A 280-mL flask contains pure helium at a pressure of 762 torr . A second flask w
ID: 923620 • Letter: A
Question
A 280-mL flask contains pure helium at a pressure of 762 torr . A second flask with a volume of 465 mL contains pure argon at a pressure of 732 torr .
A) If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of helium?
B) If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of argon?
C) If the two flasks are connected through a stopcock and the stopcock is opened, what is the total pressure?
all answers with units of torr
Explanation / Answer
According to Dalton's law of partial pressures the sum of the partial pressures of the individual gases equals that total pressure of the system.
Thereby it obeys Boyle law
So Boyle law says that for an ideal gas the pressure is inversely proportional to the volume.
For He gas :
PV = constant
PV = P'V'
Where
P = initial pressure of He gas = 762 torr
V = initial volume of He flask = 280 mL
P' = final pressure of the He gas in the mixture = ?
V' = Total volume of the flask = volume of the He flask + Volume of the Ar flask
V = 280 mL + 465 mL
= 745 mL
Plug the values we get , P' = (PV) / V'
= (762x280) / 745
= 286.4 torr
Therefore the partial pressure of He is pHe = 286.4 torr
For Ar gas :
PV = constant
PV = P'V'
Where
P = initial pressure of Ar gas = 732 torr
V = initial volume of Ar flask = 465 mL
P' = final pressure of the Ar gas in the mixture = ?
V' = Total volume of the flask = volume of the He flask + Volume of the Ar flask
V = 280 mL + 465 mL
= 745 mL
Plug the values we get , P' = (PV) / V'
= (732x465) / 745
= 456.9 torr
Therefore the partial pressure of Ar is pAr = 456.9 torr
Therefore the total pressure , P = pHe + pAr
= 286.4 + 456.9 torr
= 743.3 torr
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