Question

A $\text{ 3L }$ container at $\text{ 3 atm }$ pressure consist of a mixture of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ vapours. A solid sphere of the volume $\text{ 0}\text{.5 L }$ is also present in the same container. What will be the pressure of the gas, if the volume of the container is reduced to$\text{ 2L }$, at the same temperature (AQ. tension$\text{ = 0}\text{.2 atm }$ )?
A) $\text{ 5}\text{.4 atm }$
B) $\text{ 4}\text{.2 atm }$
C) $\text{ 4}\text{.87 atm }$
D) $\text{ 5 atm }$

Answer 1

According to Boyle’s law, at the constant temperature the pressure and the volume of a gas are inversely related to each other. That is,
$\text{ V}\propto \dfrac{\text{1}}{\text{P}}\text{ }$
Or $\text{PV =K}$ or $\text{ }{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1 }}}\text{= }{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{ }$
The $\text{AQ}$ stands for the aqueous tension. It is the pressure applied by the water vapours on the gas. Thus the total pressure is always equal to the sum of pressure of dry gas and water vapour.

Complete answer:
Here we have given the following data:
The volume of the container contains the solid sphere of the volume $\text{ 0}\text{.5 L }$. Thus the volume of the mixture of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ vapours is equal to, $\text{ }{{\text{V}}_{\text{1}}}\text{ = 3 }-\text{0}\text{.5 L = 2}\text{.5 L }$
The pressure of the mixture of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ vapours is equal to $\begin{aligned} & \text{ P = 3 atm } \\\ & \therefore {{\text{P}}_{\text{1}}}\text{ = 3 }-0.2\text{ = 2}\text{.8 atm } \\\ \end{aligned}$
The volume of the container is reduced to the $\text{ 2L }$, then the final volume of the mixture of $\text{ }{{\text{N}}_{\text{2}}}\text{ }$ and $\text{ }{{\text{H}}_{\text{2}}}\text{O }$ vapours is equal to, $\text{ }{{\text{V}}_{2}}\text{ = 2 }-\text{0}\text{.5 L = 1}\text{.5 L }$
We have given the aqueous tension equals to the $\text{ AQ = 0}\text{.2 atm }$
It is the extra pressure applied by the water vapours thus the final pressure is greater than the actual pressure on the dry gas pressure. Thus, always subtract the aqueous tension from the pressure to get the actual value of the pressure applied to the dry gas.
We are interested to find out the value of the pressure when the volume of the container is reduced to${{\text{V}}_{2}}$.
According to Boyle's law, at the constant temperature, the pressure and the volume of a gas are inversely related to each other. That is,
$\text{ V}\propto \dfrac{\text{1}}{\text{P}}\text{ }$
Or it can be also written as,
$\text{ (}{{\text{P}}_{\text{1}}}\text{+AQ)}{{\text{V}}_{\text{1 }}}\text{= (}{{\text{P}}_{\text{1}}}\text{+AQ)}{{\text{V}}_{\text{2}}}\text{ }$ (1)
Since we are dealing with the aqueous tension the equation is modified as follows,
Let's substitute the values in equation (1). We have,
$\text{ }\begin{matrix} \text{(}{{\text{P}}_{\text{1}}}-\text{AQ)}{{\text{V}}_{\text{1 }}} & = & \text{(}{{\text{P}}_{2}}-\text{AQ)}{{\text{V}}_{\text{2}}} \\\ (3\text{ }-\text{ 0}\text{.2 )}\times \text{2}\text{.5} & = & \text{(}{{\text{P}}_{2}}-\text{0}\text{.2)}\times \text{1}\text{.5} \\\ 7 & = & 1.5{{\text{P}}_{2}}-0.3 \\\ {{\text{P}}_{2}} & = & \dfrac{7.3}{1.5} \\\ \Rightarrow {{\text{P}}_{2}} & = & 4.86\simeq 4.87atm \\\ \end{matrix}$
Therefore the pressure of the mixture of the after the reduction in the volume of the container is equal to the$\text{ 4}\text{.87 atm }$.

Hence, (C) is the correct option.

Note:
Note that, then according to Dalton's law of partial pressure, the total pressure on the gas or mixture of the gas is equal to the sum of the partial pressure of the individual gas. If the mixture is made of dry gas and the water vapours, then we have to consider the water vapour or the aqueous tension produced by the water vapours. The total pressure is given as the,
$\text{ }{{\text{P}}_{\text{Total}}}\text{ = }{{\text{P}}_{\text{moist air}}}\text{ = }{{\text{P}}_{\text{dry gas}}}\text{ + }{{\text{P}}_{\text{water vapour }}}\text{ }$
Here, the ${{\text{P}}_{\text{water vapour }}}$is called the aqueous tension. When gas which is insoluble in water is collected over water, it becomes moist.