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

In summary, for the given scenarios: (a) decreasing the volume to one-fourth the original volume while holding the temperature constant increases the pressure inside the cylinder by a factor of 4; (b) reducing the Kelvin temperature to half its original value while holding the volume constant decreases the pressure inside the cylinder to half the initial pressure; and (c) reducing the amount of gas to half while keeping the volume and temperature constant also decreases the pressure inside the cylinder to half the initial pressure.

Step-by-step solution

Scenario (a) - Decreasing the volume to one-fourth the original volume while holding the temperature constant

In this scenario, we are asked to find the change in pressure when the volume is decreased to one-fourth of the original volume while keeping the temperature constant. From the ideal gas law equation, we have: \(P_1V_1 = nRT\) \(P_2(V_1/4) = nRT\) Divide the second equation by the first: \(\frac{P_2(V_1/4)}{P_1V_1} = \frac{nRT}{nRT}\), which simplifies to \(\frac{P_2}{P_1} = 4\) So, when the volume is reduced to one-fourth of its original value, the pressure inside the cylinder becomes 4 times greater than the initial pressure.

Scenario (b) - Reducing the Kelvin temperature to half its original value while holding the volume constant

In this scenario, we are asked to find the change in pressure when the Kelvin temperature is reduced to half its original value while keeping the volume constant. From the ideal gas law equation, we have: \(P_1V = nR(T_1)\) \(P_2V = nR(T_1/2)\) Divide the second equation by the first: \(\frac{P_2V}{P_1V} = \frac{nR(T_1/2)}{nR(T_1)}\), which simplifies to \(\frac{P_2}{P_1} = \frac{1}{2}\) So, when the Kelvin temperature is reduced to half of its original value, the pressure inside the cylinder becomes half the initial pressure.

Scenario (c) - Reducing the amount of gas to half while keeping the volume and temperature constant

In this scenario, we are asked to find the change in pressure when the amount of gas is reduced to half while keeping both the volume and temperature constant. From the ideal gas law equation, we have: \(P_1V = (n_1)RT\) \(P_2V = (n_1/2)RT\) Divide the second equation by the first: \(\frac{P_2V}{P_1V} = \frac{(n_1/2)RT}{(n_1)RT}\), which simplifies to \(\frac{P_2}{P_1} = \frac{1}{2}\) So, when we reduce the amount of gas to half while keeping the volume and temperature constant, the pressure inside the cylinder becomes half the initial pressure.

Key concepts

Understanding Gas Pressure

Within the study of physics and chemistry, gas pressure is a fundamental concept that describes the force exerted by gas particles when they collide with the walls of their container. To visualize this, imagine a crowd of people (the gas particles) in a room moving in all directions; as they move, they bump into the walls, pushing against them. This is similar to how gas particles behave.

When we talk about gas pressure in the context of the ideal gas law, it's important to realize that this pressure is directly related to the number of collisions against a given area and the energy of those collisions. For instance, scenario (a) of the exercise showcases how decreasing the volume of a gas, while keeping its temperature constant, will increase the number of collisions on the container's walls, thus escalating the gas pressure significantly.

The ideal gas law gives us a clear relationship to quantify this: assuming the amount of gas (number of moles, n) and the temperature (T) are constant, the pressure of a gas (P) is inversely proportional to its volume (V). If we decrease the volume of a gas, the pressure increases, as shown by the equation \(P_1V_1 = P_2V_2\), where the subscripts 1 and 2 represent the initial and final states of the gas, respectively. Reducing the volume to one-fourth causes the pressure to quadruple, proving that understanding how variables are connected is key to predicting the behavior of gases.

The Role of Kelvin Temperature

Temperature plays a pivotal role in describing the thermal state of a substance, but not all temperature scales are created equal when it comes to science. This is where the Kelvin scale becomes prominent. It is an absolute temperature scale starting at absolute zero, the theoretically lowest possible temperature where molecular motion ceases.

In the realm of gases, the Kelvin temperature is crucial because it is directly proportional to the average kinetic energy of the gas particles. As per scenario (b) from our exercise, reducing the Kelvin temperature to half of its initial value while keeping the volume constant leads to a decrease in the average kinetic energy of the particles. Consequently, the particles move more slowly and collide with the walls of the container with less force, resulting in a lower pressure.

The ideal gas law links pressure and temperature linearly, as seen in the relation \(P \propto T\) when volume and mole count remain constant. Thus, halving the Kelvin temperature of a gas (from \(T_1\) to \(T_1/2\)) will indeed halve the pressure (from \(P_1\) to \(P_1/2\)). This direct proportionality emphasizes the importance of temperature in controlling the pressure of a gas and why Kelvin is the preferred temperature scale in gas law calculations.

Gas Volume and Its Effects

Volume, in the context of gases, refers to the amount of space available for the gas particles to move in. Just like how having more room in a container can make it easier to move around, giving gas particles more space reduces the pressure they exert on the container walls. This is because, with increased volume, there are fewer collisions per unit area.

In scenario (c), the problem illustrates how reducing the amount of gas while keeping volume and temperature constant also affects pressure. By halving the amount of gas, we essentially reduce the number of particles in the same space, resulting in fewer collisions and therefore halving the pressure. This is an application of Avogadro's law, which states that at the same temperature and pressure, equal volumes of different gases contain the same number of particles.

This law is part of the ideal gas equation, \(PV = nRT\), showcasing the direct relationship between the number of moles (n) of a gas and its volume (V) at constant temperature and pressure. By modifying the amount of gas while maintaining a constant volume, you directly change the gas pressure, as lower particle counts translate to lower pressure. This reminds us that variables in the ideal gas law are closely interconnected and changes to one, assuming others are constant, will have predictable effects on the gas behavior.