Question

112 cm³ of nitrogen is collected at a pressure of 720 mm of mercury. If the temperature remains constant, find its volume when pressure is:

The pressure drops by 40 % of initial pressure.

Answer 1

Step-1: Understand the given conditions

• Temperature is constant.
• Volume of Nitrogen = $112{\mathrm{cm}}^3$
• Pressure of Nitrogen= $720\mathrm{mm}\mathrm{of}\mathrm{mercury}$
• Volume to be calculated when final pressure drops by 40 % of initial pressure

Note that mm of mercury can also be written as mm of Hg (as Hg is a symbol of Mercury)

Step-2: Calculate the pressure after 40 % drop in the initial pressure

Initial pressure ${\mathrm{P}}_1$=$720\mathrm{mm}\mathrm{of}\mathrm{mercury}$

Drop in the initial pressure= $\frac{40}{100}{\mathrm{P}}_1$

Final pressure ${\mathrm{P}}_2$ = ${\mathrm{P}}_1-\frac{40}{100}{\mathrm{P}}_1$

${\mathrm{P}}_2=720-\frac{40}{100}\times 720\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{P}}_2=720-288\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{P}}_2=432\mathrm{mm}\mathrm{of}\mathrm{Hg}$

Step-3: Apply Boyle's law equation to find out the volume at the pressure of $432\mathrm{mm}\mathrm{of}\mathrm{mercury}$

Boyle's' law: According to Boyle's law, at constant temperature, pressure is inversely proportional to volume for a given mass of the gas.
or PV= constant

Therefore, for two different conditions i.e., 1 and 2, Boyle's' law can be written as:
${\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{P}}_2{\mathrm{V}}_2$

• Let volume of Nitrogen at pressure 720 mm of mercury= ${\mathrm{V}}_1$i.e.; $112{\mathrm{cm}}^3$
• Let pressure of Nitrogen at volume ${\mathrm{V}}_1$= ${\mathrm{P}}_1$i.e.; $720\mathrm{mm}\mathrm{of}\mathrm{mercury}$
• Let pressure of Nitrogen at volume ${\mathrm{V}}_2$= ${\mathrm{P}}_2$i.e.; $432\mathrm{mm}\mathrm{of}\mathrm{mercury}$
• Let volume of Nitrogen at pressure $432\mathrm{mm}\mathrm{of}\mathrm{mercury}$= ${\mathrm{V}}_2$ (to be calculated)

Substituting the values in the above equation

$$\begin{gathered} 720\times 112=432\times {\mathrm{V}}_2 \\ {\mathrm{V}}_2=\frac{720\times 112}{432} \\ {\mathrm{V}}_2=\frac{80640}{432}=186.67{\mathrm{cm}}^3 \end{gathered}$$

${\mathrm{V}}_2=186.67{\mathrm{cm}}^3$