Question

72. When 6 moles of He gas at 27°C undergoes
reversible isothermal expansion from 30 L to
60 L, then work done will be
(1) 4.5 kJ
(2) 4.5 kJ
(3) -10.4 kJ . (4) 10.3 kJ​

Answer 1

• Work done during reversible isothermal expansion will be - 10.4 kJ

Given -

Moles of Helium gas (n) = 6 moles

Temperature (t) = 27⁰ C = 300 K

Initial volume (Vi) = 30 L

Final volume (Vf) = 60 L

We know that work done by isothermal reversible process is given by

W = - 2.303 nRT log (Vf/Vi) where n is number of moles, R is universal gas constant and T is temperature.

By putting the values we get -

W = - 2.303 × 6 × 8.314 × 300 log (60/30)

W = -10,374.95 J ≈ - 10.375 KJ

Regards