Question

0.100 moles of He are contained in a piston at equilibrium under an external pressure of 2.80 atm at 298 K. The external pressure if reduced to 1.20 atm and the piston is allowed to expand and return to equilibrium, at the same temperature. How much work is done by the system? Assume the gas is ideal.

Answer 1

Work done by the system = - external pressure x change in volume

external pressure = 1.2 atm

Initial volume has to be calculeted from ideal gas equation

pressure x volume1 = number of moles x gas constant x temperature

2.8 x volume1 = 1 x 0.082057 x 298

volume 1 = 8.733 L

Similarly the volume 2 at pressure 1.2 atm

pressure x volume2 = number of moles x gas constant x temperature

1.2 x volume2 = 1 x 0.082057 x 298

volume 2 = 20.337 L

change in volume = 11.644 L

work done = - 11.644 x 1.2 = -11.973 L-atm

1 L-atm = 101.325 J

Thus work done = -1213.16 J