Question

$2500{\mathrm{cm}}^3$ of hydrogen is taken at S.T.P. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume did hydrogen occupy now?

Answer 1

Initial conditions - ${\mathrm{P}}_1=760\mathrm{mm}\mathrm{of}\mathrm{Hg},{\mathrm{V}}_1=2500{\mathrm{cm}}^3,{\mathrm{T}}_1=273\mathrm{K}$,

Final conditions - ${\mathrm{P}}_2=760+760\times \frac{5}{2}=2660\mathrm{mm}\mathrm{of}\mathrm{Hg},{\mathrm{V}}_2=?,{\mathrm{T}}_2=273\mathrm{K}$

Using the Ideal gas equation,

$\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$

$\frac{760\mathrm{mm}\mathrm{of}\mathrm{Hf}\times 2500{\mathrm{cm}}^3}{273\mathrm{K}}=\frac{2660\mathrm{mm}\mathrm{of}\mathrm{Hg}\times {\mathrm{V}}_2}{273\mathrm{K}}$

$\frac{760\mathrm{mm}\mathrm{of}\mathrm{Hg}\times 2500{\mathrm{cm}}^3}{2660\mathrm{mm}\mathrm{of}\mathrm{Hg}}={\mathrm{V}}_2$

Hence, ${\mathrm{V}}_2=714{\mathrm{cm}}^3$