Question

A jar contains a gas and a few drops of water at $TK$. The pressure in the jar is $830mm$ of $Hg$. The temperature of the jar is reduced by $1\%$. The saturated vapour pressure of water at two temperatures are $30$ and $25$ $mm$ of $Hg$. Calculate the new pressure in the jar.
A) $917$ $mm$ of $Hg$
B) $817$ $mm$ of $Hg$
C) $1017$ $mm$ of $Hg$
D) $777$ $mm$ of $Hg$

Answer 1

Vapour pressure can be caused by evaporating the liquid and the influence of vapour press are the surface area, and the intermolecular force and the temperature. This vapour pressure can cause the molecule and that will differ at temperature.

Complete step by step solution:
Vapour pressure is defined as the pressure exerted by the vapour in which the thermodynamic equilibrium with the condensed phases of the closed system at the given temperature. It is one of the fluid substances and it can measure the tendency of the material to change its gaseous or the vapour state.
In the above question, a jar contains a gas and a few drops of water at $TK$.
Pressure in the jar is $830mm$ of $Hg$
The pressure of the moisture is $30mm$ of $Hg$
Pressure of the dry gas is $830 - 30$ $= 800$ $mm$ of $Hg$
The volume is constant and the temperature Is changing,

Using the ideal gas equation,
$\Rightarrow$ $\dfrac{{{T_1}}}{{{P_1}}} = \dfrac{{{T_2}}}{{{P_2}}}$
$\Rightarrow$ $\dfrac{{{T_1}}}{{800}} = \dfrac{{0.99{T_1}}}{{{P_2}}}$
$\Rightarrow$ ${P_2} = 0.99 \times 800$
$\Rightarrow$ ${P_2} = 792$ $mm$ of $Hg$
Then the total pressure will be,
$\Rightarrow$ $792 + 25$
$\Rightarrow$ $817$ $mm$ of $Hg$
The new pressure in the jar is $817$ $mm$ of $Hg$

Therefore Option (B) is incorrect.

Note: For the vapour pressure of a given system, every component contributes to its pressure. The contribution of each component of a mixture of the system in the total pressure in the system is called it as partial pressure. The atmospheric pressure boiling point of the liquid in which the temperature of the vapour pressure equals the ambient atmospheric pressure. If the temperature of the liquid increases then the vapour pressure also increases. This goes in accordance with Charle’s law.