The temperature of 1.00 mol of a monatomic ideal gas is raised reversibly from 300 K to 400 K , with its volume kept constant. What is the entropy change of the gas?

The number of moles, n = 1.00 mol.

The gas ismonatomic.

The initial temperature,$T_i=300K$.

The final temperature,$T_f=400K$.

The volume is constant.

If the volume is constant, then the heat absorbed is equal to the change in its internal energy. Using the equation for change in entropy from Equation 20-1, we can find the change in the entropy of the gas.

Formula:

The change in internal energy of the gas, ${\mathrm{dE}}_{\mathrm{int}}=\frac{3}{2}\mathrm{nRdT}$ (1)

The change in entropy of the gas, $∆\mathrm{S}=\int \frac{\mathrm{dQ}}{\mathrm{T}}$ (2)

Since the volume of the monatomic gas is kept constant, it does not do any work in the heating process. Therefore, the heat Q it absorbs is equal to the change in its internal energy. That is using equation (1),

$$\begin{gathered} \mathrm{dQ}={\mathrm{dE}}_{\mathrm{int}} \\ =\frac{3}{2}\mathrm{nRdT} \end{gathered}$$

Thus, the change in entropy using equation (2) and the given values is given by:

$$\begin{gathered} ∆\mathrm{S}={\int }_{{\mathrm{T}}_{\mathrm{i}}}^{{\mathrm{T}}_{\mathrm{f}}}\frac{\left({\displaystyle \frac{3\mathrm{nR}}{2}}\right)\mathrm{dT}}{\mathrm{T}} \\ =\frac{3}{2}\mathrm{nRln}\left(\frac{{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{T}}_{\mathrm{i}}}\right) \\ =\frac{3}{2}\times 1.00\mathrm{mol}\times 8.31\frac{\mathrm{J}}{\mathrm{mol}·\mathrm{K}}\times \left(\frac{400\mathrm{K}}{300\mathrm{K}}\right) \\ =3.59\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the entropy change of the gas is 3.59 J/K