Question

You have a gas confined to 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 temperature (in kelvins) to half its original value while holding the volume constant. (c) Reduce the amount of gas to one-fourth while keeping the volume and temperature constant.

Short answer

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

Step-by-step solution

Analyzing the given information

In this case, the volume is reduced to one-fourth of its original volume (V_new = V_original/4) and the temperature is held constant. So, using the Ideal Gas Law, we will keep the variables n, R, and T constant while considering the change in volume.

Calculating the new pressure

Using the Ideal Gas Law equation, with V_new = V_original/4: \(P_{new} \times \frac{1}{4}V = nRT\) Since the variables n, R, and T are constant, we can compare the initial and final equations: \(P_{new}\times \frac{1}{4}V = P_{original}\times V\) Now, we can solve for \(P_{new}\): \(P_{new} = 4\times P_{original}\)

Conclusion

As a result, when the volume of the cylinder is decreased to one-fourth of its original value, the pressure of the gas inside the cylinder increases by a factor of 4. #b) Reduce the temperature (in kelvins) to half its original value while holding the volume constant#

Analyzing the given information

In this scenario, the temperature is reduced to half of its original value (T_new = T_original/2) and the volume is held constant. We will keep the variables n, R, and V constant and consider the change in temperature.

Calculating the new pressure

Using the Ideal Gas Law equation, with T_new = T_original/2: \(P_{new} \times V = nR \times \frac{1}{2}T\) Comparing the initial and final equations: \(P_{new}\times V = P_{original}\times \frac{1}{2}\times V\) Now, we can solve for \(P_{new}\): \(P_{new} = \frac{1}{2}\times P_{original}\)

Conclusion

When the temperature of the gas is reduced to half of its original value, the pressure of the gas inside the cylinder decreases by a factor of 2. #c) Reduce the amount of gas to one-fourth while keeping the volume and temperature constant#

Analyzing the given information

In this case, the amount of gas is reduced to one-fourth of its original value (n_new = n_original/4) and the volume and temperature are held constant. We will keep the variables V, R, and T constant and consider the change in the amount of gas.

Calculating the new pressure

Using the Ideal Gas Law equation, with n_new = n_original/4: \(P_{new} \times V = \frac{1}{4}nR \times T\) Comparing the initial and final equations: \(P_{new}\times V = P_{original}\times \frac{1}{4}\times n\times V\) Now, we can solve for \(P_{new}\): \(P_{new} = \frac{1}{4}\times P_{original}\)

Conclusion

When the amount of gas in the cylinder is reduced to one-fourth of its original value, while keeping volume and temperature constant, the pressure of the gas inside the cylinder decreases by a factor of 4.

Key concepts

Gas Pressure

Gas pressure is a fundamental concept in understanding the behavior of gases and is crucial when working with the Ideal Gas Law. It is the force that a gas exerts on the walls of its container, and it's measured in units such as atmospheres (atm), Pascal (Pa), or millimeters of mercury (mmHg).

Pressure is caused by gas molecules constantly moving and colliding with each other and their container. Imagine a gas trapped inside a cylinder with a piston; when these gas molecules hit the piston, they apply a force over the piston's area, resulting in pressure. This activity at the molecular level is why changes in volume or temperature significantly impact the pressure exerted by a gas.

In the exercise given, adjusting the volume and temperature led to changes in pressure, illustrating that gas pressure isn't a static value but rather depends on other conditions of the gas.

Volume and Pressure Relationship

The relationship between volume and pressure in gases is known as Boyle's Law. It states that, for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure. This means when the volume increases, the pressure decreases, and when the volume decreases, the pressure increases, as long as the temperature remains unchanged.

Looking at the exercise solution, when the volume of the gas was reduced to one-fourth while maintaining a constant temperature, the pressure increased fourfold. This can be visualized by picturing the same number of gas molecules being compressed into a smaller space, colliding more frequently with the walls, thus increasing the pressure.

This principle can be mathematically represented as: \( P_1V_1 = P_2V_2 \) where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the pressure and volume after the change.

Temperature and Pressure Relationship

The relationship between temperature and pressure in gases is described by Gay-Lussac's Law, which indicates that the pressure of a gas is directly proportional to its absolute temperature when the volume is kept constant. Therefore, if you increase the temperature, the pressure will also increase, and if you decrease the temperature, the pressure decreases, assuming the volume does not change.

In the context of our exercise, reducing the temperature to half of its original value resulted in the pressure of the gas being halved as well. A lower temperature means the gas molecules have less kinetic energy and, consequently, they move slower, leading to fewer collisions with the container walls, hence lowering the pressure.

The mathematical relationship can be expressed as: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) where \( P_1 \) and \( T_1 \) represent the initial pressure and temperature, and \( P_2 \) and \( T_2 \) the pressure and temperature after the change, with temperature measured in Kelvins.