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 solve this problem, we need to apply Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This principle will help us determine the composition of the gases that have effused and those that remain in the container.

Understanding the Initial Conditions

We start with a mixture of 50 ml of helium (He) and sulfur dioxide (SO₂). The total volume of gas after effusion is 90 ml, which means that 10 ml of gas has effused through the orifice. Since the initial volume was 50 ml, we can denote the initial volumes of helium and sulfur dioxide as follows:

• Volume of He = V_He
• Volume of SO₂ = V_SO₂

Calculating Molar Masses

Next, we need the molar masses of the gases:

• Molar mass of He = 4 g/mol
• Molar mass of SO₂ = 64 g/mol

Applying Graham's Law

According to Graham's law, the ratio of the rates of effusion of two gases is given by:

Rate_He / Rate_SO₂ = √(M_SO₂ / M_He)

Substituting the molar masses:

Rate_He / Rate_SO₂ = √(64 / 4) = √16 = 4

This means that helium effuses four times faster than sulfur dioxide. Therefore, for every 4 parts of helium that effuse, only 1 part of sulfur dioxide effuses.

Determining the Composition of Effused Gas

Let’s denote the volume of helium that effuses as V_{effused, He} and the volume of sulfur dioxide that effuses as V_{effused, SO₂}. From the ratio we found, we can express this as:

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

Let V_{effused, SO₂} = x. Then:

V_{effused, He} = 4x

The total volume of gas that has effused is:

V_{effused, He} + V_{effused, SO₂} = 10 ml

Substituting the values gives:

4x + x = 10 ml

5x = 10 ml

x = 2 ml

Thus, the volumes of the effused gases are:

• V_{effused, He} = 4 * 2 ml = 8 ml
• V_{effused, SO₂} = 2 ml

Calculating Residual Gas Composition

Now, we can find the residual volumes of each gas in the container:

• Initial volume of He = 50 ml * (V_He / (V_He + V_SO₂))
• Initial volume of SO₂ = 50 ml * (V_SO₂ / (V_He + V_SO₂))

Assuming the initial mixture was 50 ml of gas, we can express the initial volumes as:

V_He = 50 ml - V_SO₂

After effusion, the residual volumes will be:

• Residual He = (Initial He - Effused He) = (V_He - 8 ml)
• Residual SO₂ = (Initial SO₂ - Effused SO₂) = (V_SO₂ - 2 ml)

Final Composition

To summarize, the composition of the gases after effusion is:

• Effused Gas: 8 ml He and 2 ml SO₂
• Residual Gas: (V_He - 8 ml) and (V_SO₂ - 2 ml)

This analysis gives you a clear understanding of how the gases behave during effusion and how their volumes change based on their molar masses. If you have any further questions or need clarification on any part of this process, feel free to ask!