At constant pressure :
ΔQ=nCPΔT ⇒160=nCP.50 .....(1)
At constant volume
ΔQ=nCvΔT ⇒240=nCv.100......(2)
Equation (1) divided by (2)
240160=CVCP⋅10050
CVCP=34=1+f2
f=6
JEE Main 2020 — Physics Thermodynamics
To raise the temperature of a certain mass of gas by 50∘C at a constant pressure,160 calories of heat is required. When the same mass of gas is cooled by 100∘C at constant volume,240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?
Held on 3 Sept 2020 · Verified 6 Jul 2026.
5
6
3
7
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