If 4.91 g of Ar are added to 1.47 atm of He in a 2.00 L cylinder at 27.0 °C, what is the total pressure of the resulting gaseous mixture?

[匿名]

2016-04-26 6:25 am

Could someone walk me through how to find this answer? The "hint" for the question refers to Dalton's law of partial pressures but I can't figure it out. Thanks for any help :)

回答 (1)

micatkie

2016-04-26 6:39 am

✔ 最佳答案

The molar mass of Ar is 39.95 g/mol.

Refer to the Dalton's law of partial pressures.
The addition of Ar does not affect the partial pressure of He, i.e.
P(He) = 1.47 atm

Consider Ar added :
Partial pressure, P(Ar) = ? atm
Volume, V = 2.00 L
No. of moles, n(Ar) = 4.91/39.95 mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 27) K = 300 K

P(Ar) V = n(Ar) R T
P(Ar) = n(Ar) R T / V
P(Ar) = (4.91/39.95) × 0.0821 × 300 / 2.00 atm
P(Ar) = 1.51 atm

Total pressure
= P(Ar) + P(He)
= (1.51 + 1.47) atm
= 2.98 atm

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