A gas mixture consists of 320 mg of methane, 1.75 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 66.5 Torr. Calculate (a) the volume and (b) the total pressure of the mixture.

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Solution

$P_{n e o n} = x_{n e o n} P$
$P = P_{n e o n} / x_{n e o n}$
$P = \frac{6.67}{\frac{225}{(225 + 320 + 175)}}$

$P = 6.67 / 0.3125 = 21.34$

Assuming ideal gas, we use PV=nRT to calculate for the volume,
PV = nRT
V = nRT / P

n = 225 mg ( 1 mmol / 20.18 mg) + 320 mg ( 1 mmol / 16.05 mg ) + 175 mg ( 1 mmol / 39.95 mg ) = 35.47 mg = 35467.0 g

$V = 35467.0 \times 8.314 \times 300 / 28384 = 4145447.28$

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