Question

560 cm³ of air is at a pressure of 76 cm of mercury. Find the pressure of air (assuming temperature remains constant), when its volume is:

210 cm³

Answer 1

Step-1: Given data

• Temperature is constant.
• The volume of air = $560{\mathrm{cm}}^3$
• Pressure of air= $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$
• Pressure to be calculated at $210{\mathrm{cm}}^3$

Step-2: Apply Boyle's law equation to find out the pressure at the given volume of $210{\mathrm{cm}}^3$

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

Therefore, according to Boyle's law:
${\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{P}}_2{\mathrm{V}}_2$

• Let the volume of air at pressure $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$= ${\mathrm{V}}_1$ = $560{\mathrm{cm}}^3$
• Let the pressure of air at volume ${\mathrm{V}}_1$= ${\mathrm{P}}_1$ = $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$
• Let the volume of air at pressure ${\mathrm{P}}_2$= ${\mathrm{V}}_2$ = $210{\mathrm{cm}}^3$
• Let the pressure of air at volume $210{\mathrm{cm}}^3$= ${\mathrm{P}}_2$ (to be calculated)

Substituting the values in the above equation

$$\begin{gathered} 76\times 560={\mathrm{P}}_2\times 210 \\ {\mathrm{P}}_2=\frac{76\times 560}{210} \\ {\mathrm{P}}_2=\frac{42560}{210}=202.67\mathrm{cm}\mathrm{of}\mathrm{Hg} \end{gathered}$$

${\mathrm{P}}_2=202.67\mathrm{cm}\mathrm{of}\mathrm{Hg}$

Hence, pressure (P2) will be 202.67 cm of Hg