Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Stuck here, help me understand: The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is :

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)

best_answer

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.

Posted by

perimeter

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question