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3 years ago

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A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

pressure of 727 torrs. If the two flasks are connected through a stopcock and the stopcock is opened, what is the partial pressure of helium?

1 answer:

Nadya [2.5K]3 years ago

5 0

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

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$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

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Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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