The correct option is B 5
Given,
PT=K....(1)
where, K is constant.
For one mole of an ideal gas, we can write,
PV=RT....(2)
From equation (1) and (2), we have
V=RTP
⇒V=RT2K
So,
dV=2RTKdT
∴ Work done, when temperature changes from T to (T+dT),
dW=PdV
Substituting values,
dW=KT(2RTK)dT
⇒dW=2RdT
Now from first law of thermodynamics,
dQ=dU+dW
Substituting the values for ideal equation,
CdT=CvdT+2RdT
⇒Cv=C−2R...(3)
For an idea gas,
Cv=n2R
where, n is degree of freedom of the molecules of the gas.
Substituting it in the above equation,
⇒n2R=C−2R
⇒n=2(C−2R)R
Given, C=37.35 J mol−1K−1 and we know R=8.3.
Substituting the values in the above equation, we get
n=2(37.35−2×8.3)8.3
∴n=5
Hence, degree of freedom is 5 .