Question

a container with 24litres of nitrogen at 2 atm and another container with 12litre of oxygen gas at 2atm the temprature of the both gases is 273.15k. if the both gases are mixed in 10litre container what ar the partial pressure of nitrogen and oxygen in the resulting mixture

Answer 1

Solution:
Given: V1 = 24L P1 = 2atm
V2 = 12L P2 = 2atm
V3 = 10L
T1 = T2 = T3 = T = 273.15K
Question: p(N2), p(O2) - ?
Solution:
Let’s write ideal gas law for nitrogen in the first container and for oxygen in the second container:
P1*V1 = ν(N2)*R*T
P2*V2 = ν(O2)*R*T
That’s why: ν(N2) = P1*V1/(R*T)
ν(O2) = P2*V2/(R*T)
Let’s write ideal gas law for nitrogen and for oxygen in the third container:
p(N2)*V3 = ν(N2)*R*T ⇒ p(N2) = ν(N2)*R*T/V3 = P1*V1/V3 = 2*24/10 = 4.8 (atm)
p(O2)*V3 = ν(O2)*R*T ⇒ p(O2) = ν(O2)*R*T/V3 = P2*V2/V3 = 2*12/10 = 2.4 (atm)
Answer: p(N2) = 4.8atm
p(O2) = 2.4atm