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Chapter 6: Problem 100

A quantity of 0.020 mole of a gas initially at \(0.050 \mathrm{~L}\) and \(20^{\circ} \mathrm{C}\) undergoes a constant-temperature expansion until its volume is \(0.50 \mathrm{~L}\). Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (c) If the gas in (b) is allowed to expand unchecked until its pressure is equal to the external pressure, what would its final volume be before it stopped expanding, and what would be the work done?

Short Answer

Expert verified
The work done by the gas when it expands against a vacuum is \(0\) Joules. When it expands against a constant pressure of \(0.20\) atm, the work done is approximately \(-97\) Joules. The final volume of the gas before it stops expanding when allowed to expand unchecked until its pressure is equal to the external pressure would be \(0.050\) L and the work done would be \(0\) Joules.

Step by step solution

01

Expansion Against a Vacuum

When a gas expands against a vacuum, there is no external pressure for the gas to do work against. Therefore, the work done by the gas is zero.
02

Expansion Against a Constant External Pressure

To calculate the work, use the formula for work: \(W = -P \Delta V\), where P is the constant external pressure and \( \Delta V\) is the change in volume. Remember to convert pressure from atm to Pa: \(0.20\) atm = \(20265\) Pa. After conversion, plug the values in: \(W = -20265*(0.50 - 0.050)\). The negative sign indicates that the gas is doing work on the surroundings.
03

Unchecked Expansion Until Equal External Pressure

To find the final volume before the gas stops expanding, use the Ideal Gas Law: \(P1V1=P2V2\). Here, \(P1=0.20\) atm, \(V1=0.050\) L, and \(P2=0.20\) atm. Solve for \(V2\), which will be equal to \(V1\). For the work done, use the same formula as in step 2.

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Most popular questions from this chapter

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