5 liters of nitrogen under a pressure of 2 atm , 2 l of oxygen under a pressure of 2.5 atm and 3 liters of carbon dioxide under a pressure of 5 atm of the

given:
V1(N2) = 5 lit
P1 = 2 atm
V2(О2) = 2 lit
P2 = 2,5 atm
V3(СО2) = 3 lit
P3 = 5 atm
V4(mixture) = 5 lit
find:
P4 - ? P(N2)  - ?
P(О2)  - ? P(СО2) - ?
Solution:
From the law of Boyle - Mariotte
P1 •V1 = P2 •V2
P_(N₂)=(P_1  •V_1)/V_mixture =(2 atm •5 lit)/(15 lit)= 0,67 atm.
P_( О₂)=(P_2  •V_2)/V_mixture =(2,5 atm •2 lit)/(15 lit)= 0,33 atm.
P_( СО₂)=(P_3  •V_3)/V_mixture =(5 atm •3 lit)/(15 lit)= 1,0 atm.
P4 = P(N2) + P(О2) + P(СО2) = 0,67 atm + 0,33 atm + 1,0 atm = 2 atm.
Answer: P4 = 2 atm,P(N2) = 0,67 atm, P(О2) = 0,33 atm, P(СО2) =1,0 atm.