Consider 5.5 \(\mathrm{L}\) of a gas at a pressure of 3.0 \(\mathrm{atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 \(\mathrm{L}\) . a. Calculate the work done, and indicate the correct sign. b. 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 \(\mathrm{L}\) . The second step results in a final volume of 10.5 \(\mathrm{L}\) . Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is 8.0 \(\mathrm{L}\) and the second step leads to a volume of 10.5 \(\mathrm{L}\) . Does the work differ from that in part b? Explain.

Step by step solution

01

Part a: Work done for a direct change in volume from 5.5L to 10.5L

Given the external pressure, P = 3.0 atm, initial volume, V1 = 5.5 L, and final volume, V2 = 10.5 L. Use the work formula: Work = - P * (Final Volume - Initial Volume) Plug in the given values: Work = -3.0 * (10.5 - 5.5) Work = -3.0 * 5 Work = -15 atm L (The negative sign indicates that work is done by the system, meaning the gas expanded)

02

Part b: Work done for two changes in volume (5.5L to 7.0L to 10.5L)

We need to calculate work done in two steps, V1 = 5.5 L to V2 = 7.0 L and then V2 = 7.0 L to V3 = 10.5 L: 1. Work from 5.5 L to 7.0 L: Work1 = -3.0 * (7.0 - 5.5) = -3.0 * 1.5 = -4.5 atm L (Expansion) 2. Work from 7.0 L to 10.5 L: Work2 = -3.0 * (10.5 - 7.0) = -3.0 * 3.5 = -10.5 atm L (Expansion) Total Work = Work1 + Work2 = -4.5 - 10.5 = -15 atm L

03

Part c: Work done for two changes in volume (5.5L to 8.0L to 10.5L)

Calculate work done in two steps, V1 = 5.5 L to V2 = 8.0 L and then V2 = 8.0 L to V3 = 10.5 L: 1. Work from 5.5 L to 8.0 L: Work1 = -3.0 * (8.0 - 5.5) = -3.0 * 2.5 = -7.5 atm L (Expansion) 2. Work from 8.0 L to 10.5 L: Work2 = -3.0 * (10.5 - 8.0) = -3.0 * 2.5 = -7.5 atm L (Expansion) Total Work = Work1 + Work2 = -7.5 - 7.5 = -15 atm L The work done in part c is the same as in part b. This occurs because the overall change in volume from 5.5L to 10.5L remains the same, and the external pressure remains constant at 3.0 atm throughout the process.

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