableness

   The amount of substance is

   $\dfrac{1.298 \cdot 10^{22}}{6.022 \cdot 10^{23}\,\text{1/mol}}$

   $= 0.0216\,\text{mol}$

   which fits the fact that a litre of gas at room temperature is about $0.04\,\text{mol}$ and a little less at body temperature: $0.55 \cdot 0.04 \approx 0.022\,\text{mol}$. About $21\,\%$ of the molecules are oxygen, so one breath contains about $2.7 \cdot 10^{21}$ oxygen molecules.

2. Exercise 12-17

   In new windows the space between the panes is often filled with argon, because argon conducts heat worse than air.

   The panes of an insulating glass unit are $1.20\,\text{m}$ × $1.00\,\text{m}$. According to the manufacturer, an argon filling makes the unit $6.0\,\text{g}$ heavier than an air filling when the gap is filled at a pressure of $101\,\text{kPa}$ and a temperature of $20\,{}^\circ\text{C}$.

   How thick is the layer of gas between the panes? The molar mass of argon is $40\,\text{g/mol}$ and that of air is $29\,\text{g/mol}$, and $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Give the answer in millimetres.

   mmCheck
   Answer

   $11\,\text{mm}$

   Show the solution

   Given

   insulating glass unit, $A = 1.20\,\text{m}^2$, difference in mass $\Delta m = 6.0\,\text{g}$, $p = 101\,\text{kPa}$, $\theta = 20\,{}^\circ\text{C}$, $M_\text{Ar} = 40\,\text{g/mol}$, $M_\text{air} = 29\,\text{g/mol}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   thickness $d$ of the gas layer in millimetres

   Principle

   The gap is filled at the same pressure and temperature, so the volume is the same and there is the same number of moles of gas with either filling. The difference in mass is due only to the greater molar mass of argon, so you can get the amount of substance from the difference in mass and the difference in molar masses.

   Formula

   $\Delta m = n (M_\text{Ar} - M_\text{air})$, so

   $n$$= \dfrac{\Delta m}{M_\text{Ar} - M_\text{air}}$

   From the equation of state $V = \dfrac{nRT}{p}$ and $V = Ad$, so

   $d$$= \dfrac{nRT}{pA}$

   $= \dfrac{\Delta m\,RT}{pA(M_\text{Ar} - M_\text{air})}$

   Conversions

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

   $= 293.15\,\text{K}$

   $p$$= 101\,\text{kPa}$

   $= 101\,000\,\text{Pa}$

   The mass and the molar masses are in grams, so $n$ comes out directly in moles. Finally $1\,\text{m} = 1000\,\text{mm}$.

   Substitution

   $n$$= \dfrac{6.0\,\text{g}}{40\,\text{g/mol} - 29\,\text{g/mol}}$

   $= 0.5455\,\text{mol}$

   The numerator is

   $0.5455\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 293.15\,\text{K}$

   $= 1328.9\,\text{J}$

   and the denominator is $101\,000\,\text{Pa} \cdot 1.20\,\text{m}^2 = 121\,200\,\text{N}$, because $\text{Pa}\cdot\text{m}^2 = \text{N}$. So

   $$\begin{aligned} d &= \dfrac{1328.9\,\text{J}}{121\,200\,\text{N}} \\ &= 0.01096\,\text{m} \\ &= 10.96\,\text{mm} \end{aligned}$$

   Answer

   $11\,\text{mm}$

   Two significant figures. The limiting data are the difference in mass $6.0\,\text{g}$ and the difference in molar masses $40\,\text{g/mol} - 29\,\text{g/mol} = 11\,\text{g/mol}$, each with two digits; the dimensions $1.20\,\text{m}$ and $1.00\,\text{m}$, the pressure $101\,\text{kPa}$ and $293\,\text{K}$ have three digits.

   Reasonableness

   The gap in an insulating glass unit is usually $10\text{–}20\,\text{mm}$, so the result is believable. Check the masses: the argon is $0.5455\,\text{mol} \cdot 40\,\text{g/mol} = 21.8\,\text{g}$ and the air is $0.5455\,\text{mol} \cdot 29\,\text{g/mol} = 15.8\,\text{g}$, and the difference is $6.0\,\text{g}$.

