Three moles of an ideal gas kept at a constant temperature at 300 k are compressed from a volume of 4L to 1L. The work done in the process is (A) - 10368J (B) -110368J (C) 12000J (D) 120368J


n = 3

T = 300K

V1 = 4 litre

V2 = 1 litre

W = ?

In an isothermal process, work done is given by

\(W = 2.3026nRTlog_{10}\frac{V_{2}}{V_{1}}\\=2.3026\times 3\times 8.31\times300\times log_{10}\frac{1}{4}\)

We get,

= -10368J

Therefore, the correct option is (A)

Was this answer helpful?


0 (0)


Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




Free Class