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1st Law of Thermodynamics

The figure sh...

Question

The figure shows two processes, $1$ and $2$ for a given sample of a gas. If $Δ Q_{1}, Δ Q_{2}$ are the amounts of heat absorbed by the system in the two cases and $Δ U_{1}, Δ U_{2}$ are changes in internal energies respectively, then

Δ Q_{1} = Δ Q_{2}; Δ U_{1} = Δ U_{2}

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Δ Q_{1} > Δ Q_{2}; Δ U_{1} > Δ U_{2}

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Δ Q_{1} < Δ Q_{2}; Δ U_{1} < Δ U_{2}

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Δ Q_{1} > Δ Q_{2}; Δ U_{1} = Δ U_{2}

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Solution

The correct option is D $Δ Q_{1} > Δ Q_{2}; Δ U_{1} = Δ U_{2}$ .
From indicator diagram

Initial and final states of the two processes, $1$ and $2$ are the same.
So, $Δ U_{1} = Δ U_{2}$
From $1$ st law of thermodynamics,
For process $1$ : $Δ Q_{1} = Δ U_{1} + Δ W_{1}$
For process $2$ : $Δ Q_{2} = Δ U_{2} + Δ W_{2}$
Since, area under curve $1$ $>$ Area under curve $2$
We get $Δ W_{1} > Δ W_{2}$
$\Rightarrow Δ Q_{1} > Δ Q_{2}$
Therefore, we can conclude that $(Δ U_{1}) = (Δ U_{2})$ and $(Δ Q_{1}) > (Δ Q_{2})$

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