A vessel of volume `1660 cm^(3)` contains `0.1 "mole"` of oxygen and `0.2 "mole"` of nitrogen. If the temperature of the mixture is `300 K`, find its pressure.

Options

A. `2.5 xx 10^(5) Pa`
B. `1.5 xx 10^(5) Pa`
C. `4.5 xx 10^(5) Pa`
D. `6.5 xx 10^(5) Pa`

Answer

C

Explanation

We have `pV = nRT implies p = (nRT)/(V)`
The partial pressure due to oxygen is
`p_(1) = ((0.1 mol)(8.3 J//mol-K)(300K))/((1660xx10^(-6)m^(3)) = 1.5 xx 10^(5)Pa`
Similarly, the partial pressure due to nitrogen is
`P_(2) = 3.0 xx 10^(5) Pa`
The total pressure in the vessel is `p = p_(1)+p_(2)`
= `(1.5+3.0)xx10^(5)Pa = 4.5 xx 10^(5) Pa`.