   Chapter 10, Problem 68AP

Chapter
Section
Textbook Problem

Two small containers, each with a volume of 1.00 × 102cm3, contain helium gas at 0°C and 1.00 atm pressure. The two containers are joined by a small open tube of negligible volume, allowing gas to flow from one container to the other. What common pressure will exist in the two containers if the temperature of one container is raised to 1.00 × 102 °C while the other container is kept at 0°C?

To determine
The final pressure that will exist in the two containers.

Explanation

Given info: The initial pressure ( P0 ) is 1.00 atm. The temperature is changed from T1=0οC to T2=(1.00×102)οC .

From Ideal gas equation, the number of moles of gas is,

n=PVRT

• P is the final pressure.
• V is the volume.
• R is the gas constant.
• T is the temperature.

The total mass of the gas remains constant. Therefore, the initial and final number of moles of the gas is the same.

PVRT1+PVRT2=P0VRT1+P0VRT1

On Re-arranging,

P=2P0T2T1+T2

Substitute 1.00 atm for P0 , (1

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