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

One mole of an ideal monoatomic gas is kept in a rigid vessel. The vessel is kept inside a chamber whose temperature is 87C. Initially, the temperature of the gas is 4C. The walls of the vessel have an inner surface of area 500 cm2 and thickness of 2 cm. If the temperature of the gas increases to 8C in 5.0 seconds, find the thermal conductivity of the material of the walls.

A
5 W/mK
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.05 W/mK
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.5 W/mK
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
50 W/mK
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.05 W/mK
Initial temperature difference (ΔT1)=874=83C
Final temperature difference (ΔT2)=878=79C
As the change in temperature difference is small, we find
Average temperature difference (ΔT)avg=ΔT1+ΔT22=83+792=81Cor (ΔT)avg=81 K
Rise in temperature of gas is =84=4C or 4 K
For monoatomic gas,
CV=32R=32×8.31=12.465 J/mol K
Since the walls are rigid, From first law of thermodynamics we can say that,
ΔQ=ΔU
So, heat supplied to the vessel containing gas in 5 seconds
ΔQ=nCVΔT
=1×12.465×4
=49.8650 J
Let K be the thermal conductivity of the wall.
Then, ΔQΔt=KA(T1T2)L
505=K×500×104×812×102
[T1T2=(ΔT)avg]
K=10×2×102500×104×81=0.049380.05 J s1m1K1
i.e K=0.05 W/mK

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon