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# Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If C_{A} and C_{B} be the molar heat capacities for the two processes.

A. C_{A} = C_{B}

B. C_{A} < C_{B}

C. C_{A} > C_{B}

D. C_{A} and C_{B} cannot be defined

Answer :

Q = nCdT … (i),

where Q = work done by an ideal gas

n = number of moles

C = molar heat capacity

dT = rise in temperature.

For process A, let the value of C be C_{A} and for B, let it be C_{B}.

Since the work done by the gas in process A is twice that in B, and the rise in temperature is the same in both the cases, we get two equations:

2Q = nC_{A}dT … (ii) and

Q = nC_{B}dT … (iii),

Where

Q = work done in process B

n = number of moles of gas

dT = rise in temperature

Dividing (ii) by (iii), we get

=> C_{A} = 2C_{B}

This proves that C_{A}>C_{B}.

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