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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? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.

Short Answer

Expert verified
The correct action that doubles the gas pressure is: \(\boxed{\mathrm{(c)}}\) Pushing down on the piston to halve the volume while keeping the temperature constant.

Step by step solution

01

a) Lifting up the piston to double the volume while keeping the temperature constant

In this action, since temperature (T) remains constant and the number of moles (n) and the gas constant (R) are unchanged, the Ideal Gas Law becomes: \(PV = constant\) Now, let's double the volume (V): \(P(2V) = constant\) \(2PV = constant\) Comparing the initial and final states: \(2PV = PV\) This shows that the pressure (P) remains constant in this action. Hence, lifting up the piston to double the volume does not double the pressure.
02

b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant

In this action, the volume remains constant, so we rewrite the Ideal Gas Law as: \(P \propto T\) We need to convert the temperatures from Celsius to Kelvin: Initial temperature: \(T_i = 25^{\circ}C + 273.15 = 298.15K\) Final temperature: \(T_f = 50^{\circ}C + 273.15 = 323.15K\) Now, let's find the ratio of the final pressure to the initial pressure: \(\frac{P_f}{P_i} = \frac{T_f}{T_i}\) \(\frac{P_f}{P_i} = \frac{323.15K}{298.15K} = 1.084\) This shows that the pressure increases by a factor of 1.084, which is not doubling the pressure. Hence, heating the gas with this temperature change does not double the pressure.
03

c) Pushing down on the piston to halve the volume while keeping the temperature constant

In this action, the temperature remains constant, so the Ideal Gas Law becomes: \(PV = constant\) Now, let's halve the volume (V): \(P(\frac{1}{2}V) = constant\) \(\frac{P}{2}V = constant\) Comparing the initial and final states: \(\frac{P}{2}V = PV\) This shows that the pressure (P) doubles in this action. Therefore, pushing down the piston to halve the volume while keeping the temperature constant doubles the pressure. In conclusion, the correct action that doubles the gas pressure is: \(\boxed{\mathrm{(c)}}\) Pushing down on the piston to halve the volume while keeping the temperature constant.

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Most popular questions from this chapter

Consider the combustion reaction between 25.0 \(\mathrm{mL}\) of liquid methanol (density \(=0.850 \mathrm{g} / \mathrm{mL} )\) and 12.5 \(\mathrm{L}\) of oxygen gas measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

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