Question

5 moles of helium expand isothermally and reversibly from a pressure 40 × 10^-5 Nm^-2 to 4 × 10^-5 Nm^-2 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion.(R = 8.314 JK^-1 mol^-1)

Answer 1

As the expansion takes place isothermally and reversibly, the work done is given by

W_max = -2.203 nRT log \(\frac{P_1}{P_2}\)

W_max = Maximum work done

n = Number of moles of helium = 5 moles

R = Gas constant = 8.314 JK^-1 mol^-1

T = 300 K

P₁ = Initial pressure = 40 × 10^-5 Nm^-2

P₂ = Final pressure = 4 × 10^-5 Nm^-2

∴ W_max = -2.303 × 5 × 8.314 × 300 × log\(\frac{40}{4}\)

= – 2.303 × 5 × 8.314 × 300 × log 10 

= -2.303 × 5 × 8.314 × 300 × 1 

= – 28720J

As the expansion takes place isothermally, i.e., at the same temperature, there is no change in the internal energy of the system.

Q = ΔU + W 

∴ Q = – W = + 28720 J as ΔU = 0

∴ Work done = -28720 J, 

Heat absorbed = 28720 J, 

Change in internal energy = 0