Polytropic processes and gas mixtures

A gas pushing on a spring

A gas is trapped in a cylinder. The piston presses on a spring, and the spring is slowly squashed as the gas expands. The more the gas expands, the harder the spring pushes back, so the pressure of the gas goes up as the volume goes up. Is this isothermal, adiabatic, isobaric or isochoric? It is none of them.

Most real processes are like this. A tyre, a balloon, a piston with friction and a cylinder with leaky walls all give paths that lie between the standard ones. We need a family of curves that includes the standard ones, so that one formula covers all of them. That family is the polytropic family.

The polytropic family

A polytropic process of an ideal gas is one in which

The number $x$ is the polytropic index. Many books write $n$, but we keep $n$ for the number of moles. The four standard processes are special cases.

Why is the heat capacity of an isothermal process infinite? Heat goes in, and the temperature does not change. The heat divided by the temperature change is then infinite.

Using $p = \dfrac{nRT}{V}$, the condition $pV^x =$ constant becomes

This is the same step that gave $TV^{\gamma - 1} =$ constant for the adiabatic process. In fact the adiabatic process is the polytropic one with $x = \gamma$.

Work, internal energy and heat

For $x \ne 1$, the work done by the gas is the area under the curve:

For $x = 1$ the process is isothermal, and $W = nRT\ln\dfrac{V_2}{V_1}$. The change of internal energy of an ideal gas never depends on the path, so $\Delta U = nC_v\,\Delta T$ for every process. The first law now gives the heat:

Here we used $T_1 - T_2 = -\Delta T$. We write this as $Q = nC\,\Delta T$, and so the molar heat capacity of the polytropic process is

The second form uses $C_v = \dfrac{R}{\gamma - 1}$. Test it on the table. For $x = 0$ it gives $C_v\gamma = C_p$. For $x = \gamma$ it gives $0$. For a very large $x$ it gives $C_v$. For $x = 1$ it has a zero in the denominator, which is the infinite heat capacity. The formula passes all four tests.

A heat capacity that is negative

Look at the factor $x - \gamma$ in the formula. For $1 < x < \gamma$ the molar heat capacity is negative. What does that mean? Take $x = 1.3$ for a diatomic gas, with $\gamma = 1.4$. The gas expands along this path, so it does work. It does more work than the heat it takes in, so it gives up internal energy, and its temperature falls. The heat is positive while the temperature change is negative, so the ratio is negative.

A negative heat capacity does not break any law. It only says that, on this path, taking heat in makes the gas cooler.

A worked example. 1.5 mol of nitrogen, with $f = 5$, is at 350 K. It expands along a path with $x = 1.3$ until its volume is 3 times as large. Find the final temperature, the work, the change of internal energy, the heat taken in and the molar heat capacity of the path.

1. The final temperature. $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{x-1} = 350\ \text{K} \times \left(\dfrac{1}{3}\right)^{0.3} = 251.7\ \text{K}$. The change is $\Delta T = -98.3$ K.
2. The work. First, $nR(T_1 - T_2) = 1.5\ \text{mol} \times 8.314\ \text{J/(mol K)} \times 98.3\ \text{K} = 1226\ \text{J}$. Then $W = \dfrac{1226\ \text{J}}{x - 1} = \dfrac{1226\ \text{J}}{0.3} = 4085\ \text{J}$.
3. The internal energy. $\Delta U = nC_v\Delta T = 1.5\ \text{mol} \times 20.79\ \text{J/(mol K)} \times (-98.3\ \text{K}) = -3064\ \text{J}$.
4. The heat. $Q = \Delta U + W = -3064\ \text{J} + 4085\ \text{J} = 1021\ \text{J}$. It is positive, so the gas takes in heat.
5. The heat capacity. $C = \dfrac{Q}{n\Delta T} = \dfrac{1021\ \text{J}}{1.5\ \text{mol} \times (-98.3\ \text{K})} = -6.93\ \text{J/(mol K)}$. The formula $C_v\dfrac{x - \gamma}{x - 1} = 20.79 \times \dfrac{1.3 - 1.4}{0.3}\ \text{J/(mol K)}$ gives the same, $-6.93\ \text{J/(mol K)}$.

