Question

Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O₂ and 2 g H₂ confined in a total volume of one litre at 08C is : [Given R=0.082 L atm mol⁻¹K⁻¹, T=273 K]

• A. 26.02
• B. 2.518
• C. 2.602
• D. 25.18

Answer 1

The Correct Option is D

To determine the total pressure of the mixture, we can use the ideal gas law, \(PV = nRT\), where:

• \(P\) is the pressure in atm,
• \(V\) is the volume in liters,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant \((0.082 \, \text{L atm mol}^{-1}\text{K}^{-1})\),
• \(T\) is the temperature in Kelvin.

Given:

• Mass of \(O_2\) = 4 g
• Mass of \(H_2\) = 2 g
• Volume, \(V\) = 1 L
• Temperature, \(T\) = 273 K

First, calculate the number of moles of \(O_2\) and \(H_2\):

The molar mass of \(O_2\) is 32 g/mol. Therefore, the number of moles of \(O_2\) is:

\(n_{\text{O}_2} = \frac{4 \text{ g}}{32 \text{ g/mol}} = 0.125 \text{ mol}\)

The molar mass of \(H_2\) is 2 g/mol. Therefore, the number of moles of \(H_2\) is:

\(n_{\text{H}_2} = \frac{2 \text{ g}}{2 \text{ g/mol}} = 1 \text{ mol}\)

The total number of moles, \(n_{\text{total}}\), is the sum of the moles of \(O_2\) and \(H_2\):

\(n_{\text{total}} = n_{\text{O}_2} + n_{\text{H}_2} = 0.125 + 1 = 1.125 \text{ mol}\)

Now, use the ideal gas law to find the total pressure:

\(P_{\text{total}} = \frac{n_{\text{total}} \cdot R \cdot T}{V} = \frac{1.125 \cdot 0.082 \cdot 273}{1} = 25.18 \text{ atm}\)

Therefore, the total pressure of the mixture is 25.18 atm.

The correct option is 25.18.