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Q.

Ideal gas is taken through the process as shown in the figure:

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a

in process AB, heat is rejected

b

in process AB internal energy decreases and in process BC internal energy increases

c

in process AB internal energy decreases and in process BC internal energy remains constant

d

in process AB, work done by system is positive

answer is B.

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Detailed Solution

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From A to B,

T=constant $\Rightarrow dU = 0$

$W_{A \to B} = n R T \ln (\frac{V_{B}}{V_{A}}) = n R T \ln (\frac{P_{A}}{P_{B}})$

$∵ P \propto \frac{1}{V} \Rightarrow P ↑ V ↓ \Rightarrow P_{A} < P_{B}$

$\Rightarrow W_{A \to B} < 0$

$\Rightarrow dQ = ∆ U + W = 0 + W_{A \to B} = W_{A \to B} < 0$

Hence, in the process AB, heat is rejected.

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