The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :

 

  • Option 1)

    \frac{3}{5}

  • Option 2)

    \frac{2}{3}

  • Option 3)

    \frac{3}{2}

  • Option 4)

    \frac{2}{5}

 

Answers (1)

As we learnt in

Specific heat capacity at constant pressure -

C_{p}= C_{v}+R

= \left ( \frac{f}{2}+1 \right )R

- wherein

f = degree of freedom

R= Universal gas constant

 

 Work done in isobaric process =nR\Delta T

Heat supplied =nC_{p} \Delta T

    =nR\left(\frac{f}{2}+1 \right )\Delta T=\frac{5}{2}(nR\Delta T)

Ratio =\frac{nR\Delta T}{\frac{5}{2}(nR\Delta T)}=\frac{2}{5}

Correct option is 4.


Option 1)

\frac{3}{5}

This is an incorrect option.

Option 2)

\frac{2}{3}

This is an incorrect option.

Option 3)

\frac{3}{2}

This is an incorrect option.

Option 4)

\frac{2}{5}

This is the correct option.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 6999/- ₹ 5/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions