Multiple choice

A vessel contains air and saturated vapor. The pressure of air is $\mathrm{p}{2}$ and $\mathrm{p}{1}$ is the S.V. P. On compressing the mixture to one-fourth of its original volume, what is the increase in pressure of the mixture?

  1. $2\mathrm{p}_{1}$
  2. $2\mathrm{p}_{2}$
  3. $3\mathrm{p}_{1}$
  4. $3\mathrm{p}_{2}$
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D Correct answer
Explanation

Air follows Boyle's Law (P1V1 = P2V2). If volume becomes 1/4, pressure of air becomes 4 * p2. The saturated vapor pressure p1 remains constant because it is saturated. The total pressure changes from (p1 + p2) to (p1 + 4 * p2). The increase is (p1 + 4 * p2) - (p1 + p2) = 3 * p2.

AI explanation

Using Boyle's law, the pressure of the air quadruples to 4p2 when the volume is reduced to one fourth, while the saturated vapor pressure remains constant at p1. The initial pressure of the mixture is p1 plus p2, and the final pressure is p1 plus 4p2. The increase in pressure is (p1 + 4p2) minus (p1 + p2), which equals 3p2.

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