Two moles of monoatomic gas is expanded from $(P_{0}, V_{0})$ to $(P_{0}, 2 V_{0})$ under isobaric condition. Let $Δ Q_{1}$ be the heat given to the gas $Δ W_{1}$ the work by the gas and $Δ U_{1}$ the change in internal energy Now the monoatomic gas is replaced by a diatomic gas. Other conditions remaining the same. The corresponding values in this case are $Δ Q_{2}, Δ W_{2}, Δ U_{2}$ respectively, then:

Δ Q_{1} - Δ Q_{2} = Δ U_{1} - Δ U_{2}

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Δ U_{2} + Δ W_{2} > Δ U_{1} + Δ W_{1}

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Δ U_{2} > Δ U_{1}

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All of these

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Solution

The correct option is D All of these
Since for both cases P-V curve is the same, $Δ W_{1} = Δ W_{2}$
$\Rightarrow Δ Q_{1} - Δ U_{1} = Δ Q_{2} - Δ U_{2}$
$\Rightarrow Δ Q_{1} - Δ Q_{2} = Δ U_{1} - Δ U_{2}$
So, option a is correct.

Process is isobaric
Hence $Δ Q_{1} = n C_{p 1} Δ T_{1} = \frac{5}{2} n R Δ T_{1} = \frac{5}{2} P_{0} V_{0}$
Also $Δ Q_{2} = n C_{p 2} Δ T_{2} = \frac{7}{2} n R Δ T_{2} = \frac{7}{2} P_{0} V_{0}$
Hence $Δ Q_{2} > Δ Q_{1}$
$\Rightarrow Δ U_{1} + Δ W_{1} > Δ U_{2} + Δ W_{2}$
So, option b is correct

$Δ U_{1} = n C_{v 1} Δ T_{1} = \frac{3}{2} n R Δ T_{1} = \frac{3}{2} P_{0} V_{0}$
$Δ U_{2} = n C_{v 2} Δ T_{2} = \frac{5}{2} n R Δ T_{2} = \frac{5}{2} P_{0} V_{0}$
Hence $Δ U_{2} > Δ U_{1}$
So, option c is also correct

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