wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

1 mole of an ideal gas at 25C is subjected to expand reversibly 10 times of its initial volume. Calculate the change in entropy of expansions.

A

25.15 J K-1mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

30.15 J K-1mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

19.15 J K-1mol-1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

29.15 J K-1mol-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

19.15 J K-1mol-1


Here, we know that ΔT = 0
Therefore, it is an isothermal process.
We know for an isothermal process
ΔU = 0
Now,
Entropy change for an isothermal process
ΔU=q+w
ΔU=0 (Isothermal process)
q=w
But w=2.30 nRT log V2V1 or nRT lnV2V1 [From first law of thermodynamics]
q =2.303 nRT logV2V1=nRT lnV2V1
or for a reversible process
qrev =2.303 nRT logV2V1 = nRT lnV2V1
ΔsysS=qrevT=nRT lnV2V1T=2.303nRT logV2V1T
Also V1P (Boyle's law)
Substituting this relation in above equation, we get
ΔS=2.303 nR logP1P2=nR lnP1P2
We have, ΔS =
Given, n = 1, R = 8.314 J, T = 298 K, V1 = V, V2 = 10 V
ΔS=2.303×1×8.314 log1010VV
= 19.15 J K1 mol1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Second Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon