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:

Volume increases by 25%

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 when final volume increases by 25 % of initial volume

Step-2: Calculate the volume after 25 % increase in the initial volume

Initial volume ${\mathrm{V}}_1$=$560{\mathrm{cm}}^3$

Increase in the initial volume= $\frac{25}{100}{\mathrm{V}}_1$

Final volume ${\mathrm{V}}_2$ = ${\mathrm{V}}_1+\frac{25}{100}{\mathrm{V}}_1$

${\mathrm{V}}_2=560+\frac{25}{100}\times 560{\mathrm{cm}}^3$

${\mathrm{V}}_2=560+140\mathrm{mm}\mathrm{of}\mathrm{Hg}$

${\mathrm{V}}_2=700{\mathrm{cm}}^3$

Step-3: Apply Boyle's law equation to find out the pressure at the given volume of $700{\mathrm{cm}}^3$

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.; $700{\mathrm{cm}}^3$ (calculated in Step-2)
• Let pressure of air at volume $700{\mathrm{cm}}^3$= i.e.; ${\mathrm{P}}_2$ (to be calculated)

Substituting the values in the above equation

$$\begin{gathered} 76\times 560={\mathrm{P}}_2\times 700 \\ {\mathrm{P}}_2=\frac{76\times 560}{700} \\ {\mathrm{P}}_2=\frac{42560}{700}=60.8\mathrm{cm}\mathrm{of}\mathrm{Hg} \end{gathered}$$

${\mathrm{P}}_2=60.8\mathrm{cm}\mathrm{of}\mathrm{Hg}$