Question

Find work done in the isobaric process.

• A. $W=1245J$
• B. $W=1255J$
• C. $W=1254J$
• D. $W=1240J$

Answer 1

Answer: (A)

$W=1245J$

Isochoric process:
$\Delta W=0$ as $\Delta V=0$
$\Delta U=C_v\Delta T=\frac{R}{\gamma -1}\times \Delta T$
$\Rightarrow \Delta U=\frac{\Delta \left(RT\right)}{\gamma -1}$
$\Rightarrow \frac{V}{\gamma -1}\Delta P=-\frac{1}{2}\frac{pV}{\gamma -1}$ since $\Delta P=\frac{p}{2}-p=-\frac{p}{2}$
$\Delta Q=\Delta U+\Delta W$
$\therefore \Delta Q=-\frac{1}{2}\frac{pV}{\gamma -1}+0=-\frac{1}{2}\frac{pV}{\gamma -1}=-\frac{1}{2}\frac{R{T}_0}{\gamma -1}$
In the isochoric process, temp $T_0$ is reduced to half i.e to $\frac{T_0}{2}$ since $P\alpha T$
Isobaric process:
$\Delta W={\int }_{V_i}^{V_f}PdV=\frac{P}{2}(V_f-V_i)$
Since the original temp is restored to $T_0$, $\Delta T=\frac{T_0}{2}$
$\therefore \Delta W=\frac{1}{2}R{T}_0=\frac{1}{2}\times 8.3\times 300=1245J$