# 9. Suppose two gases in a container have a total pressure of 1.20 atm. What is the pressure of gas B if the partial pressure of gas A is 0.75 atm?

Jan 26, 2017

${P}_{B} = 0.45 \cdot a t m$

#### Explanation:

Dalton's Law of partial pressure holds that in a gaseous mixture, the pressure exerted by a component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressure.

i.e. ${P}_{\text{Total}} = {P}_{1} + {P}_{2} + {P}_{3.} \ldots \ldots . .$

And here ${P}_{\text{Total}} = 1.20 \cdot a t m = {P}_{A} + {P}_{B} = 0.75 \cdot a t m + {P}_{B}$,

i.e. P_B=1.20*atm-0.75*atm=??*atm.

Note that we can use this law to access the number of moles of each gas, given that $P = \frac{n R T}{V}$,

i.e. ${P}_{\text{Total}} = \frac{R T}{V} \left\{{n}_{1} + {n}_{2} + \ldots \ldots \ldots . . {n}_{n}\right\}$, where ${n}_{i}$ is the number of moles of gas $i$.

And we can show that:

${P}_{i} / {P}_{\text{Total"=n_i/n_"Total}}$,

i.e. partial pressure is equivalent to mole fraction.