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Chapter 6: Problem 16

A gas expands in volume from \(26.7 \mathrm{~mL}\) to \(89.3 \mathrm{~mL}\) at constant temperature. Calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of \(1.5 \mathrm{~atm},\) and (c) against a constant pressure of 2.8 atm.

Short Answer

Expert verified
The work done by the gas is (a) 0 J against a vacuum, (b) -9.51 J against a constant pressure of 1.5 atm, and (c) -17.77 J against a constant pressure of 2.8 atm.

Step by step solution

01

Calculate the Change in Volume

Firstly, the change in volume needs to be calculated as it's required for the calculation of work. The formula for change in volume \( \Delta V \) is \( V_{final} - V_{initial} \), where \( V_{final} \) is the final volume and \( V_{initial} \) is the initial volume. Plugging in the given values, \( \Delta V = 89.3 m L - 26.7 m L = 62.6 mL \). Convert this volume to liters, since standard units are required for the work calculation. So, \( \Delta V = 62.6 mL * (1 L/1000 mL) = 0.0626 L \).
02

Calculate the Work against Vacuum

When a gas expands against a vacuum, it doesn't actually perform work because there's no 'opposing' pressure. Hence, the work done in this case is zero, \( W_{vacuum} = 0 J \).
03

Calculate the Work against a Constant Pressure of 1.5 atm

Now we will calculate the work done when the gas expands against a constant pressure of 1.5 atm. Use the work formula \( W = -P \Delta V \), but remember to convert the pressure from atm to Pa by multiplying by \(1.013 * 10^5 \). That gives \( P = 1.5 atm * (1.013 * 10^5 Pa/atm) = 1.5195 * 10^5 Pa \). Plugging in the values, \( W = -P \Delta V = -(1.5195 * 10^5 Pa)(0.0626 L) = -9.51 J \).
04

Calculate the Work against a Constant Pressure of 2.8 atm

Now we calculate the work done when the gas expands against a constant pressure of 2.8 atm, in a similar manner to Step 3. Convert the pressure: \( P = 2.8 atm * (1.013 * 10^5 Pa/atm) = 2.8364 * 10^5 Pa \). Compute the work: \( W = -P \Delta V = -(2.8364 * 10^5 Pa)(0.0626 L) = -17.77 J \).

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