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Difficulty: HardVapour Pressure, Boiling, Evaporation, and Relative Humidity

A closed vessel initially contains air saturated with water vapour at 27∘C27^\circ\text{C} under a total pressure of 1.04×105 Pa1.04 \times 10^5\text{ Pa}. Given that the saturated vapour pressure of water at 27∘C27^\circ\text{C} is 0.04×105 Pa0.04 \times 10^5\text{ Pa}, what is the final total pressure inside the vessel if its volume is compressed to half of its initial volume while maintaining the temperature at 27∘C27^\circ\text{C}?

  1. 2.04×105 Pa2.04 \times 10^5\text{ Pa}Answer
  2. B
    2.08×105 Pa2.08 \times 10^5\text{ Pa}
  3. C
    2.00×105 Pa2.00 \times 10^5\text{ Pa}
  4. D
    1.08×105 Pa1.08 \times 10^5\text{ Pa}

Answer

The final total pressure inside the vessel is 2.04×105 Pa2.04 \times 10^5\text{ Pa}.
The total pressure is the sum of the partial pressures of dry air and saturated water vapour. Under isothermal compression to half volume, dry air follows Boyle's law and its partial pressure doubles from 1.00×105 Pa1.00 \times 10^5\text{ Pa} to 2.00×105 Pa2.00 \times 10^5\text{ Pa}. The saturated water vapour pressure remains constant at 0.04×105 Pa0.04 \times 10^5\text{ Pa} because liquid condenses out. Summing these yields 2.04×105 Pa2.04 \times 10^5\text{ Pa}.

Step-by-Step Solution

1
Calculate the initial partial pressure of the dry air.
Pair, 1=Ptotal, 1−Pvapour, 1=1.04×105 Pa−0.04×105 Pa=1.00×105 PaP_{\text{air, 1}} = P_{\text{total, 1}} - P_{\text{vapour, 1}} = 1.04 \times 10^5\text{ Pa} - 0.04 \times 10^5\text{ Pa} = 1.00 \times 10^5\text{ Pa}.
According to Dalton's law of partial pressures, the total pressure of a gas mixture is the sum of the partial pressures of its individual components.
2
Determine the final partial pressure of dry air after isothermal compression.
Pair, 2=Pair, 1×V1V2=1.00×105 Pa×2=2.00×105 PaP_{\text{air, 2}} = P_{\text{air, 1}} \times \frac{V_1}{V_2} = 1.00 \times 10^5\text{ Pa} \times 2 = 2.00 \times 10^5\text{ Pa}.
Dry air behaves as an ideal gas and follows Boyle's law (P1V1=P2V2P_1 V_1 = P_2 V_2) at constant temperature.
3
Determine the final partial pressure of the saturated water vapour.
Pvapour, 2=0.04×105 PaP_{\text{vapour, 2}} = 0.04 \times 10^5\text{ Pa}.
Saturated vapour pressure depends strictly on temperature. When compressed at constant temperature, excess vapour condenses into liquid, keeping the partial pressure constant at the saturated value.
4
Sum the final partial pressures to find the new total pressure.
Ptotal, 2=Pair, 2+Pvapour, 2=2.00×105 Pa+0.04×105 Pa=2.04×105 PaP_{\text{total, 2}} = P_{\text{air, 2}} + P_{\text{vapour, 2}} = 2.00 \times 10^5\text{ Pa} + 0.04 \times 10^5\text{ Pa} = 2.04 \times 10^5\text{ Pa}.
The total final pressure is the sum of the new dry air pressure and the unchanged saturated vapour pressure.

Key Concept

Saturated Vapour Pressure and Gas Law Applications
Estimated Time:2m 0s
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Topics

Vapour Pressure, Boiling, Evaporation, and Relative Humidity

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