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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2025
  • Hard
  • Numerical

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 _____

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Answer & Solution

Correct Answer

(A) 15

Step-by-step Solution

Detailed explanation

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.
\(\begin{aligned}
& \frac{5}{3}=\frac{x}{9} \\
& x=15
\end{aligned}\)
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