Question

A controlled environment for guinea pigs is made by constructing a 294 liter box to be maintained at \(21^{\circ} \mathrm{C}\) and \(40 \%\) relative humidity, and in which the entire air volume is changed every minute. In order to maintain this relative humidity, what weight of water must be added to the dry air flow every minute. The equilibrium vapor pressure for water at \(21^{\circ} \mathrm{C}\) is about 19 torr.

Short answer

To maintain a 40% relative humidity in a 294-liter box at \(21^{\circ}C\), approximately 2.2 grams of water must be added to the dry air flow every minute.

Step-by-step solution

Calculate the total pressure in the box

We will first need to find the total pressure inside the box. Based on the ideal gas law and assuming standard atmospheric pressure, the total pressure (P) is approximately 760 torr.

Calculate partial pressure of water vapor

Now, we can calculate the partial pressure of water vapor (Pw) by multiplying the equilibrium vapor pressure for water at 21°C (19 torr) by the relative humidity (40%). Pw = 19 torr * 0.40 Pw = 7.6 torr

Calculate the number of moles of water vapor in the box

Using the ideal gas law, we can calculate the number of moles (n) of water vapor in the box at the given partial pressure. The ideal gas law formula is: \(PV = nRT\) Where: - P is the partial pressure of water vapor (7.6 torr) - V is the volume of the box (294 L) - n is the number of moles we want to find - R is the ideal gas constant, which in this case is 62.363 L·torr/mol·K - T is the temperature in Kelvin, which is 21°C + 273.15 K = 294.15 K Rearranging the formula to calculate the number of moles of water vapor (n): \(n = \frac{PV}{RT}\) Plugging in the values we already have, we get: \(n = \frac{7.6 \text{ torr } \times 294 \text{ L}}{62.363 \text{ L·torr/mol·K} \times 294.15 \text{ K}}\) \(n = \frac{2234.4 \text{ torr·L}}{18300.6 \text{ torr·L/mol·K}}\) \(n \approx 0.122 \text{ mol}\)

Calculate the weight of water vapor in the box

Now we can calculate the weight of water vapor (W) in the box using the molar mass of water (Mw) which is 18.015 g/mol. W = n * Mw W = 0.122 mol * 18.015 g/mol W ≈ 2.2 g

Determine the weight of water needed to be added per minute

Since we have calculated the weight of water vapor present in the box and the volume of the box is changed every minute, the weight of water that must be added per minute to maintain the 40% relative humidity is the same as the weight of water vapor in the box. Weight of water to be added per minute ≈ 2.2 g

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that helps us understand and predict the behavior of gases under different conditions. It is expressed with the formula:\[PV = nRT\]where:

• \(P\) stands for pressure,
• \(V\) represents volume,
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant,
• \(T\) is the temperature in Kelvin.

This equation connects these variables, providing a relationship between them. It assumes that the gas behaves ideally, meaning its particles have no volume and do not interact. Though no gas perfectly matches these assumptions, many real gases approximate this behavior under normal conditions.
Understanding the ideal gas law aids in tasks like calculating the number of moles in a gas mixture or determining how changes in temperature or pressure affect gas volume.

Partial Pressure

Partial pressure is the pressure that a gas exerts in a mixture of gases. It is critical to understanding how different gases interact in a shared space. Dalton's Law of Partial Pressures helps explain the concept, stating that the total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas.
To find the partial pressure of a single gas, you would use the initial conditions of the mixture. For example, if the equilibrium vapor pressure of water at a certain temperature is known and you have the relative humidity, you can find its partial pressure by:

• Multiplying the equilibrium vapor pressure by the relative humidity (in decimal form).

In our scenario, the partial pressure of water vapor is essential in calculating the required weight of water vapor needed to maintain specific humidity conditions.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid form in a closed container. It signifies the tendency of a liquid to evaporate; higher vapor pressures indicate a greater propensity for evaporation.
When the vapor pressure equals the surrounding atmospheric pressure, the liquid boils. For temperature-specific conditions, such as the temperature in the guinea pig enclosure, knowing water's vapor pressure allows us to determine its partial pressure alongside the relative humidity. Vapor pressure is a pivotal concept in many applications, ranging from weather predictions to designing chemical processes that require precise humidity control.

Relative Humidity

Relative humidity is a measure of how much moisture is in the air compared to the maximum amount of moisture the air can hold at a given temperature. Expressed as a percentage, it is calculated by:\[\text{Relative Humidity} = \frac{\text{Actual Vapor Pressure}}{\text{Equilibrium Vapor Pressure}} \times 100\%\]In the guinea pig enclosure exercise, maintaining 40% relative humidity means the air only holds 40% of the water vapor it could potentially hold at that temperature.
Understanding relative humidity is crucial, as it affects comfort, health, and processes like drying and cooling. It also plays a significant role in calculating how much water must be added to an environment to achieve desired humidity levels.