at what temperature will the rate of diffusion of N2 will be 1.6 times that of so2 at 27 degrees celsius

To determine the temperature at which the rate of diffusion of nitrogen gas (N2) is 1.6 times that of sulfur dioxide (SO2) at 27 degrees Celsius, we can use Graham's law of effusion. This law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Let's break this down step by step.

Understanding Graham's Law

Graham's law can be expressed mathematically as:

Rate1 / Rate2 = √(Molar Mass2 / Molar Mass1)

Where:

• Rate1 is the rate of diffusion of the first gas (N2).
• Rate2 is the rate of diffusion of the second gas (SO2).
• Molar Mass1 is the molar mass of the first gas (N2).
• Molar Mass2 is the molar mass of the second gas (SO2).

Calculating Molar Masses

The molar masses of the gases are as follows:

• N2 (Nitrogen) has a molar mass of approximately 28 g/mol.
• SO2 (Sulfur Dioxide) has a molar mass of approximately 64 g/mol.

Setting Up the Equation

According to the problem, we want the rate of diffusion of N2 to be 1.6 times that of SO2:

Rate(N2) = 1.6 × Rate(SO2)

Substituting this into Graham's law gives us:

1.6 = √(Molar Mass(SO2) / Molar Mass(N2))

Plugging in the molar masses:

1.6 = √(64 / 28)

Calculating the Ratio

Now, let's calculate the right side:

√(64 / 28) ≈ √(2.2857) ≈ 1.51

This means that at 27 degrees Celsius, the rate of diffusion of N2 is about 1.51 times that of SO2, which is less than 1.6. To find the temperature at which N2 diffuses 1.6 times faster, we need to adjust the temperature.

Using the Temperature-Rate Relationship

The rate of diffusion is also affected by temperature, which can be described by the equation:

Rate ∝ √(T)

Where T is the absolute temperature in Kelvin. Therefore, we can express the rates at two different temperatures:

Rate(N2) / Rate(SO2) = √(T(N2) / T(SO2))

Finding the Required Temperature

Let’s denote the temperature of SO2 as T1 (which is 27 degrees Celsius or 300 K) and the temperature of N2 as T2. We want:

1.6 = √(T2 / 300)

Squaring both sides gives:

2.56 = T2 / 300

Now, solving for T2:

T2 = 2.56 × 300 = 768 K

Converting Back to Celsius

Finally, to convert the temperature back to Celsius:

T2 (°C) = T2 (K) - 273.15

T2 (°C) = 768 - 273.15 ≈ 494.85°C

In summary, the temperature at which the rate of diffusion of nitrogen gas will be 1.6 times that of sulfur dioxide at 27 degrees Celsius is approximately 494.85 degrees Celsius. This illustrates how temperature significantly influences the rate of diffusion of gases, as described by Graham's law.