Question

The volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure will be :- [Use R = 0.083 bar L $K^{-1}mol^{-1}$]

• A. 96.66 L
• B. 55.87 L
• C. 3.10 L
• D. 5.37 L

Answer 1

The Correct Option is D

To determine the volume occupied by 1.8 g of water vapour at 374 °C and 1 bar pressure, we will apply the ideal gas equation:

PV = nRT

Where:

• P is the pressure in bars.
• V is the volume in liters.
• n is the number of moles.
• R is the ideal gas constant = 0.083 \, \text{bar L K}^{-1} \text{mol}^{-1}.
• T is the temperature in Kelvin.

First, calculate the number of moles of water vapour (H_2O):

Molecular weight of water (H_2O) = 18 g/mol

Number of moles (n) = \frac{\text{mass}}{\text{molar mass}} = \frac{1.8 \, \text{g}}{18 \, \text{g/mol}} = 0.1 \, \text{mol}

Convert the temperature from Celsius to Kelvin:

T = 374 \, \degree C + 273.15 = 647.15 \, \text{K}

Substitute the known values into the ideal gas equation:

1 \, \text{bar} \cdot V = 0.1 \, \text{mol} \times 0.083 \, \text{bar L K}^{-1} \text{mol}^{-1} \times 647.15 \, \text{K}

Solve for V:

V = \frac{0.1 \times 0.083 \times 647.15}{1} = 5.37 \, \text{L}

Hence, the volume occupied by 1.8 g of water vapor at the given conditions is 5.37 L. This matches the answer in the options provided.