At 298K, a flask 'A' of unknown volume (V) contains oxygen at 5 atm. Another flask 'B' of volume 2L contains helium at 3 atm. Two flasks are connected together by a small tube of zero volume. After the two gases are completely mixed, if the resulting mixture is found to have the mole fraction of oxygen as 0.2, the volume of flask 'A' (in L) is (Assume oxygen and helium as ideal gases)
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For problems involving mixing ideal gases at constant temperature, you can use the fact that the number of moles, $n$, is directly proportional to the product $PV$. Use these proportional quantities ($n \propto PV$) to calculate mole fractions.
According to the ideal gas law, $PV=nRT$. Since temperature $T$ is constant, the number of moles $n$ is proportional to the product $PV$.
Let's find the number of moles (or a quantity proportional to it) for each gas before mixing.
For Oxygen (O$_2$) in flask A:
$P_{O_2} = 5$ atm.
$V_{O_2} = V$ L.
Number of moles of O$_2$, $n_{O_2} \propto P_{O_2}V_{O_2} = 5V$.
For Helium (He) in flask B:
$P_{He} = 3$ atm.
$V_{He} = 2$ L.
Number of moles of He, $n_{He} \propto P_{He}V_{He} = 3 \times 2 = 6$.
After mixing, the total number of moles is $n_{total} = n_{O_2} + n_{He}$.
The mole fraction of oxygen in the final mixture is given by:
$X_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{n_{O_2}}{n_{O_2} + n_{He}}$.
We are given that the mole fraction of oxygen is 0.2.
$0.2 = \frac{5V}{5V + 6}$.
Now, we solve this equation for V.
$0.2(5V + 6) = 5V$.
$1V + 1.2 = 5V$.
$1.2 = 4V$.
$V = \frac{1.2}{4} = 0.3$ L.
The volume of flask 'A' is 0.3 L.
