The temperature of 1 mole of an ideal monoatomic gas is increased by $50^{\circ} \mathrm{C}$ at constant pressure. The total heat added and change in internal energy are $E_1$ and $E_2$, respectively. If $\frac{E_1}{E_2}=\frac{x}{9}$ then the value of $x$ is _____

Question

Aicharya · Accepted Answer

Given that process is isobaric $\Delta \mathrm{T}=50^{\circ} \mathrm{C}$
Q in isobaric process $=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{E}_1$
$\Delta \mathrm{U}$ in isobaric process $=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{E}_2$
$\therefore \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$
Given, gas is monoatomic
$\begin{aligned}
\therefore \gamma & =1+\frac{2}{\mathrm{f}} \\ & =1+\frac{2}{3} \\ & =\frac{5}{3}
\end{aligned}$
Now, as per question.
$\begin{aligned}
& \frac{5}{3}=\frac{x}{9} \\ & x=15
\end{aligned}$

The temperature of 1 mole of an ideal monoatomic gas is increased by $50^{\circ} \mathrm{C}$ at constant…

Solution