Question

Choose the correct option for the total pressure (in atm.) in a mixture of $4 \,g\, O _{2}$ and $2 \,g \,H _{2}$ confined in a total volume of one litre at $0^{\circ} C$ is : $\left[\right.$ Given $\left.R = 0.082 \,L \,atm\, mol ^{-1} K ^{-1}, T =273\, K \right]$

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

Answer 1

The Correct Option is C

To determine the total pressure in a mixture of gases, we use the Ideal Gas Law and Dalton's Law of Partial Pressures. The Ideal Gas Law is given by:

\(PV = nRT\)

where:

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

First, we need to determine the number of moles of each gas:

1. For \(O_2\):

Molecular weight of \(O_2\) = 32 g/mol

Moles of \(O_2 = \frac{4 \,g}{32 \,g/mol} = 0.125 \,mol\)

1. For \(H_2\):

Molecular weight of \(H_2\) = 2 g/mol

Moles of \(H_2 = \frac{2 \,g}{2 \,g/mol} = 1.0 \,mol\)

According to Dalton's Law of Partial Pressures:

\(P_{\text{total}} = P_{O_2} + P_{H_2}\)

Using the Ideal Gas Law for each gas at \(V = 1 \,L\) and \(T = 273 \,K\):

1. Pressure of \(O_2\): \(P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{0.125 \cdot 0.082 \cdot 273}{1} = 2.7975 \,atm\)
2. Pressure of \(H_2\): \(P_{H_2} = \frac{n_{H_2}RT}{V} = \frac{1.0 \cdot 0.082 \cdot 273}{1} = 22.383 \,atm\)

Total pressure:

\(P_{\text{total}} = 2.7975 + 22.383 = 25.18 \,atm\)

Hence, the correct option is \(25.18\).