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NEET ChemistryMedium

A mixture of N2N_2 and Ar gases in a cylinder contains 7 g of N2N_2 and 8 g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2N_2 is : [Use atomic masses (in g mol−1mol^{-1}) : N = 14, Ar = 40]

1

9 bar

2

12 bar

3

15 bar

4

18 bar

Step-by-Step Solution

nN2=728=14=0.25n_{N_2} = \frac{7}{28} = \frac{1}{4} = 0.25. nAr=840=15=0.20n_{Ar} = \frac{8}{40} = \frac{1}{5} = 0.20. Now, Applying Dalton's law of partial pressure, pN2=(χN2)PTotal=0.250.45×27 bar=59×27=15 barp_{N_2} = (\chi_{N_2}) P_{Total} = \frac{0.25}{0.45} \times 27 \text{ bar} = \frac{5}{9} \times 27 = 15 \text{ bar}.

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