The heat input (Q) required to raise the temperature of n moles of gas from T_{1} to T_{2} depends not only on ΔT but also on how the pressure and volume of the gas are changed.

**There are two important heat capacities:**

Heating or cooling at constant volume: Q_{v} = nC_{v}ΔT, where C_{v} is the molar heat capacity at constant volume.

Heating or cooling at constant pressure: Q_{p} = nC_{p}ΔT, where C_{p} is the molar heat capacity at constant pressure.

### Heat capacity at constant volume (C_{v}) for gases

**Significance of C _{v}:** The heat input to the gas at the constant volume goes solely into increasing the internal energy (E

_{int}) of the gas, without doing any mechanical work.

$$Q_{V} = n C_{V} \Delta T = \Delta E_{int}$$

$\Delta E_{int}$ comprises of the change in translational KE, rotational KE, vibrational energies of the gas particles. The significance of $E_{int}$ will be discussed later.

**For mono-atomic ideal gases:**

The change in internal energy is given by the change in translational kinetic energy of the atoms:

$$\Delta E_{int} = \Delta E_{trans} = \frac{3}{2} n R \Delta T$$

Hence, the heat capacity at constant volume per mole of gas:

$$C_{v} = \frac{3}{2} R$$

which = 12.5 $\text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ for monatomic ideal gas

Therefore, the heat capacity at constant volume per particle:

$$C_{v} = \frac{3}{2} k_{B} $$

**For diatomic ideal gases:**

The change in internal energy at room temperature is given by the change in translational kinetic energy of the molecules plus the change in rotational kinetic energy:

$$\begin{aligned} \Delta E_{int} &= \Delta E_{trans} + \Delta E_{rot} \\ &= \frac{3}{2} nR \Delta T + \frac{2}{2} nR \Delta T \end{aligned}$$

Note that there are 2 rotational degrees of freedom for a diatomic gas.

hence,

$$C_{v} = \frac{5}{2} R$$

which = 20.8 $\text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ for diatomic ideal gas at room temp.

Only at sufficiently high temperatures is the vibrational mode also activated, and so:

$$\begin{aligned} \Delta E_{int} &= \Delta E_{trans} + \Delta E_{rot} + \Delta E_{vib} \\ &= \frac{3}{2} n R \Delta T + nR \Delta T + nR \Delta T \end{aligned} $$

Note that there is 1 vibrational mode for diatomic gas.

hence,

$$C_{v} = \frac{7}{2} R$$

which = 29.1 $\text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ for diatomic ideal gas at high temp.

**Polyatomic gases: with increasing number of atoms, we can expect C _{v} to increase.**

### Heat Capacity at constant pressure

The heat input to the gas at constant pressure goes into increasing the internal energy (E_{int}) of the gas and also into doing mechanical expansion work (-W).

$$Q_{p} = n C_{p} \Delta T = \Delta E_{\text{int}} – W$$

Note that we use the sign convention that work done on gas corresponds to positive W, while work done by the gas corresponds to negative W.

The expansion work is given by -W = PΔV = nRΔT (Ideal gas) which contributes R to C_{p}.

**For a mono-atomic ideal gas,**

$$C_{p} = \frac{5}{2} R$$

which = 20.8 $\text{J} \, \text{K}^{-1} \, \text{mol}^{-1}$ for monatomic ideal gas

The ΔE_{int} term contributes 1.5 R, while the -W term contributes another R to C_{p}.

This gives C_{p} – C_{v} = R = 8.314 J K^{-1} mol^{-1} (for all ideal gases) and heat capacity ratio $\gamma = \frac{C_{p}}{C_{v}} = 1.667$ (for all mono-atomic gases).

** Molar heat capacities of some gases (at 20°C and 1 atm)**

Next: Crisis for equipartition theorem