Question

Air inside a closed container is saturated with water vapor. The air pressure is $p$, and the saturated vapor pressure of water is $\bar p$. If the mixture is compressed to one-half of its volume by maintaining temperature constant, the pressure becomes:
A. $2\left( {p + \bar p} \right)$
B. $2p + \bar p$
C. $\dfrac{{\left( {p + \bar p} \right)}}{2}$
D. $p + 2\bar p$

Answer 1

In the question, we need to determine the pressure inside the closed container such that the mixture of air water is compressed to half of its volume. For this we will use the first law of thermodynamics which establishes a relation pressure and volume of the liquid (or air) as $PV = {\text{Constant}}$.

Complete step by step answer:
Given air pressure$= p$
The saturated vapor pressure of water is$= \bar p$
We know the equation of ideal gas is given as $Pv = nRT - - (i)$
As given in the question, the mixture is compressed by maintaining temperature; hence we can write equation (i) as
$Pv = {\text{constant}}$[As T is constant]
Now when the mixture is compressed to one-half its volume and pressure changes so by comparing the vapor and pressure for the two conditions, we can write
${P_2}{v_2} = {P_1}{v_1} - - (ii)$
Since the mixture is compressed to one-half of its volume so we can write
${v_2} = \dfrac{{{v_1}}}{2} - - (iii)$
Now by substituting the value of equation (iii) in equation (ii), we get
${P_2}\left( {\dfrac{{{v_1}}}{2}} \right) = {P_1}{v_1}$
Hence we get the new pressure as
${P_2} = 2{P_1} - - (iv)$
Since total pressure in the container consisting of air and water vapor is
${P_T} = p + \bar p$
Hence after the compression, the total pressure inside the container using equation (iv) becomes
${P_T} = 2p + \bar p$
Hence option (B) is correct.

Note: Students must note that if the temperature of a system is maintained constant and the volume of the system becomes inversely proportional to the pressure, and if the pressure on the system increases, then volume decreases.