Question

Hydrogen gas generated in laboratory experiments is usually collected over water. It is called a “wet gas” when collected in this manner because it contains water vapor. A sample of “wet” hydrogen at 258C fills a 125-mL flask at a pressure of 769 mm Hg. If all the water is removed by heating, what volume will the dry hydrogen occupy at a pressure of 722 mm Hg and a temperature of 378C? (The vapor pressure of water at 258C is 23.8 mm Hg.)

Short answer

Answer: The volume of the dry hydrogen gas is approximately 125.35 mL.

Step-by-step solution

Understand the problem and list the given values

We are given the volume (V1), pressure (P1), and temperature (T1) of the wet hydrogen gas. We need to find the volume (V2) of the dry hydrogen gas at a new temperature (T2) and pressure (P2). The vapor pressure of water at T1 is also provided. V1 = 125 mL P1 = 769 mm Hg T1 = 25°C (or 298 K, when converted to Kelvin using T(K) = T(°C) + 273) Vapor pressure of water at T1 = 23.8 mm Hg T2 = 37°C (or 310 K, when converted to Kelvin) P2 = 722 mm Hg

Calculate the pressure of dry hydrogen gas

We will use the concept of partial pressure. Since the sample contains both hydrogen gas and water vapor, its total pressure P1 can be written as the sum of their individual partial pressures (PH2 and Pwater): P1 = PH2 + Pwater We are given the vapor pressure of water at T1 (i.e., Pwater = 23.8 mm Hg). We can now calculate the partial pressure of hydrogen gas (PH2): PH2 = P1 - Pwater PH2 = 769 mm Hg - 23.8 mm Hg PH2 ≈ 745.2 mm Hg

Use the Ideal Gas Law formula to solve for the initial and final states

We will use the Ideal Gas Law formula, which is: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. The number of moles (n) of dry hydrogen gas remains constant during the process, so we can write an equation for the initial and final states: P1*V1 / T1 = P2*V2 / T2 Since we are looking for the volume of dry hydrogen gas (V2), we can rearrange the equation: V2 = (P1*V1*T2) / (T1*P2)

Plug in the given values and solve for V2

We can now plug in the values we've found or were given: V2 = (745.2 mm Hg * 125 mL * 310 K) / (298 K * 722 mm Hg) V2 ≈ 125.35 mL So, the volume of the dry hydrogen gas at a pressure of 722 mm Hg and a temperature of 37°C is approximately 125.35 mL.

Key concepts

Understanding Partial Pressure

Partial pressure is a fundamental concept in understanding gas mixtures. Consider the air around us—it's a mixture of different gases like nitrogen, oxygen, and others, each contributing to the total pressure of the atmosphere. For any single gas in a mixture, the partial pressure is the pressure that gas would exert if it alone occupied the entire volume of the mixture at the same temperature.

When dealing with wet gases like in the exercise, calculating the partial pressure of the gas itself (excluding the water vapor) is essential to applying gas law calculations accurately. The wet hydrogen gas has water vapor mixed in, exerting its own pressure. To find the dry hydrogen gas volume, it's necessary to subtract the vapor pressure of the water from the total pressure to obtain the partial pressure of hydrogen alone.

Demystifying Vapor Pressure

Vapor pressure is a term that can confuse many students, but it's really just a measure of a liquid's tendency to evaporate. When a liquid is in a closed container, the molecules escape from the liquid into the gas phase and eventually, an equilibrium is reached where the rate of evaporation equals the rate of condensation. The pressure exerted by the vapor at this point is called the vapor pressure.

Vapor pressure is influenced by temperature; as temperature increases, vapor pressure rises. This explains why we have a specific value for the vapor pressure of water at 25°C in the exercise. Knowing this allows us to correct the pressure of the hydrogen gas to account for the presence of water vapor. It's a key step in gas law calculations involving wet gases.

Performing Gas Law Calculations

Applying the Ideal Gas Law

The Ideal Gas Law, represented as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of substance in moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvins, is a cornerstone of gas law calculations. It relates the four variables of a gas. If we keep the amount of gas and the gas constant fixed, as in our problem, the relationship simplifies to \(P_1V_1/T_1 = P_2V_2/T_2\).

By rearranging this equation to solve for the unknown, you can connect the initial and final states of a gas. The use of partial pressure and accurate temperature conversion is crucial to finding the correct volume, as demonstrated in the step-by-step solution.

Mastering Temperature Conversion

Temperature conversion is an indispensable skill in chemistry, especially when working with gases. The Ideal Gas Law requires temperatures to be in Kelvins, so converting from Celsius to Kelvin is a necessary step. The Kelvin scale is the SI base unit for temperature and is favored in scientific calculations to avoid negative numbers, as its zero point is absolute zero.

The conversion is simple: \(T(K) = T(°C) + 273.15\). But remember, it's 273.15 and not 273; the extra precision can matter in scientific work. In our exercise, converting the given Celsius temperatures to Kelvins allowed us to insert the proper values into the Ideal Gas Law equation and solve for the unknown volume of dry hydrogen gas.