3. Exercise 12-18

   The compressed air tank of a car workshop's compressor has a volume of $200.0\,\textsf{l}$. In the evening the pressure gauge of the tank shows $10.0\,\text{bar}$, but in the morning only $7.0\,\text{bar}$, because air leaks from a joint.

   The gauge shows gauge pressure, that is, how much greater the pressure in the tank is than atmospheric pressure ($1\,\text{bar}$ = $100\,\text{kPa}$). The temperature of the workshop is $18\,{}^\circ\text{C}$ the whole time.

   How many kilograms of air leaked from the tank during the night? The molar mass of air is $29\,\text{g/mol}$, and $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. Justify why in this calculation the readings of the gauge can be used without adding atmospheric pressure to them.

   kgCheck
   Answer

   $0.72\,\text{kg}$

   Show the solution

   Given

   compressed air tank, $V = 200.0\,\textsf{l}$, gauge pressure before $10.0\,\text{bar}$ and after $7.0\,\text{bar}$, $\theta = 18\,{}^\circ\text{C}$, $M = 29\,\text{g/mol}$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   mass $\Delta m$ of the air that leaked out, in kilograms

   Principle

   The volume and the temperature of the tank stay the same, so the drop in pressure is due only to the decrease in the amount of substance: $\Delta p V = \Delta n RT$. The reading of the gauge is the total pressure minus atmospheric pressure, and atmospheric pressure is the same at the start and at the end. It cancels in the pressure difference, so the readings can be used as they are.

   Formula

   $\Delta n = \dfrac{\Delta p\,V}{RT}$ and $\Delta m = \Delta n\,M$, so $\Delta m = \dfrac{\Delta p\,V M}{RT}$

   Conversions

   $\Delta p$$= 10.0\,\text{bar} - 7.0\,\text{bar}$

   $= 3.0\,\text{bar}$

   $= 3.0 \cdot 10^5\,\text{Pa}$

   $V$$= 200.0\,\textsf{l}$

   $= 0.2000\,\text{m}^3$

   $M$$= 0.029\,\text{kg/mol}$

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

   $= 291.15\,\text{K}$

   Substitution

   $\Delta n$$= \dfrac{3.0 \cdot 10^5\,\text{Pa} \cdot 0.2000\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 291.15\,\text{K}}$

   $= 24.80\,\text{mol}$

   $\Delta m$$= 24.80\,\text{mol} \cdot 0.029\,\tfrac{\text{kg}}{\text{mol}}$

   $= 0.7192\,\text{kg}$

   Answer

   $0.72\,\text{kg}$

   Two significant figures. The limiting datum is the pressure difference $10.0\,\text{bar} - 7.0\,\text{bar} = 3.0\,\text{bar}$: the difference of the readings has the decimals of the less precise reading, so it has two digits. The volume $200.0\,\textsf{l}$, the molar mass $29\,\text{g/mol}$ and $291\,\text{K}$ have at least two digits and do not limit the result.

   Reasonableness

   At the start the tank contained, at a total pressure of $11.0\,\text{bar}$, about $2.6\,\text{kg}$ of air, so the leaked $0.72\,\text{kg}$ is about $27\,\%$. The gauge pressure fell $10.0 \to 7.0\,\text{bar}$, which is $30\,\%$ of the gauge pressure and a little more than $27\,\%$ of the total pressure, because atmospheric pressure does not leak out. The amount of air that leaked was $24.8\,\text{mol}$, and that is a sensible amount for a drop in pressure of three bars.

