1
Question
At constant volume, the specific heat of a gas is 0.025 cal K−1 g−1 and its molecular weight is 120 g mol−1. The atomicity of the gas is:
Given: R = 2 cal K−1mol−1
At constant volume, the specific heat of a gas is 0.025 cal K−1 g−1 and its molecular weight is 120 g mol−1. The atomicity of the gas is:
Given: R = 2 cal K−1mol−1
Open in App
Solution
The correct option is C Monoatomic
We know that molar heat capacity at constant volume,
(Cv,m) = Specific heat at constant volume × Molecular weight.
= 0.025×120=3 cal K−1 mol−1....(i)
Also, Cp–Cv=R
⇒Cp=R+Cv
= Cp=2+3=5 cal K−1 mol−1.....(ii)
Now, CpCv=γ
Using equation (i) and (ii)
γ=53=1.67
This value shows that the gas is monoatomic.
We know that molar heat capacity at constant volume,
(Cv,m) = Specific heat at constant volume × Molecular weight.
= 0.025×120=3 cal K−1 mol−1....(i)
Also, Cp–Cv=R
⇒Cp=R+Cv
= Cp=2+3=5 cal K−1 mol−1.....(ii)
Now, CpCv=γ
Using equation (i) and (ii)
γ=53=1.67
This value shows that the gas is monoatomic.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
