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(B) In the given graph, $V \propto T$, which implies $\frac{V}{T} = 	ext{constant}$.This indicates that the process is an isobaric process (constant

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$8.4$

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(B) In the given graph, $V \propto T$, which implies $\frac{V}{T} = 	ext{constant}$.This indicates that the process is an isobaric process (constant pressure process).For an isobaric process, the heat supplied is given by $\Delta Q = n C_p \Delta T$.The work done is given by $\Delta W = p \Delta V = n R \Delta T$.Since $C_p = \frac{5}{2} R$ for a monoatomic gas like helium, we have:$\Delta W = n R \Delta T = n R \left( \frac{\Delta Q}{n C_p} ight) = \frac{\Delta Q R}{C_p} = \frac{\Delta Q R}{\frac{5}{2} R} = \frac{2}{5} \Delta Q$.Given $\Delta Q = 5 	ext{ cal}$, we get:$\Delta W = \frac{2}{5} 	imes 5 	ext{ cal} = 2 	ext{ cal}$.Converting to Joules, $\Delta W = 2 	imes 4.2 	ext{ J} = 8.4 	ext{ J}$.

$A$ graph is drawn between absolute temperature and volume of $3$ moles of helium gas as shown in the figure. If $5 \text{ cal}$ of heat is used in the process, then the work done is (in $\text{ J}$)

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