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Chapter 10: Problem 25

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

Short Answer

Expert verified
In conclusion, pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

Step by step solution

01

Action (a): Doubling the Volume

For this action, we'll lift the piston to double the volume of the gas while keeping the temperature constant. Using the Ideal Gas Law, we can write the initial and final states as follows: Initial state: \(PV = nRT\) Final state: \(P_{final} (2V) = nR(298.15)\) We want to determine the relationship between the initial and final pressures: \(\frac{P_{final}}{P} = \frac{nRT}{2nR(298.15)} = \frac{1}{2}\) The pressure in this case is halved, not doubled. So, action (a) does not double the gas pressure.
02

Action (b): Heating the Gas

In this action, we'll heat the gas so that its temperature rises from \(25^{\circ} C\) to \(50^{\circ} C\) while keeping the volume constant. First, convert the final temperature to Kelvin: \(50^{\circ}C + 273.15 = 323.15 K\) Now, let's use the Ideal Gas Law for the initial and final states: Initial state: \(PV = nR(298.15)\) Final state: \(P_{final}V = nR(323.15)\) Using the relationship between the initial and final pressures, we get: \(\frac{P_{final}}{P} = \frac{nR(323.15)}{nR(298.15)} = \frac{323.15}{298.15}\) The pressure does not double in this case. Therefore, action (b) does not double the gas pressure.
03

Action (c): Halving the Volume

For this action, we'll push down on the piston to halve the volume of the gas while keeping the temperature constant. Using the Ideal Gas Law, we can write the initial and final states as follows: Initial state: \(PV = nRT\) Final state: \(P_{final} \frac{V}{2} = nRT\) To determine the relationship between the initial and final pressures, we get: \(\frac{P_{final}}{P} = \frac{2nRT}{nRT} = 2\) The pressure doubles in this case. Thus, action (c) doubles the gas pressure. In conclusion, pushing down on the piston to halve the volume while keeping the temperature constant (action c) would double the gas pressure.

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