Equal weights of two gases of molecular weight 4 and 40 are mixed....

Equal weights of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is

Correct answer

Step-by-step solution

Number of mol of gas

1 = \dfrac{w}{4}

Number of mol of gas

2 = \dfrac{w}{40}

Mole fraction of gas 1,

{X}_{1} = \dfrac{\dfrac{w}{4}}{\dfrac{w}{4} + \dfrac{w}{40}} = \dfrac{10}{11}

Partial pressure of gas 1 = mol fraction of gas

1 \times

Total pressure

= \dfrac{10}{11} \times 1.1

atm = 1 atm

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