Question

Minimum amount of work done required to compress 5.00 moles of an ideal gas isothermally from 200 L to 40 L is

(a) + 20.1 kJ

(b) - 20.1 kJ

(c) - 20.1 J

(d) + 20.1 J

Answer 1

Answer is : (a) + 20.1 kJ

Work done in isothermal compression can be calculated as follows:

\(W=-nRT\,In\frac{V_2}{V_1}\)

= - 5 x 8.314 x 300 x 2.303 \(log\frac{40}{200}\)

= - 1500 x 8.314 x 2.303 x (- log 5)

= + 20.1 kJ