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A vessel of volume $$20\ L$$ contains a mixture of hydrogen and helium at temperature of $$27^{\circ}C$$ and pressure $$2\ atm$$. The mass of mixture is $$5\ g$$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of the helium in the given mixture will be


A
1:2
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B
2:3
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C
2:1
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D
2:5
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Solution

The correct option is C $$2 : 5$$
Let there are $$n_{1}$$ moles of hydrogen and $$n_{2}$$ moles of helium in the given mixture. As $$Pv = nRT$$
Then the pressure of the mixture
$$P = \dfrac {n_{1}RT}{V} + \dfrac {n_{2}RT}{V} = (n_{1} + n_{2}) \dfrac {RT}{V}$$
$$\Rightarrow 2\times 101.3\times 10^{3} = (n_{1} + n_{2}) \times \dfrac {(8.3\times 300)}{20\times 10^{-3}}$$
or, $$(n_{1} + n_{2}) = \dfrac {2\times 101.30 \times 10^{3} \times 20\times 10^{-3}}{(8.3)(300)}$$
or, $$n_{1} + n_{2} = 1.62 .... (1)$$
The mass of the mixture is (in grams)
$$n_{1}\times 2 + n_{2} \times 4 = 5$$
$$\Rightarrow (n_{1} + 2n_{2}) = 2.5 ..... (2)$$
Solving the eqns. $$(1)$$ and $$(2)$$, we get
$$n_{1} = 0.74$$ and $$n_{2} = 0.88$$
Hence, $$\dfrac {m_{H}}{m_{He}} = \dfrac {0.74\times 2}{0.88\times 4} = \dfrac {1.48}{3.52} = \dfrac {2}{5}$$.

Physics

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