Question

A container of \(1 \mathrm{~L}\) capacity contains a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) at \(0^{\circ} \mathrm{C}\). What will be the total pressure of the mixture? (a) \(50.42 \mathrm{~atm}\) (b) \(25.21 \mathrm{~atm}\) (c) \(15.2 \mathrm{~atm}\) (d) \(12.5 \mathrm{~atm}\)

Short answer

The total pressure of the mixture is approximately \(25.23 \text{ atm}\). The closest answer is (b) \(25.21 \text{ atm}\).

Step-by-step solution

- Convert mass to moles using the molar mass

First, convert the given masses of \(\mathrm{O}_2\) and \(\mathrm{H}_2\) to moles. The molar mass of oxygen (O) is approximately 16 g/mol, so for \(\mathrm{O}_2\), it is 32 g/mol. For hydrogen (H), the molar mass is approximately 1 g/mol, so for \(\mathrm{H}_2\), it is 2 g/mol. Using the formula \(n = \frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass, we can calculate the moles:\[n(\mathrm{O}_2) = \frac{4 \text{ g}}{32 \text{ g/mol}}\]\[n(\mathrm{H}_2) = \frac{2 \text{ g}}{2 \text{ g/mol}}\]

- Apply Ideal Gas Law to find partial pressures

Next, using the Ideal Gas Law \(PV=nRT\), where \(P\) is the pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. We'll use this to find the partial pressures of \(\mathrm{O}_2\) and \(\mathrm{H}_2\). At \(0^\circ\text{C}\) or 273.15 K, assuming that the volume of the container is 1 L (or 1 dm³) and using \(R = 0.0821 \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})\), we can find the partial pressures:\[P(\mathrm{O}_2) = n(\mathrm{O}_2)RT/V\]\[P(\mathrm{H}_2) = n(\mathrm{H}_2)RT/V\]

- Calculate moles and partial pressures

We calculate the number of moles for each gas and then the partial pressures:\[n(\mathrm{O}_2) = \frac{4}{32} = 0.125 \text{ moles}\]\[n(\mathrm{H}_2) = \frac{2}{2} = 1 \text{ mole}\]\[P(\mathrm{O}_2) = 0.125 \cdot 0.0821 \cdot 273.15\]\[P(\mathrm{H}_2) = 1 \cdot 0.0821 \cdot 273.15\]

- Find the total pressure of the mixture

The total pressure of the mixture is the sum of the partial pressures of the gases. Using Dalton's Law of Partial Pressures, we can add them up:\[P_{\text{total}} = P(\mathrm{O}_2) + P(\mathrm{H}_2)\]Calculate the numerical values of the partial pressures and then add them together for the total pressure.

- Calculate the final answer

Calculate the final partial pressures:\[P(\mathrm{O}_2) = 0.125 \cdot 0.0821 \cdot 273.15 = 2.81 \text{ atm}\]\[P(\mathrm{H}_2) = 1 \cdot 0.0821 \cdot 273.15 = 22.42 \text{ atm}\]Now sum the partial pressures to find the total pressure:\[P_{\text{total}} = 2.81 \text{ atm} + 22.42 \text{ atm} = 25.23 \text{ atm}\]

Key concepts

Dalton's Law of Partial Pressures

When dealing with a mixture of gases, each component in the mixture exerts its own pressure as if it was alone in the container. This pressure is known as the partial pressure. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual component in the gas mixture.

Using this principle, one can predict the behavior of a gas mixture by analyzing each component separately. In the context of the textbook exercise, Dalton's Law is crucial for determining the total pressure in a container holding a mixture of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \), by simply adding the pressure contributed by each gas.

Conversion of Mass to Moles

Understanding material quantities in chemical reactions involves knowing how to convert the mass of a substance to moles. The mole is a fundamental unit in chemistry that provides a bridge between the atomic or molecular scale and the macroscopic scale. To convert mass to moles, the formula \( n = \frac{m}{M} \) is used, where \( n \) represents the number of moles, \( m \) is the mass of the substance in grams, and \( M \) is the molar mass in grams per mole.

In the exercise, this conversion allows us to work out how many moles of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) are present in the container, which is then used to calculate the partial pressures using the Ideal Gas Law. This conversion is vital as most gas-related calculations require the number of moles rather than the mass.

Universal Gas Constant

The universal gas constant, denoted as \( R \), is a pivotal figure in gas-related equations. This constant provides the necessary link between the gas macroscopic properties like pressure, volume, and temperature, with the quantity of gas in moles. It appears in the Ideal Gas Law \( PV=nRT \) and has the value of 0.0821 \( \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \) when pressure is measured in atmospheres, volume in liters, and temperature in Kelvin.

Consistent use of proper units is essential to ensure the correct application of the Ideal Gas Law, which we utilize in the exercise to find the partial pressures of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \).

Partial Pressures

The partial pressure of a gas is the pressure that gas would exert if it occupied the entire volume of the mixture at the same temperature. In our exercise scenario, each gas — \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) — has its own partial pressure, depending on the number of moles and the temperature (through the Ideal Gas Law).

After calculating the moles of each gas, we proceed to apply the Ideal Gas Law separately for \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) to find their respective partial pressures. The total pressure can then be determined by summing the partial pressures, which reflects the physical reality that in a mixture of gases, each component contributes to the total force per unit area on the container's walls.