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.

Question

Philoid · Accepted 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_AdT … (ii) and

Q = nC_BdT … (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.

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 CA and CB be the molar heat capacities for the two processes.

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.