Question

A 2.00-L sample of was collected over water at a total pressure of 750. torr and 22 °C. When the was dried (water vapor removed), the gas had a volume of 1.95 L at 22 °C and 750. torr. Calculate the vapor pressure of water at 22 °C.

Answer 1

Answer : The vapor pressure of water at $22^oC$ is 1.01 atm.

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = ?

$P_2$ = final pressure = 750 torr = 0.987 atm    (1 atm = 760 torr)

$V_1$ = initial volume = 1.95 L

$V_2$ = final volume =  2.00 L

Now put all the given values in the above equation, we get:

$P_1\times 1.95L=0.987atm\times 2.00L$

$P_1=1.01atm$

Therefore, the vapor pressure of water at $22^oC$ is 1.01 atm.