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:

840 cm³

Answer 1

Step-1: Understand the given conditions

• Temperature is constant.
• Volume of air = $560{\mathrm{cm}}^3$
• Pressure of air= $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$
• Pressure to be calculated at $840{\mathrm{cm}}^3$

Step-2: Apply Boyle's law equation to find out the pressure at the given volume of $840{\mathrm{cm}}^3$

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 air at pressure $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$= ${\mathrm{V}}_1$i.e.; $560{\mathrm{cm}}^3$
• Let pressure of air at volume ${\mathrm{V}}_1$= ${\mathrm{P}}_1$i.e.; $76\mathrm{cm}\mathrm{of}\mathrm{mercury}$
• Let volume of air at pressure ${\mathrm{P}}_2$= ${\mathrm{V}}_2$ i.e.; $840{\mathrm{cm}}^3$
• Let pressure of air at volume $840{\mathrm{cm}}^3$= i.e.; ${\mathrm{P}}_2$ (to be calculated)

Substituting the values in the above equation

$$\begin{gathered} 76\times 560={\mathrm{P}}_2\times 840 \\ {\mathrm{P}}_2=\frac{76\times 560}{840} \\ {\mathrm{P}}_2=\frac{42560}{840}=50.67\mathrm{cm}\mathrm{of}\mathrm{Hg} \end{gathered}$$

${\mathrm{P}}_2=50.67\mathrm{cm}\mathrm{of}\mathrm{Hg}$