Specific Heats of Gases and Thermodynamic Processes: Quick Revision

A brief note on Specific Heats of Gases and Thermodynamic Processes. Read it before a test. Try each question at the end before you open its answer.

The big ideas

• Two heat capacities. At a steady volume $Q = nC_v\,\Delta T$; at a steady pressure $Q = nC_p\,\Delta T$, which is more because the gas also does work.
• Mayer's relation: $C_p - C_v = R$. With $f$ ways to move, $C_v = \tfrac{f}{2}R$ and $\gamma = \dfrac{C_p}{C_v} = 1 + \dfrac{2}{f}$.
• $\Delta U = nC_v\,\Delta T$ in every process of an ideal gas.
• Four processes: steady temperature, steady pressure, steady volume, and no heat (adiabatic).
• An adiabat is steeper than an isotherm, because the gas also cools as it expands.

The facts at a glance

Monatomic: $C_v = 12.5\ \text{J mol}^{-1}\text{K}^{-1}$, $C_p = 20.8\ \text{J mol}^{-1}\text{K}^{-1}$, $\gamma = 1.67$. Diatomic: $20.8\ \text{J mol}^{-1}\text{K}^{-1}$, $29.1\ \text{J mol}^{-1}\text{K}^{-1}$, $\gamma = 1.40$. Adiabatic: $pV^\gamma$ and $TV^{\gamma - 1}$ stay fixed.

A worked example

$0.75\ \text{mol}$ of a diatomic gas at $310\ \text{K}$ doubles its volume (a) at a steady pressure, (b) at a steady temperature and (c) adiabatically. Find the work each time.

1. Steady pressure. $T$ doubles to $620\ \text{K}$: $W = nR\,\Delta T = 0.75\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 310\ \text{K} = 1932\ \text{J}$.
2. Steady temperature. $W = 0.75\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 310\ \text{K} \times \ln 2 = 1339\ \text{J}$.
3. Adiabatic. $T_2 = 310\ \text{K} \times 2^{-0.40} = 234.9\ \text{K}$, so $W = 0.75\ \text{mol} \times 20.8\ \text{J mol}^{-1}\text{K}^{-1} \times 75.1\ \text{K} = 1170\ \text{J}$.
4. So the steady pressure path does the most work; the adiabatic one the least.

Practice with new numbers

A program to try

The program works out the example, with $\Delta U$ and $Q$ too.

Mistakes to avoid

• $C_p$ for $\Delta U$. $\Delta U$ always uses $C_v$, even at a steady pressure.
• 'Isothermal means no heat'. It means no change of $U$; the heat equals the work.
• Boyle's law in an adiabatic change. Use $pV^\gamma$, not $pV$.
• Celsius in $TV^{\gamma - 1}$. Use kelvin.
• The sign of the logarithm. A squeeze has $V_2 < V_1$, so $\ln\dfrac{V_2}{V_1}$ and $W$ are negative.

Check yourself

1. In which process does a gas do no work?

2. A gas has $C_p = 4.5R$. What are its $C_v$ and $\gamma$?

3. A monatomic gas is squeezed adiabatically to $\tfrac{1}{27}$ of its volume. By what factor does its temperature grow?

4. Why does a gas cool when it expands adiabatically?

5. A diatomic gas takes in $560\ \text{J}$ at a steady pressure. How much of it is work?