Question #100720

A 250 centimeters cube flask contain Krypton at 500mmHg and 450 centimeters cube flask contain helium at 950mmHg. The content are mixed by opening a stock cup connecting them, assuming all all operations are carried out a constant temperature and the volume of the stock cup is negligible. Calculate the partial pressure of each gases in the mixture.

Expert's answer

**Solution.**

"p(Kr)(part.) = p(Kr) \\times \\frac{V(Kr)}{V(Kr) + V(He)}"

"p(He)(part.) = p(He) \\times \\frac{V(He)}{V(Kr) + V(He)}"

"p(Kr)(part.) = 66661 \\ Pa \\times \\frac{0.00025}{0.00025+0.00045} = 23807.45 \\ Pa"

"p(He)(part.) = 126655.9 \\ Pa \\times \\frac{0.00045}{0.00025+0.00045} = 81421.65 \\ Pa"

**Solution.**

"p(Kr)(part.) = 23807.45 \\ Pa"

"p(He)(part.) = 81421.65 \\ Pa"

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