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Question
Solve the following.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If the pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?
Solution
Given:
mO2 = 70.6 g,
mNe = 167.5 g,
PTotal = 25 bar
To find: Partial pressure of each gas
Formula: P1 = x1 × PTotal
Calculation:
Determine number of moles (n) of each gas using formula:
n = `"m"/"M"`
∴ nO2 = `(70.6 "g")/(32 "g mol"^-1)` = 2.206 mol
nNe = `(167.5 "g")/(20 "g mol"^-1)` = 8.375 mol
Determine the mole fraction of each gas using the formula:
x = `"n"/("n"_"Total")`
`""^"x""O"_2=(""^"n""O"_2)/(""^"n""O"_2+"n"_"Ne")=(2.206 "mol")/((2.206+8.375) "mol")=(2.206 "mol")/(10.581 "mol")=0.208`
xNe = `("n"_"Ne")/("n"_"Total")=(8.375 "mol")/(10.581 "mol")` = 0.792
Calculate the partial pressure of each gas:
PO2 = xO2 × PTotal = 0.208 × 25 bar = 5.2 bar
PNe = xNe × PTotal = 0.792 × 25 bar = 19.8 bar
The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.
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