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

Two moles of an ideal gas expended isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is:

A
11.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
-11.4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C -11.4
Work done in a reversible isothermal process is:

W=2.303nRTlogVfVi.....(1)

Given:-
n=2 moles
T=300K
Vf=10L
Vi=1L
R= Gas constant =8.314J/Kmol

Substituting these values in eqn(1), we have

W=2.303×2×8.314×300×log101

W=11488.285J=11.4kJ

Now as we know that,

ΔH=ΔU+W

For an isothermal process,

ΔU=0

ΔH=W=11.4kJ

Hence the enthalpy change for the given process is 11.4kJ.

Hence, the correct option is B

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