The gas took in 1021 J and its temperature fell by 98.3 K.

Find the value of $x$ at which the heat is zero. Then move $x$ above that value. Now the gas expands and gives out heat. Move $x$ below $1$. What do you notice about the temperature at the end of the expansion?

A straight line on the p-V diagram

A line that is straight on a p-V graph is not a polytropic curve, because its index changes from point to point. But at each point the line behaves like a polytropic curve with a local index

This is the steepness of the line, written in a form that does not depend on the units.

When is heat taken in? During an expansion, heat enters while the local index is below $\gamma$, which means that the line is less steep than the adiabat through the same point. Heat leaves while the line is steeper than the adiabat. At the point where the line just touches an adiabat, the heat changes sign.

Take the line $p = a - bV$. Heat enters while $b < \dfrac{\gamma p}{V}$. Put $p = a - bV$ in this, and the heat stops entering at the volume

The temperature is highest where $pV$ is largest, which is at $V = \dfrac{a}{2b}$. Because $\dfrac{\gamma}{\gamma + 1}$ is larger than $\dfrac{1}{2}$, the heat goes on entering after the temperature has peaked. In that part, the gas takes in heat and cools.

Gas mixtures

Two or more ideal gases in one container, without any reaction, share the same volume and the same temperature. Each gas pushes with its own partial pressure, and the pressures add (Dalton's law). The energy also adds: $U = U_1 + U_2$. So the heat capacities add too:

The same rule holds for $C_p$, and $\gamma_{\text{mix}} = \dfrac{C_p}{C_v}$. Do not average the values of $\gamma$. The values of $\dfrac{1}{\gamma - 1}$ are the ones that add with the moles:

The mean molar mass is $M = \dfrac{n_1M_1 + n_2M_2}{n_1 + n_2}$. An adiabatic change of the mixture follows $pV^{\gamma_{\text{mix}}} =$ constant, with the exponent of the mixture.

A worked example. A tank holds 2.0 mol of helium and 3.0 mol of oxygen at 300 K. The gas is compressed adiabatically until its volume is one quarter of the starting volume.

1. The heat capacity. $C_v = \dfrac{2.0 \times \frac{3}{2} + 3.0 \times \frac{5}{2}}{2.0 + 3.0}R = 2.10R$ and $C_p = 3.10R$.
2. The exponent. $\gamma_{\text{mix}} = \dfrac{3.10}{2.10} = 1.476$. It lies between $\dfrac{7}{5}$ and $\dfrac{5}{3}$, as it must.
3. The final temperature. $T_2 = T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma - 1} = 300\ \text{K} \times 4^{0.476} = 581\ \text{K}$.
4. The work done on the gas. It all goes into internal energy, $\Delta U = nC_v\Delta T = 5.0\ \text{mol} \times 2.10 \times 8.314\ \text{J/(mol K)} \times 281\ \text{K} = 24488\ \text{J}$.
5. The mean molar mass. $M = \dfrac{2.0 \times 4 + 3.0 \times 32}{5.0}\ \text{g/mol} = 20.8\ \text{g/mol}$.

Two gases at two temperatures. Suppose that two gases at different temperatures are mixed in a rigid, insulated box. No work is done and no heat leaves, so the total internal energy stays the same. Then $n_1C_{v1}(T - T_1) + n_2C_{v2}(T - T_2) = 0$, and

For 1.0 mol of argon at 400 K and 2.0 mol of hydrogen ($C_v = \frac{5}{2}R$) at 250 K, the numerator is $1.0 \times 1.5 \times 400 + 2.0 \times 2.5 \times 250 = 1850$ and the denominator is $1.0 \times 1.5 + 2.0 \times 2.5 = 6.5$, both in units of $R$. So $T = \dfrac{1850}{6.5}\ \text{K} = 284.6\ \text{K}$. The factor $C_v$ weights the temperatures, not only the moles.

Degrees of freedom that change

The kinetic theory gives $C_v = \dfrac{f}{2}R$ and $\gamma = 1 + \dfrac{2}{f}$, where $f$ is the number of degrees of freedom that hold energy.

