Question

5 mole of an ideal gas change it's volume from 10 L to 20 Lit. reversible isothermally at 300 K then work done by gas is.......KCal. [Given R = 2 Cal]

Answer 1

Correct answer is (2)

For isothermal reversible process

W = -2.303 nRT log (v₂/v₁)

= -2.303 x 5 x 2 x 300 log (20/10)

= -2072.7 cal = -20727 x 10³ cal 

= -2 K Cal = |W| = 2 K cal