CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

One mole of an ideal gas Cv,m=52R at 300 K and 5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature of the gas is:

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

The correct option is C 248.5 K
Solution:- (C) 248.5K
Given that:-
Pext.=2atm
P1=5atm
P2=2atm
T1=300K
T2=T(say)=?
n=1
CV=52R
For an adiabatic process,
W=nCVdT=PΔV
nCV(T2T1)=Pext(V2V1)
1×52R(T300)=2(nRT2P2nRT1P1)
52R(T300)=2R(T23005)
5T1500=2T+240
7T=1740T=248.57K
T2=T=248.57K

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