Question

A vessel of volume 20L contains a mixture of hydrogen and helium at a temperature of 27 degrees Celsius and pressure 2 atm. The mass of the mixture is 5g. Assuming the gas is to be ideal the ratio of the mass of hydrogen to that of helium in the given mixture will be:
A. 1:2
B. 2:3
C. 2:1
D. 2:5

Answer 1

Hint: Use the ideal gas law and find the total pressure of the mixture in terms of the individual pressures exerted by each gas and then find the total number of moles of the two gases. Write the mass of the mixture in terms of the moles of each gas. Solve the equations to find the ratio of masses.

Complete step-by-step answer:
Ideal gas law is given by PV = nRT
Where P is equal to the pressure exerted by the gas, Vis equal to the volume occupied by the gas and n is equal to the number of moles and T is equal to the temperature
Now we can write the total pressure of the mixture is the sum of the pressure exerted by each gas as $ P = \dfrac{n_1 RT}{V} + \dfrac{n_2 RT}{V} = (n_1 + n_2) \dfrac{RT}{V} $
From the above relation, we can find: $2 \times 101.3 \times 10^3 = (n_1 + n_2) \dfrac{(8.3)(300)}{(20 \times 10^{-3})} = 1.62$
Now the mass of the mixture is given as 5 grams
In terms of moles of individual gases, we can write the mass of the mixture as: $n_1 \times 2 + n_2 \times 4 = 5$
Solving the two equations in terms of number of moles of individual gases we get: $n_1 = 0.74 $ and $n_2 = 0.88$
Now we can write the mass of the Helium in the mixture as $0.88 \times 4$
We can write the mass of the hydrogen gas in the mixture as $0.74 \times 2$
So the ratio of the mass of the hydrogen and the mass of the Helium in the mixture becomes:
$\dfrac{H_m}{He_m} = \dfrac{0.74 \times 2}{0.88 \times 4} = \dfrac{1.48}{3.52} = \dfrac{2}{5} $
Thus we have found the ratio of the mass of the two gases in the mixture using the ideal gas law as 2:5

Note: One of the possible mistakes that one can make in solving the numerical is that we tend to take pleasure as 2 units which is wrong. The given pressure is 2 atm which we have to convert into the SI system before solving the problem. Another mistake that we can make in this problem is that we may take weight in place of mass of the mixture. Taking care of these two things we can solve this kind of problems easily.