Question

The entropy change when two moles of ideal monoatomic gas is heated from 200${}^{o}C$ to 300${}^{o}C$ reversibly and isochorically is:

• A. $\frac{3}{2}R\ln \left(\frac{300}{200}\right)$
• B. $\frac{5}{2}R\ln \left(\frac{573}{273}\right)$
• C. $3R\ln \left(\frac{573}{473}\right)$
• D. $\frac{3}{2}R\ln \left(\frac{573}{473}\right)$

Answer 1

Answer: (C)

$3R\ln \left(\frac{573}{473}\right)$

Solution:- (C) $3R\ln \left(\frac{573}{473}\right)$

As we know that,

$\Delta S=n{C}_V\ln \frac{T_2}{T_1}+nR\ln V_2{V}_1$

$\because$ The process is isochoric, i.e., volume is constant.

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

Given:-

$n=2\text{moles}$

$C_P=\frac{3}{2}R[\because \text{the gas is monoatomic}]$

$T_2=200℃=(200+273)K=473K$

$T_1=300℃=(300+273)K=573K$

$\therefore \Delta S=2\times \frac{3}{2}R\times \ln \left(\frac{573}{473}\right)$

$\Rightarrow \Delta S=3R\ln \left(\frac{573}{473}\right)$

Hence the change in entropy of the gas is $3R\ln \left(\frac{573}{473}\right)$.