A gas undergoes change from state $\mathrm{A}$ to state $\mathrm{B}$. In this process, the heat absorbed and…

A gas undergoes change from state $\mathrm{A}$ to state $\mathrm{B}$. In this process, the heat absorbed and work done by the gas is $5 \mathrm{~J}$ and $8 \mathrm{~J}$, respectively. Now gas is brought back to $\mathrm{A}$ by another process during which $3 \mathrm{~J}$ of heat is evolved. In this reverse process of $\mathrm{B}$ to $\mathrm{A}$ :
  1. $10 \mathrm{~J}$ of the work will be done by the gas.
  2. $6 \mathrm{~J}$ of the work will be done by the gas.
  3. $10 \mathrm{~J}$ of the work will be done by the surrounding on gas.
  4. $6 \mathrm{~J}$ of the work will be done by the surrounding on gas.

Solution

$A \stackrel{q=+5, w=-8 J}{\uparrow} B$
$\Delta \mathrm{U}_{\mathrm{AB}}=\mathrm{q}+\mathrm{w}=+5+(-8)=-3$
$q=-3, \Delta \mathrm{U}_{\mathrm{BA}}=+3$
$\Delta \mathrm{U}_{\mathrm{BA}}=\mathrm{q}+\mathrm{w}$
$\Rightarrow 3=-3+\mathrm{w} \Rightarrow \mathrm{w}=+6 \mathrm{~J}$ (work done on
the system).

Asked in: JEE-TOPICTESTS-CHEMISTRY

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