4. Exercise 12-19

   At a party, balloons are filled from a helium cylinder whose volume is $20.0\,\textsf{l}$. The cylinder was first in a cold storeroom and was then used indoors in two stages. The table shows the total pressure indicated by the pressure gauge in bars ($1\,\text{bar} = 100\,\text{kPa}$) at three moments (example data set). The temperature of the gas in the cylinder is always the same as that of the surroundings, and the cylinder does not leak. Treat the helium as an ideal gas and use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$. The amount of helium that goes into one filled balloon is such that the balloon takes up a volume of $5.0\,\textsf{l}$ at atmospheric pressure $101\,\text{kPa}$ and temperature $21.0\,{}^\circ\text{C}$.

   a) Calculate the amount of substance of helium in the cylinder at moments 1, 2 and 3.

   b) Calculate the mass of helium used between moments 1 and 3. The molar mass is $M = 4.0\,\tfrac{\text{g}}{\text{mol}}$. Explain why the amount used cannot be calculated directly from the difference of the pressures.

   c) How many balloons were filled with the helium used? How many more balloons can be obtained from the cylinder when it is used until the pressure has fallen to atmospheric pressure $101\,\text{kPa}$?

   The answer box takes the mass of part b).

   Temperature and total pressure of the helium cylinder (example data set)

   Moment	$\theta\,({}^\circ\text{C})$	$p\,(\text{bar})$
   1 (storeroom, before use)	5.0	14.2
   2 (indoors, after stage 1)	21.0	11.0
   3 (indoors, after stage 2)	21.0	6.5

   gCheck
   Answer

   $28\,\text{g}$

   Show the solution

   Given

   helium, $V = 20.0\,\textsf{l}$; from the table moment 1: $\theta_1 = 5.0\,{}^\circ\text{C}$, $p_1 = 14.2\,\text{bar}$, moment 2: $\theta_2 = 21.0\,{}^\circ\text{C}$, $p_2 = 11.0\,\text{bar}$, moment 3: $\theta_3 = 21.0\,{}^\circ\text{C}$, $p_3 = 6.5\,\text{bar}$; balloon $p_\text{b} = 101\,\text{kPa}$, $\theta_\text{b} = 21.0\,{}^\circ\text{C}$, $V_\text{b} = 5.0\,\textsf{l}$; $M = 4.0\,\tfrac{\text{g}}{\text{mol}}$, $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   the mass $m$ of helium used in grams (part b); the amounts of substance $n_1$, $n_2$, $n_3$ (a) and the numbers of balloons (c)

   Principle

   The volume of the cylinder is constant and no gas leaks, so the equation of state $pV = nRT$ gives the amount of substance at each moment separately. The temperature changes between moments 1 and 2, so the amount of substance is not directly proportional to the pressure: the same pressure corresponds to a greater amount of substance in the cold. The amount used is the difference of the amounts of substance.

   Formula

   $n = \dfrac{pV}{RT}$, the amount used $\Delta n = n_1 - n_3$ and the mass $m = \Delta n \cdot M$. The amount of substance in one balloon is

   $n_\text{b}$$= \dfrac{p_\text{b}V_\text{b}}{RT_\text{b}}$

   and the number of balloons is $\dfrac{\Delta n}{n_\text{b}}$.

   Conversions

   $V$$= 20.0\,\textsf{l}$

   $= 0.0200\,\text{m}^3$

   $V_\text{b} = 0.0050\,\text{m}^3$;

   $p_1$$= 14.2\,\text{bar}$

   $= 1.42 \cdot 10^6\,\text{Pa}$

   $p_2 = 1.10 \cdot 10^6\,\text{Pa}$ and $p_3 = 6.5 \cdot 10^5\,\text{Pa}$;

   $T_1$$= (5.0 + 273.15)\,\text{K}$

   $= 278.15\,\text{K}$

   $T_2$$= T_3$

   $= T_\text{b}$

   $= 294.15\,\text{K}$

   Substitution

   a)

   $n_1$$= \dfrac{1.42 \cdot 10^6\,\text{Pa} \cdot 0.0200\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 278.15\,\text{K}}$

   $= 12.29\,\text{mol}$

   $n_2$$= \dfrac{1.10 \cdot 10^6\,\text{Pa} \cdot 0.0200\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 294.15\,\text{K}}$

