Question

A vessel has \[{N_{2\;}}\] gas saturated with water vapours at a total pressure of 1 atm. The vapour pressure of water is $0.3$ atm. The contents of this vessel are completely transferred to another vessel having one-third of the capacity of the original volume, at the same temperature. The total pressure of this system in the new vessel is?

Answer 1

Hint: Boyle’s law states that the pressure $\left( {\text{P}} \right)$ of a given quantity of gas varies inversely with its volume $\left( {\text{V}} \right)$at constant temperature; i.e., in equation form, ${\text{PV = k}}$, a constant.

Complete step by step answer:
From the definition of Boyle’s law ${\text{PV = k}}$. Now to find the correct answer to this question we need to apply this law two times as it is mentioned in the question that the gas is transferred to another vessel.
Total pressure given = 1 atm.
We know vapour pressure of \[{H_2}O\] = $0.3$ atm. Out of the total pressure the vapour pressure of \[\;{N_2} = 1 - 0.3 = 0.7\;{\text{atm}}\]. We know that \[{N_2}\] is compressible unlike \[{H_2}O\].
$\therefore $ Using Boyle’s Law as temperature and moles are constant
\[{\text{}}{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ }}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}\] ---(i)
${{\text{P}}_{\text{1}}}$ = Initial pressure = $0.7$ atm, ${{\text{V}}_{\text{1}}}$= Initial volume = V. We need to calculate the final pressure ${{\text{P}}_{\text{2}}}$ .
It is given in the question that ${{\text{V}}_{\text{2}}}$ = Final Volume = $\dfrac{{\text{V}}}{3}$
Substituting these values in equation (i)
\[ \Rightarrow 0.7 \times {\text{V}} = {{\text{P}}_{\text{2}}} \times \dfrac{{\text{V}}}{3}\]
Solving this, we get:
\[ \Rightarrow {{\text{P}}_{\text{2}}} = 0.7 \times 3 = 2.1\] atm.
Hence, the total pressure can be calculated using the calculated pressure of \[{N_{2\;}}\] and initially known pressure of \[{H_2}O\].
$\therefore $Total pressure = \[2.1 + 0.3\;\; = 2.4\;\] atm.

Thus, the total pressure of the system is $2.4$ atm.

Note:
At 'higher temperature' and 'lower pressure', a gas behaves like an ideal gas, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the space between them. Thus, real gases obey Boyle’s law at low pressures, although the product ${\text{PV}}$ generally decreases slightly at higher pressures, where the gas begins to depart from ideal behaviour.