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Question

In an isobaric reversible process, the ratio of heat supplied to the system (dq) and work done by the system (dW) for a ideal monoatomic gas is:

A
52
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B
72
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C
32
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D
12
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Solution

The correct option is A 52
At constant pressure, (isobaric process) heat supplied is given as:
qp=nCpdT
Cp for a monoatomic gas = 52R
So, qp=52nR dT
Work done by system is given as:
dw=nR dT
Hence, the ratio is: dqdw=(5/2)nR dTnR dT
=52

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