Thermodynamics

JEE Main level, numerical answer, medium

27 of 190

Question 27 of 190

The temperature of 1 mole of an ideal monoatomic gas is increased by 50∘C50^{\circ} \mathrm{C} at constant pressure. The total heat added and change in internal energy are E1E_1 and E2E_2, respectively. If E1E2=x9\frac{E_1}{E_2}=\frac{x}{9} then the value of xx is _____

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Answer: 15

Given that process is isobaric ΔT=50∘C\Delta \mathrm{T}=50^{\circ} \mathrm{C}
Q in isobaric process =nCpΔT=E1=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T}=\mathrm{E}_1
ΔU\Delta \mathrm{U} in isobaric process =nCvΔT=E2=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{E}_2
∴E1E2=CPCv=γ\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.
$\begin{aligned}
& \frac{5}{3}=\frac{x}{9} \\
& x=15
\end{aligned}$

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