Question

A quantity of an ideal gas is collected in a graduated tube over the mercury in a barometer type arrangement. The volume of gas at $20{}^\circ C$ is 50 ml and the level of mercury is 100 mm above the outside of the mercury. The atmospheric pressure is 750 mm. The volume of gas at STP
(Take$R=0.083\text{ lt atm/K/mole}$)

Answer 1

Combining Boyle’s law, Charle’s law, Gay-Lussac’s law, and Avagadro’s law gives us the Combined Gas law which can combine into one proportion as-
$V\propto \dfrac{T}{P}$
Removing the proportionality and inserting a constant,
$\dfrac{PV}{T}=C$
This clearly says that as the pressure rises, the temperature also rises and vice versa.
Therefore, we can also say that, $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{P}_{3}}{{V}_{3}}}{{{T}_{3}}}...$

Complete step by step answer:
-Robert Boyle in 1662 formulated a law known as Boyle’s law or Pressure-Volume law which states that ‘the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are kept constant. That is,
$V\propto \dfrac{1}{P}$
Removing the proportionality by inserting a constant-
$PV=C$
This clearly says that when the pressure rises, the volume goes down and when volume rises, the pressure goes down.
Therefore, we can also say that, ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}={{P}_{3}}{{V}_{3}}...$

-Jacques Charles in the year 1787 formulated a law known as Charle’s law which states that ‘the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. That is,
$V\propto T$
Removing the proportionality by inserting a constant,
$\dfrac{V}{T}=C$
This clearly says that as the volume goes up, the temperature also goes up and vice versa.

Therefore, we can also say that, $\dfrac{{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{V}_{3}}}{{{T}_{3}}}...$

-Joseph Gay-Lussac in the year 1808 gave a law known as Gay-Lussac’s law which states that ‘the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. That is,
$P\propto T$
Removing the proportionality by inserting a constant,
$\dfrac{P}{T} = C$
This clearly says that an increase in temperature pressure will also go up and vice versa.

Therefore, we can also say that, $\dfrac{{{P}_{1}}}{{{T}_{1}}} = \dfrac{{{P}_{2}}}{{{T}_{2}}} = \dfrac{{{P}_{3}}}{{{T}_{3}}}...$

-Amedeo Avagadro in 1811 gave a law known as Avagadro’s law, which gives the relationship between volume and amount when pressure and temperature are held constant. This law states that ‘if the amount of gas in a container is increased, the volume increases and vice versa. That is,
$V\propto n$
Removing the proportionality and inserting a constant,
$\dfrac{V}{n} = C$
This means that the volume and amount fraction will always be the same if the pressure and temperature remain constant.

Therefore, we can also say that, $\dfrac{{{V}_{1}}}{{{n}_{1}}} = \dfrac{{{V}_{2}}}{{{n}_{2}}} = \dfrac{{{V}_{3}}}{{{n}_{3}}}...$

-According to the question-
$\begin{aligned} & {{P}_{1}}=(750-100)=650\text{ mm Hg = 650 torr} \\\ & {{V}_{1}}=50\text{ ml} \\\ & {{T}_{1}}=20{}^\circ C=20+273=293K \\\ & {{P}_{2}}=750\text{ torr} \\\ & {{V}_{2}}=? \\\ & {{T}_{2}}=273K \\\ \end{aligned}$
Inserting the values in the formula of Combined Gas Law,
$\dfrac{650\text{ torr}\times \text{50ml}}{293K}=\dfrac{750\text{ torr}\times {{\text{V}}_{2}}}{273K}$ $\Rightarrow {{V}_{2}}=39.84\text{ ml}$

Note: The application of Combined Gas law is that it can be used to explain to the mechanics where pressure, temperature, and volume are affected. For example, in air conditioners and refrigerators, and the formation of clouds.