   $= 9.000\,\text{mol}$

   $n_3$$= \dfrac{6.5 \cdot 10^5\,\text{Pa} \cdot 0.0200\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 294.15\,\text{K}}$

   $= 5.318\,\text{mol}$

   b)

   $\Delta n$$= 12.29\,\text{mol} - 5.318\,\text{mol}$

   $= 6.968\,\text{mol}$

   $m$$= 6.968\,\text{mol} \cdot 4.0\,\tfrac{\text{g}}{\text{mol}}$

   $= 27.87\,\text{g}$

   c)

   $n_\text{b}$$= \dfrac{101\,000\,\text{Pa} \cdot 0.0050\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 294.15\,\text{K}}$

   $= 0.2066\,\text{mol}$

   so the helium used filled

   $\dfrac{6.968\,\text{mol}}{0.2066\,\text{mol}}$$= 33.7$

   balloons. When the pressure is $101\,\text{kPa}$, the cylinder contains

   $n_\text{end}$$= \dfrac{101\,000\,\text{Pa} \cdot 0.0200\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 294.15\,\text{K}}$

   $= 0.8264\,\text{mol}$

   so $5.318\,\text{mol} - 0.8264\,\text{mol} = 4.492\,\text{mol}$ is still available, enough for

   $\dfrac{4.492\,\text{mol}}{0.2066\,\text{mol}}$$= 21.7$

   balloons.

   Answer

   b) $28\,\text{g}$, two significant figures, because the molar mass $4.0\,\tfrac{\text{g}}{\text{mol}}$ is given with two digits. a) $n_1 = 12.3\,\text{mol}$, $n_2 = 9.00\,\text{mol}$ and $n_3 = 5.32\,\text{mol}$. c) The helium used filled about $34$ balloons (about $16$ in the first stage and about $18$ in the second), and $21$ more full balloons can still be filled. From the pressure difference $14.2 - 6.5 = 7.7\,\text{bar}$ the mass would have been calculated as about $25\,\text{g}$, which is too small, because at the first moment the cylinder contained cold gas: the same pressure corresponds to a greater amount of substance in the cold gas, so the initial amount of substance is greater than the pressure difference would suggest.

   Reasonableness

   At the start the cylinder contained about $14 \cdot 20 : (0.083 \cdot 278) \approx 12\,\text{mol}$ and at the end about $5\,\text{mol}$, so the difference is about $7\,\text{mol}$ and $7 \cdot 4\,\text{g} \approx 28\,\text{g}$. One litre of gas at room temperature is about $0.041\,\text{mol}$, so a $5\,\textsf{l}$ balloon needs about $0.21\,\text{mol}$, and $7 : 0.21 \approx 33$ agrees with the result.

5. Exercise 12-20

   A weather balloon is filled with helium at ground level, and its volume is then $2.50\,\text{m}^3$. The table shows the pressure and temperature of the atmosphere at different heights (ICAO standard atmosphere, rounded). Assume that the pressure and temperature of the gas in the balloon are always the same as those of the surrounding air, that no gas escapes and that the balloon rises. Use the value $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$.

   a) Calculate the amount of substance of helium in the balloon at ground level.

   b) Calculate the volume of the balloon at a height of $10\,\text{km}$.

   c) The envelope of the balloon withstands a volume of at most $12.0\,\text{m}^3$. Calculate the volume at a height of $15\,\text{km}$ and estimate up to what height the balloon withstands: assume the volume is linear with respect to height between the heights $10\,\text{km}$ and $15\,\text{km}$. Explain why the volume increases even though the gas cools.

   The answer box takes the volume of part b).

