GATE 2023 ME – Question 61
Consider a mixture of two ideal gases, X and Y, with molar masses $\bar{M}_X = 10$ kg/kmol and $\bar{M}_Y = 20$ kg/kmol, respectively, in a container. The total pressure in the container is 100 kPa, the total volume of the container is 10 m$^3$ and the temperature of the contents of the container is 300 K. If the mass of gas-X in the container is 2 kg, then the mass of gas-Y in the container is ____ kg. (Rounded off to one decimal place)
Assume that the universal gas constant is 8314 J kmol$^{-1}$K$^{-1}$.
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Correct answer: 3.98 to 4.02
Explanation
The total amount of gas is $n = \frac{PV}{\bar{R}T} = \frac{100000 \times 10}{8314 \times 300} = 0.4009$ kmol. The amount of X is $\frac{2}{10} = 0.2$ kmol, so $n_Y = 0.4009 - 0.2 = 0.2009$ kmol. The mass of Y is $0.2009 \times 20 = 4.02$ kg, which is 4.0 kg.