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Question

A gas has molar heat capacity C=37.35 J mol−1K−1 in the process PT=constant. Find the number of degrees of freedom of molecules in the gas.

A
4
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B
5
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C
2
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D
3
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Solution

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=C2R...(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=C2R

n=2(C2R)R

Given, C=37.35 J mol1K1 and we know R=8.3.
Substituting the values in the above equation, we get

n=2(37.352×8.3)8.3

n=5
Hence, degree of freedom is 5 .

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