Question

# Find out the change in entropy ΔS in J K−1 when 4 mol of Helium gas is isobarically heated from 250 K to 350 K. Assume Cp,m=5.19 kJ mol−1 K−1

A
4812
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B
3950
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C
6989
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D
7050
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Solution

## The correct option is C 6989Given here, T1=250 KT2=350 Kn=4 molCp,m=5.19 kJ mol−1 K−1 Change in entropy for an isobaric process is given by: ΔS= n×Cp,m×lnT2T1 putting the values, ΔS=[4×5.19×ln350250] kJ K−1ΔS=6.9886 kJ K−1ΔS=6988.6 J K−1 Theory: Entropy Change in an Isobaric Process : we know, ΔSsys=nCv,mlnT2T1+nRlnV2V1 When pressure is constant; V2V1=T2T1 Taking common, n lnT2T1 ΔSsys=n lnT2T1(Cv,m+R) We know, Cp,m−Cv,m=R Cv,m+R=Cp,m ΔSsys=nCp,m lnT2T1 Also from , ΔSsys=nCp,mlnT2T1+nRlnP1P2 at constant P ΔSsys=nCp,m lnT2T1

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