The value of $f$ is not fixed for a gas. A diatomic gas at very low temperature holds energy only in the motion of the whole molecule, so $f = 3$. As the temperature rises, the rotation switches on and $f$ becomes $5$. At a few thousand kelvin, the atoms also vibrate and $f$ rises to $7$. A mode switches on when the typical energy of a collision, about $kT$, is large enough to excite it. Hydrogen is the clearest case. Its $C_v$ is $\tfrac{3}{2}R$ well below $100$ K and is close to $\tfrac{5}{2}R$ at room temperature.

A gas can also change its number of particles. In a hot diatomic gas, a fraction $\alpha$ of the molecules may split into atoms. Take one mole of molecules at the start, and ignore the energy that it takes to break the bonds. Then there are $1 - \alpha$ moles of molecules with $f = 5$ and $2\alpha$ moles of atoms with $f = 3$. The total is $1 + \alpha$ moles, and

For $\alpha = 0.20$, this gives $C_v = 2.167R$ and $\gamma = 1.462$. The gas has moved a little towards a monatomic gas.

Practise with new numbers

Try it in Python

The program takes 2 mol of oxygen through four polytropic paths, all from 300 K to twice the volume. For each path, it finds the final temperature, the work, the change of internal energy and the heat. It then finds the molar heat capacity in two ways and prints both.

Common mistakes

• Using the formula for work at $x = 1$. The expression $\dfrac{p_1V_1 - p_2V_2}{x-1}$ has a zero in the denominator. Use $nRT\ln\dfrac{V_2}{V_1}$ for the isothermal case.
• Using $Q = nC_v\Delta T$ for a polytropic path. That gives only $\Delta U$. The heat needs the heat capacity of the path, $C = C_v - \dfrac{R}{x-1}$.
• Believing that heat taken in always raises the temperature. For $1 < x < \gamma$ an expanding gas takes in heat and cools.
• Averaging the values of $\gamma$ for a mixture. Average $C_v$ over the moles and then find $\gamma$.
• Using $\tfrac{5}{2}R$ for a diatomic gas at every temperature. It holds near room temperature only.
• Mixing the two meanings of $n$. Here $n$ is the number of moles and $x$ is the polytropic index.

Remember these

• Polytropic process: $pV^x =$ constant and $TV^{x-1} =$ constant. The index is $0$ for constant pressure, $1$ for isothermal, $\gamma$ for adiabatic and very large for constant volume.
• Work: $W = \dfrac{p_1V_1 - p_2V_2}{x-1} = \dfrac{nR(T_1 - T_2)}{x-1}$. Internal energy: $\Delta U = nC_v\Delta T$.
• Heat: $Q = nC\Delta T$ with $C = C_v - \dfrac{R}{x-1} = C_v\dfrac{x-\gamma}{x-1}$. It is negative for $1 < x < \gamma$.
• On a straight line $p = a - bV$, heat enters up to $V^* = \dfrac{\gamma a}{(\gamma+1)b}$, and the temperature peaks at $\dfrac{a}{2b}$.
• Mixture: $C_v = \dfrac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2}$, then $\gamma = \dfrac{C_v + R}{C_v}$. Never average the values of $\gamma$.
• Degrees of freedom: $C_v = \dfrac{f}{2}R$ and $f$ rises with temperature as rotation and vibration switch on.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A monatomic gas is compressed along a path with $x = 1.5$. Does it take in heat or give out heat? Does its temperature rise or fall?

2. A mixture of helium and oxygen has a value of $\gamma$ between $\dfrac{7}{5}$ and $\dfrac{5}{3}$. Why must it lie between them?

3. The heat capacity at constant volume of hydrogen falls to $\tfrac{3}{2}R$ at very low temperatures. Why?

4. A gas expands along the straight line $p = 400 - 50V$, with $p$ in kPa and $V$ in L. At what volume is its temperature highest, and at what volume does a diatomic gas stop taking in heat?

References

• OpenStax University Physics Volume 2, Chapter 2 The Kinetic Theory of Gases and Chapter 3 The First Law of Thermodynamics, on heat capacities and processes of an ideal gas.
• Irodov, Problems in General Physics, the chapter on thermodynamics and molecular physics, for the polytropic process and its heat capacity.