Multiple choice

Air inside a closed container is saturated with water vapour. The air pressure is $p$ and the saturated vapour pressure of water is $\overline { p } $. If the mixture is compressed to one-half of its volume by maintaining temperature constant, the pressure becomes :

  1. $2(p+\overline { p } )$
  2. $2p+\overline { p } $
  3. $\dfrac{(p+\overline { p } )}{2}$
  4. $p+2\overline { p } $
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B Correct answer
AI explanation

The ideal gas law states that when temperature is constant, the pressure of a gas is inversely proportional to its volume. Compressing the dry air to half its volume doubles the air pressure from p to 2p. The saturated vapor pressure of water depends solely on temperature, so it remains unchanged at p-bar when maintained at a constant temperature. The total pressure of the mixture becomes the sum of the new air pressure and the constant vapor pressure, which equals 2p plus p-bar.

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