Easy MCQ +4 / -1 PYQ · JEE Mains 2020

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of
$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$ are given, respectively by:

  1. A U = ${5 \over 2}RT$ and $\gamma = {7 \over 5}$ Correct answer
  2. B U = 5RT and $\gamma = {6 \over 5}$
  3. C U = 5RT and $\gamma = {7 \over 5}$
  4. D U = ${5 \over 2}RT$ and $\gamma = {6 \over 5}$

Solution

Total degree of freedom (f) = 3 + 2 = 5 <br><br>U = ${{nfRT} \over 2}$ = ${{5RT} \over 2}$ <br><br>$\gamma$ = ${{{C_p}} \over {{C_v}}}$ = $1 + {2 \over f}$ = $1 + {2 \over 5}$ = ${7 \over 5}$

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →