Question

A vessel of volume 0.02 m3 contains a mixture of hydrogen and helium at 20c and 2 atmospheric pressure. The mass of mixture is 5 gms. Find the ratio of mass of hydrogen to that of helium in the mixture.

Answer 1

The ratio of mass of hydrogen to that of helium in the mixture is 1 : 2

Explanation:

From ideal gas equation PV=nRT,

Here, assume mass of helium in mixture is x, then mass of hydrogen is (5-x) So, 2 × 10^5  × 0.02  = ((5−x) / 2 + x / 4) 8.314 × 293

x = 3.43 and 5 − x = 1.57

So ratio of hydrogen to helium is 1 : 2

Then x+ y = 3

Thus the ratio of mass of hydrogen to that of helium in the mixture is 1 : 2

Answer 2

The ratio of mass of hydrogen to that of helium in the mixture is 1:2.

Explanation:

According to the ideal gas equation, PV = nRT

Let us assume that mass of helium the given mixture is x and mass of hydrogen is (5 - x).

Putting these values into the above ideal gas equation as follows.

          PV = nRT

  $2 \times 10^{5} \times 0.02 = (\frac{(5 - x)}{2} + \frac{x}{4}) \times 8.314 \times 293$

   x = 3.43

and,   (5 - x) = (5 - 3.43)

                 = 1.57

This means that the ratio of hydrogen to helium is 1:2.

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