A vessel of volume 49.8 L contains a mixture of H ₂ and He gases at a pressure of 150 kPa and a temperature of 300 K . If the mass of H ₂ in the mixture is equal to the mass of He , the total mass of the gaseous mixture is: (Take R = 8.3 J mol ⁻¹ K ⁻¹ )

Step-by-step solution

Using the ideal gas equation, PV = nRT The total number of moles n in the mixture is given by: n = PV RT Substituting the given values ( P = 150 10^3 Pa , V = 49.8 10⁻³ m ^3 , R = 8.3 J mol ⁻¹ K ⁻¹ , T = 300 K ): n = 150 10^3 49.8 10⁻³ 8.3 300 = 7470 2490 = 3 moles Let the mass of each gas in the mixture be m grams. Number of moles of H ₂ = m 2 Number of moles of He = m 4 Total number of moles n = m 2 + m 4 = 3m 4 Equating the total moles: 3m 4 = 3 m = 4 g The total mass of the mixture is 2m = 2 4 = 8 g . Answer: 8