. KTG_01_09_21_JEE (MAIN)-2021_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2021

01-09-21 S2

Question

The temperature of 3.00 mol of an ideal diatomic gas is increased by $40.0^{\circ} \mathrm{C}$ without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $\frac{x}{10}$. Then the value of $x$ (round off to the nearest integer) is $\_\_\_\_$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Pressure is not changing ⇒ isobaric process $$ \Rightarrow \quad \Delta U=n C_{\vee} \Delta T=\frac{5 n R \Delta T}{2} $$ and $W=n R \Delta T$ $$ \frac{\Delta U}{W}=\frac{5}{2}=\frac{x}{10} \Rightarrow x=25.00 $$

Question Tags

JEE Main

Physics

Easy

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