Question

Pressure of a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) confined in a bulb of \(1.0 \mathrm{~L}\) capacity at \(0^{\circ} \mathrm{C}\) is (a) \(25.18 \mathrm{~atm}\) (b) \(31.205\) atm (c) \(40.215 \mathrm{~atm}\) (d) \(15.210 \mathrm{~atm}\)

Short answer

The pressure of the gas mixture is approximately 25.18 atm, which corresponds to option (a).

Step-by-step solution

Calculate moles of each gas

To find the number of moles, use the formula \( n = \frac{m}{M} \), where \( m \) is the mass and \( M \) is the molar mass. For \( \mathrm{O}_2 \), \( m = 4 \mathrm{~g} \) and \( M = 32 \mathrm{~g/mol} \), so \( n_{O_2} = \frac{4}{32} = 0.125 \mathrm{~mol} \). For \( \mathrm{H}_2 \), \( m = 2 \mathrm{~g} \) and \( M = 2 \mathrm{~g/mol} \), so \( n_{H_2} = \frac{2}{2} = 1 \mathrm{~mol} \).

Calculate total moles in the mixture

Add the moles of \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \) to find the total moles in the mixture. So, \( n_{\text{total}} = n_{O_2} + n_{H_2} = 0.125 + 1 = 1.125 \mathrm{~mol} \).

Use ideal gas law to find total pressure

The ideal gas law is given by \( PV = nRT \). We need to find \( P \). Given \( V = 1.0 \mathrm{~L} \), \( n = 1.125 \mathrm{~mol} \), \( R = 0.0821 \mathrm{~L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1} \), and \( T = 273 \mathrm{~K} \) (since \( 0^{\circ} \mathrm{C} = 273 \mathrm{~K} \)). Plug these values into the equation to find \( P \):\[P \times 1 = 1.125 \times 0.0821 \times 273 \]Thus,\[P = \frac{1.125 \times 0.0821 \times 273}{1} \approx 25.18 \mathrm{~atm} \].

Key concepts

Moles Calculation

Moles, often symbolized as \(n\), are a measure of quantity in chemistry, similar to how we use dozens for counting. Calculating the moles of a substance requires the use of the formula:
\[ n = \frac{m}{M} \]
where:

• \(n\) is the number of moles,
• \(m\) is the mass of the gas in grams, and
• \(M\) is the molar mass of the gas in grams per mole.

To determine the moles of a gas, you divide the mass of the gas by its molar mass. For instance, in our example, with \(4 \mathrm{~g}\) of \(\mathrm{O}_2\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_2\), calculating moles helps us find out how much of each gas we're dealing with on a molecular level. Having accurate molar values is crucial because it directly affects calculations related to pressure, volume, and temperature using the ideal gas law.

Partial Pressure

Partial pressure is the pressure exerted by each individual gas in a mixture. When you have a mixture of gases like \(\mathrm{O}_2\) and \(\mathrm{H}_2\), each gas contributes to the total pressure based on its proportion in the mixture.
The ideal gas law expression \(PV = nRT\) ties into this, as the partial pressure of a gas can be calculated using:
\[ P_i = \frac{n_iRT}{V} \]
where:

• \(P_i\) is the partial pressure of gas \(i\),
• \(n_i\) is the moles of gas \(i\),
• \(R\) is the ideal gas constant,
• \(T\) is the temperature in Kelvin,
• \(V\) is the volume of the container.

When combining the partial pressures, you obtain the total pressure for the gas mixture. Understanding partial pressures is important, especially when you're analyzing a compound gas mixture, to know how each component behaves within a certain volume.

Molar Mass

Molar mass, denoted \(M\), is a critical concept when analyzing gases. It represents the mass of one mole of a substance, typically expressed in grams per mole (\(\mathrm{g/mol}\)).
For example, oxygen \(\mathrm{O}_2\) has a molar mass of \(32 \mathrm{~g/mol}\), calculated from two oxygen atoms each having an atomic mass of \(16 \mathrm{~g/mol}\). Similarly, hydrogen \(\mathrm{H}_2\) has a molar mass of \(2 \mathrm{~g/mol}\), derived from two hydrogen atoms each with an atomic mass of \(1 \mathrm{~g/mol}\).
Understanding molar mass is essential because it allows us to convert between the mass of a substance and the amount of substance (in moles). It's particularly handy in the ideal gas equation to predict or manipulate conditions such as pressure, temperature, and volume of gases. Keeping molar mass in mind, we can confidently handle problems involving gas mixtures and predict their behavior under varying environmental factors.