Equal masses of the two gases oxygen (O₂) and sulfur dioxide (SO₂) are placed in separate vessels - VCE - SSCE Chemistry - Question 15 - 2005 - Paper 1

Since both gases are in the same conditions (same volume and temperature), we can apply the ideal gas law, which states that pressure is proportional to the number of moles:

$P = \frac{nRT}{V}$ Where:

• P = pressure
• n = number of moles
• R = ideal gas constant
• T = temperature
• V = volume

If equal masses of each gas are used, let the mass of each be 'm'. The number of moles for each gas can be calculated as:

• For O₂: $n_{O₂} = \frac{m}{32}$
• For SO₂: $n_{SO₂} = \frac{m}{64}$

Using the relationship of pressures and number of moles derived from the ideal gas law, we find:

• The pressure exerted by oxygen is given as 100 kPa:

$P_{O₂} = \frac{n_{O₂}RT}{V}$ Putting in values, we get:

$100 = \frac{(m/32)RT}{V}$ Now, for SO₂ using the moles calculated earlier:

$P_{SO₂} = \frac{(m/64)RT}{V}$ Using the relationship:

$P_{SO₂} = P_{O₂} \cdot \frac{\frac{m}{64}}{\frac{m}{32}} = 100 \cdot \frac{32}{64} = 50 \, \text{kPa}$

Therefore, the pressure exerted by SO₂ is approximately 50 kPa, which corresponds to option B.