When 2moles of an ideal monoatomic gas are heated from 300K to 600K at constant pressure. The change in entropy of gas (△S) is:
A
3Rln2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
32Rln2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5Rln2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
52Rln2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C5Rln2 The entropy change for an isobaric process is given as: △S=nCp,mlnT2T1 where T2=600K T1=300K n=2 for ideal monoatomic gas we know Cp,m=52R
Putting the values we get: △S=2×52R×ln600300 = 5Rln2