Question

A glass tumbler containing \(243 \mathrm{~cm}^{3}\) of air at \(1.00 \times\) \(10^{2} \mathrm{kPa}\) (the barometric pressure) and \(20^{\circ} \mathrm{C}\) is turned upside down and immersed in a body of water to a depth of \(25.5 \mathrm{~m}\). The air in the glass is compressed by the weight of water above it. Calculate the volume of air in the glass, assuming the temperature and barometric pressure have not changed.

Short answer

The new volume of air is approximately \(69.4 \,\mathrm{cm}^3\).

Step-by-step solution

Understand the problem

We have a closed container of air that is submerged in water. We need to find the new volume of the air, considering it is compressed due to the pressure from the water.

Identify the known variables

Initial volume of air, \(V_1 = 243 \,\mathrm{cm}^3\), initial pressure, \(P_1 = 100 \,\mathrm{kPa}\), and depth of water \(h = 25.5 \,\mathrm{m}\). The density of water, \(\rho = 1000 \,\mathrm{kg/m}^3\), and acceleration due to gravity, \(g = 9.81 \,\mathrm{m/s}^2\).

Calculate the pressure due to water

The pressure exerted by the water can be found using the formula \(P_{\text{water}} = \rho g h\). Substituting the values: \(P_{\text{water}} = 1000 \times 9.81 \times 25.5 = 250155 \,\mathrm{Pa}\) or approximately \(250.2 \,\mathrm{kPa}\).

Calculate the total pressure on the air

The total pressure on the air is the sum of the barometric pressure and the water pressure: \(P_2 = P_1 + P_{\text{water}}\). Therefore, \(P_2 = 100 \,\mathrm{kPa} + 250.2 \,\mathrm{kPa} = 350.2 \,\mathrm{kPa}\).

Apply the ideal gas law: Boyle’s Law

According to Boyle's Law, which states \(P_1 V_1 = P_2 V_2\), we can rearrange it to find \(V_2 = \frac{P_1 V_1}{P_2}\).

Substitute the known values into Boyle’s Law

We substitute our known values into the rearranged equation: \(V_2 = \frac{100 \,\mathrm{kPa} \times 243 \,\mathrm{cm}^3}{350.2 \,\mathrm{kPa}}\).

Calculate the new volume of the air

Perform the calculation: \(V_2 = \frac{100 \times 243}{350.2} \approx 69.4 \,\mathrm{cm}^3\). This is the volume of the air in the glass when submerged.

Key concepts

Pressure Calculation

Pressure is essentially the force exerted per unit area. When a glass tumbler is submerged in water, the weight of the water above it adds extra pressure on the air inside the glass.
This pressure increase is important to calculate because it affects the volume of air within the tumbler. To calculate the additional pressure exerted by the water, we use the formula:\[ P_{\text{water}} = \rho g h \]Where:

• \( \rho \) is the density of water, typically about \(1000 \, \mathrm{kg/m}^3\).
• \( g \) is the acceleration due to gravity, approximately \(9.81 \, \mathrm{m/s}^2\).
• \( h \) is the depth of water the tumbler is submerged in.

By multiplying these values, you can find the pressure exerted by the column of water. This value is then added to the atmospheric pressure to find the total pressure on the submerged air.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that describes how gases behave under varying conditions of pressure, volume, and temperature. For our exercise, we will consider a simplified version called Boyle's Law which applies when the temperature and the amount of gas do not change.Boyle's Law states:\[ P_1 V_1 = P_2 V_2 \]This shows that the initial pressure \(P_1\) and volume \(V_1\) are equal to the final pressure \(P_2\) and volume \(V_2\).
This equation helps us understand how the air's volume will decrease when the pressure increases, assuming constant temperature and amount of gas. By rearranging the formula, you can solve for the new volume \(V_2\).

Volume of Air Under Pressure

When compressing air under water, as pressure increases, the volume decreases. This is a practical illustration of Boyle's Law. Initially, the volume of air in the tumbler is \(243 \, \mathrm{cm}^3\). Using the calculated pressures from our previous sections, we can easily determine the compression effect on this volume.To find the new volume \(V_2\):\[ V_2 = \frac{P_1 V_1}{P_2} \]Where the new pressure \(P_2\) includes both barometric and water pressures which causes a decrease in volume when submerged. This compressed volume shows the adaptability and changeability of gases under new conditions.

Compression by Water Depth

Compression by water depth explains a real-world application of atmospheric and hydrostatic pressure principles. When a glass tumbler is submerged in water, the depth of submersion directly influences how much the air inside is compressed.As depth (\( h \)) increases, the pressure exerted by the water (\( P_{\text{water}}\)) rises too, as calculated using the formula from the earlier section. This scenario mirrors how divers experience pressure changes the deeper they dive. Every additional meter underwater adds more pressure to the air inside, compressing its volume further.
By understanding this, you can see how divers and underwater vessels need to be designed to withstand such pressures, keeping materials from collapsing.