What is the entropy change for 3.20 molof an ideal monatomic gas undergoing a reversible increase in temperature from 380 K to 425 Kat constant volume?

The number of moles of ideal monatomic gas, n = 3.20

The initial temperature of the gas, $T_i=380K$

The final temperature of the gas, $T_f=425K$

The ideal monatomic gas undergoes a reversible process. In this, the temperature of the gas increases but the volume remains constant. Hence, we use the specific heat of the gas at constant volume to calculate the entropy change of the gas.

Formula:

The entropy change of the gas at constant volume, $∆S=n{C}_v\ln \left(\frac{T_f}{T_i}\right)$ (1)

For an ideal monatomic gas,

The entropy change for a constant volume process is given by the equation (1) and the given values as follows:

$$\begin{gathered} ∆S=3.20\times \frac{3}{2}\times 8.31\times \ln \left(\frac{425K}{380K}\right) \\ =\frac{3.20\times 3\times 8.31J/K\times 0.11}{2} \\ =4.46J/K \end{gathered}$$

Hence, the value of the change in entropy of an ideal monatomic gas is 4.46 J/K