Question

When two mole of an ideal gas $(C_{p.m}=\frac{5}{2}R)$ heated from $300K$ to $600K$ at constant pressure. The change in entropy of gas ($\Delta S$) 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

The correct option is C $5R\ln 2$

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

$\Delta S=n{C}_P\ln \frac{T_2}{T_1}+nR\ln \frac{P_1}{P_2}$

$\because$ 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 $5R\ln 2$.