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A 2.48 L sample of He at 739 torr is mixed with 6.08 L of Ar...

A 2.48 L sample of He at 739 torr is mixed with 6.08 L of Ar at 325 torr both at 25oC is placed in a 4.80 L container at 25oC.

What is the total pressure of the mixture? (units of torr are assumed)

(Atomic weights: He = 4.00, Ar = 39.95)

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Answer #1

Step 1: find partial pressure of He

Given:

Pi = 739 torr

Vi = 2.48 L

Vf = 4.80 L

use:

Pi*Vi = Pf*Vf

739 torr * 2.48 L = Pf * 4.8 L

Pf = 382 torr

Step 2: find partial pressure of Ar

Given:

Pi = 325 torr

Vi = 6.08 L

Vf = 4.80 L

use:

Pi*Vi = Pf*Vf

325 torr * 6.08 L = Pf * 4.8 L

Pf = 412 torr

Step 3: find total pressure

total pressure = p(He) + p(Ar)

= 382 torr + 412 torr

= 794 torr

Answer: 794 torr

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