Q 11-12-114JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
A cylindrical container of volume $4.0\times10^{-3}$ m$^3$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is $400$ K. The pressure of the mixture of gases is: [Take gas constant as $8.3$ J mol$^{-1}$ K$^{-1}$]
Answer: (C) $24.9\times10^5$ Pa
$$P = \frac{nRT}{V} = \frac{3\times8.3\times400}{4.0\times10^{-3}} = 2.49\times10^6\ \text{Pa} = 24.9\times10^5\ \text{Pa}$$
Solution by Sreeraj P, M.Sc Physics