Question

560 cm^3 of air is at preessure of 76 cm of mercury.Find the pressure of air(temp. remains constant) ,when its volume is

1. 210cm^3
2. 840cm^3
3. volume increases by 25%

Answer 1

As the temperature remains constant so, let us use Boyle's law to find the required pressure.
Initial pressure, P₁ = 76 cm of Hg
Initial Volume, V₁ = 560 cm³
By using Boyle's law,
$$\begin{gathered} 1.P_1{V}_1=P_2{V}_2 \\ \Rightarrow 76\times 560=P_2\times 210 \\ \Rightarrow P_2=\frac{76\times 560}{210}=202.67\mathrm{cm}\mathrm{of}\mathrm{Hg} \\ 2.P_1{V}_1=P_3{V}_3 \\ \Rightarrow 76\times 560=P_3\times 8400 \\ \Rightarrow P_3=\frac{76\times 560}{840}=50.67\mathrm{cm}\mathrm{of}\mathrm{Hg} \\ 3.P_1{V}_1=P_4{V}_4 \\ \Rightarrow 76\times 560=P_4\times 210 \\ \Rightarrow P_4=\frac{76\times 560}{1.25\times 210}=162.13cmofHg \end{gathered}$$