Question

0.25mole of qn ideal monoactomic gas undergoes isothermal expansion from a volume of 2.0dm to 10dm qt 27c calculate the max work that can be obtained frm this process

Answer 1

Given:

n = 0.25

V1 = 2 dm³

V2 = 10 dm³

T = 27 C = 300 K

To Find:

Maximum work obtained from the process.

Calculation:

- We know that for an isothermal process, work done is given by:

W = - 2.303nRT log (V2/V1)

⇒ W = (-2.303 × 0.25 × 8.314 × 300) × log (10/2)

⇒ W = (-1436.03) × log 5

⇒ W = (-1436.03) × 0.6989

⇒ W = - 1003.64 J

- So the max work that can be obtained from this process is 1003.64 J (or 1.003 kJ approx.)