Question

Find out the change in entropy $\Delta SinJ{K}^{-1}$ when 4 mol of Helium gas is isobarically heated from $250Kto350K$.
Assume $C_{p,m}=5.19kJmo{l}^{-1}{K}^{-1}$

• A. 4812
• B. 3950
• C. 6989
• D. 7050

Answer 1

The correct option is C 6989

Given here,
$T_1=250K{T}_2=350Kn=4mol{C}_{p,m}=5.19kJmo{l}^{-1}{K}^{-1}$

Change in entropy for an isobaric process is given by:
$\Delta S=n\times C_{p,m}\times ln\frac{T_2}{T_1}$

putting the values,
$\Delta S=[4\times 5.19\times ln\frac{350}{250}]kJ{K}^{-1}\Delta S=6.9886kJ{K}^{-1}\Delta S=6988.6J{K}^{-1}$


Theory:

Entropy Change in an Isobaric Process :
we know,
$\Delta S_{sys}=n{C}_{v,m}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
When pressure is constant;
$\frac{V_2}{V_1}=\frac{T_2}{T_1}$

Taking common, $nln\frac{T_2}{T_1}$
$\Delta S_{sys}=nln\frac{T_2}{T_1}(C_{v,m}+R)$
We know,
$C_{p,m}-C_{v,m}=R$
$C_{v,m}+R=C_{p,m}$

$\Delta S_{sys}=n{C}_{p,m}ln\frac{T_2}{T_1}$

Also from , $\Delta S_{sys}=n{C}_{p,m}ln\frac{T_2}{T_1}+nRln\frac{P_1}{P_2}$
at constant P
$\Delta S_{sys}=n{C}_{p,m}ln\frac{T_2}{T_1}$