Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A gas of amount 0.2 mole undergoes a reversible isothermal compression at a constant temperature of <img alt="300 \mathrm{~K}" src="https://entrancecorner.oncodecogs.com/gif.latex?300%20%5Cmathrm%7

A gas of amount 0.2 mole undergoes a reversible isothermal compression at a constant temperature of 300 \mathrm{~K}. The gas is compressed from an initial volume of 10 liters to a final volume of 5 liters. Calculate the work done on the gas during the process.

Option: 1

168 J


Option: 2

-200 J


Option: 3

345.8 J


Option: 4

200 J


Answers (1)

best_answer

Given data: Initial volume \mathrm{\left(V_i\right)=10} liters, Final volume \mathrm{\left(V_f\right)=5} liters, Temperature \mathrm{ (T)=300 \mathrm{~K}}

The work done (W) on the gas during isothermal compression is given by:

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

where n is the number of moles of the gas.

                            \mathrm{ \begin{gathered} W=0.2(8 . .314)(300) \ln \left(\frac{10}{5}\right) \\ W=345.8 \mathrm{~J} \end{gathered} }

Therefore, the work done on the gas during isothermal compression is 345.8

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 2025: Hardwired to excel? Know BTech electrical engineering cut-offs at NITs
anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22

Latest Question