Question

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the following? (a) Decrease the volume to one-fourth the original volume while holding the temperature constant. (b) Reduce the Kelvin temperature to half its original value while holding the volume constant. (c) Reduce the amount of gas to half while keeping the volume and temperature constant.

Short answer

(a) If the volume is decreased to one-fourth the original volume while holding the temperature constant, the pressure inside the cylinder increases by a factor of 4. (b) When the Kelvin temperature is reduced to half its original value while holding the volume constant, the gas pressure inside the cylinder decreases by a factor of 2. (c) If the amount of gas is reduced to half while keeping the volume and temperature constant, the gas pressure inside the cylinder decreases by a factor of 2.

Step-by-step solution

Solution for Part (a)

Decrease the volume to one-fourth the original volume while holding the temperature constant. Since the temperature is constant, we can apply Boyle's Law, which states that \[P_1V_1 = P_2V_2\] where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume. To find the new pressure, divide the original volume by 4 and substitute into the formula: \[P_2 = \frac{P_1V_1}{(1/4)V_1}\] The \(V_1\) values cancel out, leaving: \[P_2 = 4P_1\] So when the volume is decreased to one-fourth of the original volume while holding the temperature constant, the pressure inside the cylinder increases by a factor of 4.

Solution for Part (b)

Reduce the Kelvin temperature to half its original value while holding the volume constant. Since the volume is constant, we can apply Gay-Lussac's Law, which states that \[\frac{P_1}{T_1} = \frac{P_2}{T_2}\] where \(P_1\) and \(T_1\) are the initial pressure and temperature, and \(P_2\) and \(T_2\) are the final pressure and temperature. To find the new pressure, divide the original temperature by 2 and substitute into the formula: \[P_2 = P_1 \frac{T_2}{T_1} = P_1 \frac{(1/2)T_1}{T_1}\] The \(T_1\) values cancel out, leaving: \[P_2 = \frac{1}{2}P_1\] So when the Kelvin temperature is reduced to half its original value while holding the volume constant, the gas pressure inside the cylinder decreases by a factor of 2.

Solution for Part (c)

Reduce the amount of gas to half while keeping the volume and temperature constant. Since both volume and temperature are constant, we can rearrange the Ideal Gas Law formula to solve for the pressure as follows: \[P = \frac{nRT}{V}\] To find the new pressure after reducing the amount of gas to half, substitute the new value for n and solve for P: \[P_2 = \frac{(1/2)nRT}{V}\] Comparing this to the initial pressure, \(P_1 = \frac{nRT}{V}\), we can determine the relationship between the new and initial pressure: \[P_2 = \frac{1}{2}\times (P_1)\] So when the amount of gas is reduced to half while keeping the volume and temperature constant, the gas pressure inside the cylinder decreases by a factor of 2.