   Standard atmosphere (ICAO, rounded)

   $h\,(\text{km})$	$p\,(\text{kPa})$	$\theta\,({}^\circ\text{C})$
   0	101	15.0
   5	54.0	−17.5
   10	26.5	−50.0
   15	12.1	−56.5

   m³Check
   Answer

   $7.4\,\text{m}^3$

   Show the solution

   Given

   helium, $V_0 = 2.50\,\text{m}^3$; from the table ground level $p_0 = 101\,\text{kPa}$, $\theta_0 = 15.0\,{}^\circ\text{C}$, height $10\,\text{km}$: $p = 26.5\,\text{kPa}$, $\theta = -50.0\,{}^\circ\text{C}$ and height $15\,\text{km}$: $p = 12.1\,\text{kPa}$, $\theta = -56.5\,{}^\circ\text{C}$; $V_\text{max} = 12.0\,\text{m}^3$; $R = 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}}$

   Find

   volume $V$ at a height of $10\,\text{km}$ (part b)

   Principle

   The balloon is closed, so the amount of substance $n$ stays the same. The pressure and temperature of the atmosphere fall, and the pressure of the gas in the balloon is the same as that of the surroundings. The equation of state $pV = nRT$ gives the volume at each height.

   Formula

   a) $n = \dfrac{p_0V_0}{RT_0}$. b) and c) $V = \dfrac{nRT}{p}$. In the linear estimate the height is

   $h$$= h_{10} + (h_{15} - h_{10}) \cdot \dfrac{V_\text{max} - V_{10}}{V_{15} - V_{10}}$

   Conversions

   $p_0$$= 101\,000\,\text{Pa}$

   $T_0$$= (15.0 + 273.15)\,\text{K}$

   $= 288.15\,\text{K}$

   $T_{10}$$= (-50.0 + 273.15)\,\text{K}$

   $= 223.15\,\text{K}$

   $T_{15} = 216.65\,\text{K}$ and $p_{10} = 26\,500\,\text{Pa}$, $p_{15} = 12\,100\,\text{Pa}$.

   Substitution

   a)

   $n$$= \dfrac{101\,000\,\text{Pa} \cdot 2.50\,\text{m}^3}{8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 288.15\,\text{K}}$

   $= 105.4\,\text{mol}$

   b)

   $V_{10}$$= \dfrac{105.4\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 223.15\,\text{K}}{26\,500\,\text{Pa}}$

   $= 7.379\,\text{m}^3$

   c)

   $V_{15}$$= \dfrac{105.4\,\text{mol} \cdot 8.31\,\tfrac{\text{J}}{\text{mol}\cdot\text{K}} \cdot 216.65\,\text{K}}{12\,100\,\text{Pa}}$

   $= 15.69\,\text{m}^3$

   $h$$= 10\,\text{km} + 5\,\text{km} \cdot \dfrac{(12.0 - 7.379)\,\text{m}^3}{(15.69 - 7.379)\,\text{m}^3}$

   $= 12.8\,\text{km}$

   Answer

   b) $7.4\,\text{m}^3$, two significant figures, because the ICAO values are rounded and the gas of the balloon is not exactly at the temperature of the surrounding air. The amount of helium is a) $105\,\text{mol}$. c) At a height of $15\,\text{km}$ the volume would be $15.7\,\text{m}^3$, which is over the limit of the envelope, $12.0\,\text{m}^3$. According to the linear estimate the balloon withstands up to a height of about $12.8\,\text{km}$ and bursts after that; the real limit is slightly higher, because the volume grows at an accelerating rate. The volume increases because the pressure falls by the factor $26.5 : 101 \approx 0.26$ but the temperature in kelvins only by the factor $223.15 : 288.15 \approx 0.77$. The volume is directly proportional to the temperature and inversely proportional to the pressure, so the fall in pressure wins over the fall in temperature.

   Reasonableness

   The ratio of the volumes is

   $V_{10} : V_0$$= 0.77 : 0.26$

   $\approx 3.0$

   $3.0 \cdot 2.50\,\text{m}^3 \approx 7.4\,\text{m}^3$. The mass of the helium is

   $105.4\,\text{mol} \cdot 4.0\,\tfrac{\text{g}}{\text{mol}}$

   $\approx 0.42\,\text{kg}$

   which is reasonable for a weather balloon. The limit $12.0\,\text{m}^3$ lies between the volumes $7.4\,\text{m}^3$ and $15.7\,\text{m}^3$, so the height $12.8\,\text{km}$ lies between the heights $10\,\text{km}$ and $15\,\text{km}$.