Question

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He. What is the total pressure inside the cylinder? What is the mole fraction of O2 and Ar?

Answer 1

The total pressure is 6.7 atm.  The mole fractions of O₂ and Ar are 0.42 and 0.037.

p(Tot) = p(N₂) + p(O₂) + p(Ar) + p(He) = 3.5 atm + 2.8 atm + 0.25 atm + 0.15 atm = 6.7 atm

χ(O₂) = p(O₂)/p(Tot) = 2.8 atm/6.7 atm = 0.42

χ(Ar) = p(Ar)/p(Tot) = 0.25 atm/6.7 atm = 0.037

Answer 2

Answer:

total pressure inside the cylinder is 6.7 atm

molar fraction of $O_2$ is 0.46

mole fraction of Ar is 0.041

Explanation:

To calculate the total pressure inside the vessel we must add the partial pressures of each gas

Partial pressures

$3.5 atm N_2\\ 2.8 atm O_2\\ 0.25 atm Ar\\0.15 atm He$

$P_t= 3.5+2.8+0.25+0.15= 6.7$

total pressure inside the cylinder is 6.7 atm

One way to calculate the molar fraction of a gas is with the following formula

$X_a=\frac{P_a}{P_t}$

$X_O_2=\frac{P_O_2}{P_t}\\ X_O_2=\frac{2.8atm}{6.1atm}\\X_O_2=0.46$

$X_{Ar} =\frac{P_{Ar} }{P_t}\\ X_{Ar} =\frac{0.25atm}{6.1atm}\\X_{Ar} =0.041$