Question

Consider 5.5 L of a gas at a pressure of 3.0 atm in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 L. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 L. The second step results in a final volume of 10.5 L. Calculate the work done, and indicate the correct sign.

Answer 1

The total work done by the gas on the surroundings is -1273437.5 Joules as the negative value confirms that the gas did work against the external pressure during the expansion.

W = -PΔV

3.0 atm * 101325 Pa/atm = 303975 Pa

Initial volume (V1) = 5.5 L = 0.0055 m³

Pressure (P1) = 303975 Pa

Final volume after step 1 (V2) = 7.0 L = 0.0070 m³

Change in volume (ΔV1) = V2 - V1 = 0.0070 m³ - 0.0055 m³

= 0.0015 m³

W1 = -P1 * ΔV1

= -303975 Pa * 0.0015 m³

= -455962.5

Initial volume after step 1 (V2) = 0.0070 m³

Pressure (P2) can be calculated using the Boyle's Law principle:

P2 * V2 = P1 * V1

P2 = (P1 * V1) / V2 =

(303975 Pa * 0.0055 m³) / 0.0070 m³

= 233550 Pa

Final volume (V3) = 10.5 L = 0.0105 m³

Change in volume (ΔV2) = V3 - V2

= 0.0105 m³ - 0.0070 m³

= 0.0035 m³

W2 = -P2 * ΔV2 = -233550 Pa * 0.0035 m³

= -817475 J

The Total Work Done: W1 + W2 = -455962.5 J - 817475 J

= -1273437.5 J