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=20 JK1mol1) initially at STP is heated at constant volume to twice the initial temperature. For the process, work done (w) and q will be :

A
w=0; q=5.46 kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
w=0; q=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
w=5.46 kJ; q=5.46 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
w=5.46 kJ; q=5.46 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A w=0; q=5.46 kJ
As volume is constant so work done is zero,
w=PV=P×0=0
heat absorbed at constant volume is given as
qv=nCv(T)
n=1
Cv=20 JK1mol1
now T=T2T1=546273=273 K
(as at STP, T=273 K)
so,
qv=20×273=5460 J=5.46 kJ

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon