Psychrometry application (saturated air at 20 °C): A rigid vessel of volume 1000 m^3 contains air that is saturated with water vapour at a total pressure of 100 kPa and a temperature of 20 °C. Given that the saturation vapour (partial) pressure of water at 20 °C is 2.34 kPa, estimate the mass of water vapour (in kg) present in the vessel.

Introduction / Context:
In psychrometry and gas–vapour calculations, the water vapour in a closed, saturated air–water system can be quantified using the ideal-gas relationship applied to the vapour phase. At a given temperature, the partial pressure of water vapour equals the saturation vapour pressure, independent of the total pressure. This question asks for the mass of water vapour in a large rigid vessel at 20 °C when the air is saturated.

Given Data / Assumptions:

• Vessel volume V = 1000 m^3 (rigid).
• Total pressure P_total = 100 kPa at T = 20 °C.
• Saturation vapour pressure of water at 20 °C: p_sat = 2.34 kPa.
• Water vapour behaves ideally (reasonable at 20 °C and low partial pressure).
• Specific gas constant for water vapour: R_v ≈ 461.5 J/kg·K.

Concept / Approach:
For a saturated mixture, the partial pressure of water vapour equals the saturation pressure at the specified temperature. The vapour mass then follows from the ideal-gas equation in the form m = p_v * V / (R_v * T_abs). Only the vapour partial pressure is used, not the total pressure, because Dalton’s law applies and each component obeys the ideal-gas law with its own partial pressure.

Step-by-Step Solution:

Verification / Alternative check:
A quick order-of-magnitude check: at a few kPa and room temperature, ideal-gas density ρ ≈ p / (R_v * T) ≈ 2340 / (461.5 * 293) ≈ 0.0173 kg/m^3; multiplying by 1000 m^3 gives ≈ 17.3 kg, confirming the result.

Why Other Options Are Wrong:

• 20, 25, 34 kg: these overestimate the vapour mass; they would imply higher p_v or lower T than given.
• 12 kg: underestimates the mass; inconsistent with the ideal-gas calculation at the stated saturation pressure.

Common Pitfalls:
Using total pressure instead of vapour partial pressure; forgetting to convert °C to K; mixing up the universal gas constant with the specific gas constant for water vapour; rounding too early.

Final Answer:
17