a mixture of 50ml of helium gas and sulphur dioxide gas is allowed to effuse through a small orifice of area 0.2mm2 till the residual gas occupies 90ml. what is the composition of the effused gas and the residual gas?

To determine the composition of the effused gas and the residual gas after the effusion process, we can apply Graham's law of effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, we have two gases: helium (He) and sulfur dioxide (SO₂).

Understanding the Gases Involved

First, let's look at the molar masses of the gases:

• Helium (He): approximately 4 g/mol
• Sulfur Dioxide (SO₂): approximately 64 g/mol

Applying Graham's Law

According to Graham's law, the ratio of the rates of effusion of two gases can be expressed as:

Rate of He / Rate of SO₂ = √(Molar Mass of SO₂ / Molar Mass of He)

Substituting the molar masses:

Rate of He / Rate of SO₂ = √(64 / 4) = √16 = 4

This means that helium effuses four times faster than sulfur dioxide.

Calculating the Effusion Volumes

Initially, we have a total volume of 50 ml of gas, which consists of helium and sulfur dioxide. Let's denote the volume of helium as V_He and the volume of sulfur dioxide as V_SO₂. Therefore:

V_He + V_SO₂ = 50 ml

Let’s assume that after effusion, the volume of gas remaining is 90 ml. This means that the total volume of gas that has effused is:

50 ml - 90 ml = -40 ml

This indicates that there is a misunderstanding in the problem statement, as the residual gas cannot exceed the initial volume. However, let's assume that the initial volume was 140 ml (50 ml + 90 ml) for the sake of calculation.

Setting Up the Equations

Let’s denote the volumes of helium and sulfur dioxide in the original mixture as:

• V_He = x ml
• V_SO₂ = (50 - x) ml

Using the ratio of effusion rates, we can express the volumes of the gases that effuse:

V_{effused, He} / V_{effused, SO₂} = 4

Let V_{effused, SO₂} = y ml. Then, V_{effused, He} = 4y ml. The total volume of gas that has effused is:

y + 4y = 5y

Finding the Residual Gas Composition

Now, if we assume that the total volume of gas that has effused is equal to the initial volume minus the residual volume, we can set up the equation:

5y = 50 ml - 90 ml

Again, this leads to an inconsistency. If we take the total volume of gas as 140 ml, we can recalculate:

5y = 140 ml - 90 ml = 50 ml

Thus, y = 10 ml. This means:

• V_{effused, SO₂} = 10 ml
• V_{effused, He} = 40 ml

Final Composition

Now, we can find the residual volumes:

• V_{residual, SO₂} = (50 - 10) ml = 40 ml
• V_{residual, He} = (0 - 40) ml = 0 ml

In summary, after the effusion process, the composition of the effused gas is:

• Helium: 40 ml
• Sulfur Dioxide: 10 ml

And the residual gas consists of:

• Helium: 0 ml
• Sulfur Dioxide: 40 ml

This analysis shows how the differences in molar mass affect the rates of effusion and ultimately the composition of the gases remaining after effusion.