Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Question : Reversible Process and Work A gas expands reversibly and isothermally from an initial volume of 0.04 m3 to a final volume of 0.08 m3 at a temperature of 300 K.

Question : Reversible Process and Work

A gas expands reversibly and isothermally from an initial volume of 0.04 m3 to a final volume of 0.08 m3 at a temperature of 300 K. Calculate the work done by the gas.

 

Option: 1

1946J
 


Option: 2

1740 J

 


Option: 3

500 J
 


Option: 4

2540 J


Answers (1)

best_answer

For a reversible isothermal expansion, the work done by the gas is given by:

\mathrm{W=n R T \ln \left(\frac{V_f}{V_i}\right)}

Given

\mathrm{\begin{aligned} V_i & =0.04 \mathrm{~m}^3 \\ V_f & =0.08 \mathrm{~m}^3 \\ T & =300 \mathrm{~K} \end{aligned}}

Assuming the number of moles and the gas constant remain constant:

\mathrm{\begin{aligned} & W=n R T \ln \left(\frac{V_f}{V_i}\right) \\ & W=R T \ln \left(\frac{0.08 \mathrm{~m}^3}{0.04 \mathrm{~m}^3}\right) \\ & W=(8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K}) \cdot(300 \mathrm{~K}) \cdot \ln (2) \\ & W \approx 1946 \mathrm{~J} \end{aligned}}

Therefore, the correct option is A.

 

Posted by

Anam Khan

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 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question