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 find the total pressure of the gas mixture, we need to use the ideal gas law given by:

\(PV = nRT\), 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, given as \(0.082 \,L \,atm\, mol ^{-1} K ^{-1}\).
• \(T\) is the temperature in Kelvin, given as \(273 \,K\).

First, we calculate the number of moles for each gas using their respective molecular weights:

• Molecular weight of \(O_2 = 32 \, g/mol\)
• Moles of \(O_2 = \frac{4 \,g}{32 \,g/mol} = 0.125 \,mol\)
• Molecular weight of \(H_2 = 2 \, g/mol\)
• Moles of \(H_2 = \frac{2 \,g}{2 \,g/mol} = 1 \,mol\)

Total moles \((n_{\text{total}})\) in the mixture = \(0.125 + 1 = 1.125 \,mol\).

Substituting into the ideal gas equation for total pressure:

\(P = \frac{nRT}{V} = \frac{1.125 \times 0.082 \times 273}{1} = 25.18 \, atm\).

Thus, the correct option is \(25.18 \, atm\).