Question

At ${30}^0$C dry air[$75\%N_2+25\%O_2$] is placed over $H_{2}O$ (I) at $800$ torr combined pressure of all $3$ gases. If pressure is gradually increased isothermally to $1560$ torr, then calculate partial pressure of $O_2$ at high pressure in torr.
$[V.P_{H_{2}O}=40$ torr at ${30}^{o}C]$

Answer 1

We know that $P_{N_2}+P_{O_2}=P_{Total}$ at ${30}^{o}C,800torr$

Total pressure increased at ${30}^{o}C$ to $1560torr$ .

$\Rightarrow$ It was given that= $P_{Total}=1560torr,P_{H_{2}O}=40torr$

$\therefore P_{N_2}+P_{O_2}+P_{H_{2}O}=P_{Total}$

$\Rightarrow P_{N_2}+P_{O_2}+40=1560$

$\Rightarrow P_{N_2}+P_{O_2}=1520$

$\Rightarrow \frac{75}{100}\times 1520=1140=P_{N_2}$

$\Rightarrow \frac{25}{100}\times 1520=380=P_{O_2}$