Question

Two mole of an ideal gas $(C_{p,m}=\frac{5}{2}R)$ is heated from 300 K to 600 K at constant pressure. The change in entropy of the gas $\left(\Delta S\right)$ is:

• A. $\frac{3}{2}R\ln 2$
• B. $-\frac{3}{2}R\ln 2$
• C. $5R\ln 2$
• D. $\frac{5}{2}R\ln 2$

Answer 1

Answer: (C)

$5R\ln 2$

Solution:- (C) $5R\ln 2$

$\Delta S=n{C}_P\ln \frac{T_2}{T_1}+nR\ln P_1{P}_2$

$\because$ Pressure is constant,

$\Delta S=n{C}_P\ln \frac{T_2}{T_1}$

Given:-

$n=2\text{moles}$

$C_P=\frac{5}{2}R$

$T_2=600K$

$T_1=300K$

$\therefore \Delta S=2\times \frac{5}{2}R\times \ln \frac{600}{300}$

$\Rightarrow \Delta S=5R\ln 2$

Hence the change in entropy of the gas is $5R\ln 2$.