Question

A sample of air occupies \(3.8 \mathrm{~L}\) when the pressure is \(1.2 \mathrm{~atm} .\) (a) What volume does it occupy at \(6.6 \mathrm{~atm} ?\) (b) What pressure is required in order to compress it to \(0.075 \mathrm{~L} ?\) (The temperature is kept constant.)

Short answer

For part (a), the final volume is approximately 0.69 L, and for part (b), the final pressure is approximately 60.8 atm.

Step-by-step solution

Applying Ideal Gas Law - Part (a)

For part (a), the relationship between the initial and final states of pressure and volume can be expressed using the ideal gas law. Thus, the equation is \(P_1V_1 = P_2V_2\). Plugging in the given initial pressure \(P_1 = 1.2 \, \mathrm{atm}\) and volume \(V_1 = 3.8 \, \mathrm{L}\), as well as the final pressure \(P_2 = 6.6 \, \mathrm{atm}\), we can solve for the final volume \(V_2\). That gives us \(V_2 = \frac{P_1V_1}{P_2}\).

Calculation of Final Volume - Part (a)

Substituting the given values into the equation from Step 1, we find \(V_2 = \frac{(1.2 \, \mathrm{atm})(3.8 \, \mathrm{L})}{6.6 \, \mathrm{atm}}\). This results in \(V_2 \approx 0.69 \, \mathrm{L}\). This is the volume the air would occupy at a pressure of 6.6 atm.

Applying Ideal Gas Law - Part (b)

For part (b), the same ideal gas law translates to \(P_1V_1 = P_2V_2\). Given the initial pressure \(P_1 = 1.2 \, \mathrm{atm}\) and volume \(V_1 = 3.8 \, \mathrm{L}\), as well as the final volume \(V_2 = 0.075 \, \mathrm{L}\), we can find the final pressure \(P_2\). Therefore \(P_2 = \frac{P_1V_1}{V_2}\).

Calculation of Final Pressure - Part (b)

Substituting the given values into the equation from Step 3, we find \(P_2 = \frac{(1.2 \, \mathrm{atm})(3.8 \, \mathrm{L})}{0.075 \, \mathrm{L}}\). This results in \(P_2 \approx 60.8 \, \mathrm{atm}\). This is the pressure required in order to compress the air to a volume of 0.075 L.

Key concepts

Volume

Volume is a measure of the amount of space an object or substance occupies. In the context of gases, the volume is usually measured in liters (L). When dealing with gases, the concept of volume is crucial because gases have the ability to expand and fill the container they are in.

For example, if a balloon is filled with air, the gas particles inside will spread out evenly to fill the entire space within the balloon. This is due to the fact that gas particles are always in motion, constantly colliding and bouncing off the walls of their container.

In our given exercise, the initial volume of the gas is 3.8 liters. When the pressure changes, the volume also changes, but the amount of gas remains constant. This shows how volume and pressure are interconnected, which will lead us to the concept of Boyle's Law.

Pressure

Pressure refers to the force exerted by the gas per unit area. It's how much the gas "pushes" against the surfaces of its container. In a more scientific sense, it is defined as the force applied perpendicular to the surface of an object over a particular area.

The standard unit of pressure used in many scientific contexts, including this exercise, is atmospheres (atm). Other commonly used units include pascals (Pa), bars, and mmHg.

In the exercise, the initial pressure is given as 1.2 atm. As the volume decreases, the pressure increases, because the same number of molecules are confined to a smaller space, causing more frequent collisions with the walls. This behavior exemplifies how pressure and volume are related in gas laws.

Boyle's Law

Boyle's Law is one of the fundamental principles that explain the relationship between pressure and volume for a given amount of gas at constant temperature. According to this law, the pressure and volume of a gas are inversely proportional.

Mathematically, it can be expressed 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 final pressure and volume. This means if you increase the pressure, the volume decreases, provided the temperature stays constant, and vice versa.

In the exercise, Boyle's Law helps us determine how changing the pressure affects the volume or how altering the volume impacts the pressure. By rearranging the formula, you can solve for the unknown variable using the known values. For example, when solving for the final volume in Part (a) of the exercise, we use \(V_2 = \frac{P_1V_1}{P_2}\). Conversely, to find the final pressure in Part (b), we adjust the equation to \(P_2 = \frac{P_1V_1}{V_2}\). This clarity in the mathematical relationship helps us predict how a gas will behave under different conditions.