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A graph drawn between absolute temperature and volume of 3 moles of helium gas as shown in the figure. If $5 \mathrm{cal}$ of heat is used in the process, then the work done is
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Verified Answer
The correct answer is:
8.4 J
In given graph,
$$
V \propto T \Rightarrow \frac{V}{T}=\text { constant }
$$
Hence, process is a constant pressure process.
$$
\text { So, } \begin{aligned}
\Delta Q= & n C_p \Delta T \text { and } \Delta W=p \Delta V=n R \Delta T=n R \cdot \frac{\Delta Q}{n C_p} \\
= & \frac{\Delta Q R}{\frac{5}{2} R}=\frac{2}{5} \times \Delta Q=2 \mathrm{cal}=2 \times 4.2 \mathrm{~J}=8.4 \mathrm{~J}
\end{aligned}
$$
$$
V \propto T \Rightarrow \frac{V}{T}=\text { constant }
$$
Hence, process is a constant pressure process.
$$
\text { So, } \begin{aligned}
\Delta Q= & n C_p \Delta T \text { and } \Delta W=p \Delta V=n R \Delta T=n R \cdot \frac{\Delta Q}{n C_p} \\
= & \frac{\Delta Q R}{\frac{5}{2} R}=\frac{2}{5} \times \Delta Q=2 \mathrm{cal}=2 \times 4.2 \mathrm{~J}=8.4 \mathrm{~J}
\end{aligned}
$$
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