22. A 6.2L sample of N2 at 738Torr is mixed (at constant temperature) with 15.2L of O2 at 325Torr. The gaseou

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22. A 6.2L sample of N2 at 738Torr is mixed (at constant temperature) with 15.2L of O2 at 325Torr. The gaseous mixture is placed in a 12.0L container. (a) What is the pressure of the mixture? (b) What is the composition of the mixture in mole fractions?


The pressure of the mixture is dependent on the number of moles of each gas.
The Ideal gas law equation:
P * V = n* R * T
Pressure in atmospheres * Volume in liters = number of moles * 22.4/273 * Temperature in ˚K
(The ideal gas constant = 22.4/273 = 0.08205)

1 atmosphere = 760 torr
The Ideal gas law equation (with pressure in Torr)
P/760 * V = (number of moles) * 0.08205 * T
Since the temperature is constant, let the temperature = T

For N2:
738/760 * 6.2 = (Number of moles of N2) * 0.08205 * T

For O2:
325/760 * 15.2 = (Number of moles of O2) * 0.08205 * T

The ratio of moles of N2 / O2 = (738/760 * 6.2) / (325/760 * 15.2) = 0.926


For N2:
738/760 * 6.2 = (Number of moles of N2) * 0.08205 * T
Number of moles of N2 = 73.4 ÷ T
For O2:
325/760 * 15.2 = (Number of moles of O2) * 0.08205 * T
Number of moles of O2 = 79.2 ÷ T
Total number of moles = (73.4+ 79.2) ÷ T = 152.6 ÷ T

P/760 * 12 = (152.6 ÷ T)* 0.08205 * T
P/760 * 12 = 12.521
Multiply both sides by 760
P * 12 = 760 * 12.521
Divide both sides by 12
P = 760 * 12.521 ÷ 12
P = 793 Torr