A mixture consisting of 2 moles of hydrogen , some number the of moles of oxygen and 1 mol of nitrogen at 20 ° C and a pressure of 4 atm , occupies

given:
ν1(Н2) = 2 mol
ν2(N2) = 1 mol
P = 4 atm
Т = 200С = 293 К
V(mixture) = 40 lit
find:
ν3(О2)  - ?  P(Н2)  - ?
 P(N2)  - ?P(О2) - ?
Solution:
From equation Mendeleev - Clapeyron total number of moles of all the gases making up the mixture
ν_mixture=(P_mixture  •V_mixture)/(R•Т)=(4 atm •40 lit )/(0,082 л•atm/(mol•К) •293 К)=  6,66 mol
The number of moles of oxygen in the mixture is equal to:
ν3(О2)  = νmixture - ν1(Н2)  - ν2(N2)   = 6,66 mol  - 2 mol  - 1 mol = 3,66 mol.
The partial pressure each of gas calculated by the equation :
P_gasа=(P_mixture  •ν_gasa)/ν_mixture .
P_(N₂)=(4 atm •1 mol)/(6,66 mol)= 0,6 atm.
P_( О₂)=(4 atm •3,66 моль)/(6,66 mol)= 2,2 atm.
P_( Н₂)=(4 atm •2 mol)/(6,66 mol)= 1,2 atm.
Answer:  ν3(О2) = 3,66 mol;P(N2) = 0,6 atm, P(О2) = 2,2 atm, P(Н2) =1,2 atm.