Answer to Question #100720 in Inorganic Chemistry for Olaibi Gbenga

Question

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.

Answer 1

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.

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

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