Molar Specific Heat of an Ideal Gas

For gases, even with the same amount of heat applied, the temperature increase varies depending on how much the volume changes. Thus, to calculate the molar specific heat, one variable must be held constant. Here, we will examine how to determine the molar specific heat at constant volume and at constant pressure for an ideal gas.

Molar Specific Heat at Constant Volume

If we define \(C_V\) as the molar specific heat at constant volume,

$$Q=nC_V\Delta T \cdots (1)$$

$$C_V=\frac{Q}{n \cdot \Delta T}$$

However, if the number of atoms of the gas is known, the specific heat can be determined without knowing \(Q\) and \(\Delta T\). For example, the process of finding the specific heat for a monatomic gas is as follows.

As this process occurs at constant volume,

$$\Delta E_{int}=Q+W$$

$$W=0$$

$$\Delta E_{int}=Q \cdots (2)$$

By inserting equation (1) into equation (2), we obtain:

$$\Delta E_{int}=nC_V\Delta T$$

Solving for \(C_V\), we obtain:

$$
C_V=\frac{1}{n} \frac{dE_{int}}{dT} \cdots (3)
$$

Meanwhile, if the gas is monatomic, only translational kinetic energy exists.

$$E_{int}=K_{tot \space trans}=\frac{3}{2}nRT \cdots (4)$$

By inserting equation (4) into equation (3), we obtain:

$$C_V=\frac{3}{2}R=12.5J/mol \cdot K$$

Molar Specific Heat at Constant Pressure

If we define \(C_P\) as the molar specific heat at constant pressure,

$$Q=nC_P \Delta T$$

$$C_P=\frac{Q}{n \cdot \Delta T}$$

Relationship between Molar Specific Heat at Constant Volume and Molar Specific Heat at Constant Pressure

In Terms of Molar Specific Heat at Constant Volume

As stated earlier

$$\Delta E_{int}=nC_V \Delta T$$

In Terms of Molar Specific Heat at Constant Pressure

$$\Delta E_{int}=Q+W=nC_P \Delta T+(-P\Delta V) \cdots (1)$$

As it is ideal gas,

$$PV=nRT$$

As the pressure is constant,

$$PV_i=nRT_i$$

$$PV_f=nRT_f$$

$$PV_f-PV_i=nRT_f-nRT_i$$

$$P\Delta V=nR \Delta T \cdots (2)$$

By inserting equation (2) into equation (1), we obtain:

$$\Delta E_{int}=nC_P \Delta T-nR \Delta T$$

Combining

Since the change in internal energy is the same if the temperature change is identical, regardless of whether the process occurs at constant volume or constant pressure,

$$\Delta E_{int}=nC_V \Delta T=nC_P \Delta T-nR \Delta T$$

By simplifying the above equation, we obtain:

$$C_V=C_P-R$$

That is, the molar specific heat at constant volume is obtained by subtracting \(R\) from the molar specific heat at constant pressure.