Gas Law Chemistry

Question

1) A weather balloon at Earth’s surface has a volume of 3.9 L at 296 K and 792 mm Hg. If the balloon is released and the volume reaches 4.07 L at 718 mm Hg, what is the temperature?

2) A flask containing 172 cm3 of hydrogen was collected under a pressure of 21.1 kPa. What pressure would have been required for the volume of the gas to have been 80 cm3 , assuming the same temperature?

Answer 1

a.)

Assuming the air in the balloon remains the same, implying n₁ = n₂, we can set the initial and final conditions equal to each other:

P₁V₁/T₁ = P₂V₂/T₂

Now we will solve for T₂:

T₂ = (P₂V₂)T₁/P₁V₁ = (718 mmHg x 4.07 L) 296 K / ( 3.9 L x 792 mmHg) = 280 K

Note: Since the units of mmHg are used in the initial and final conditions, their units cancel out.

b.)

Because the temperature is constant and the number of molecules of hydrogen are constant, we will set the Pressure and Volume of the initial and final conditions equal to each other as follows:

P₁V₁ = P₂V₂

Solve for P₂ :

P₂ = P₁V₁/V₂ = 21.2 kPa(172 cm3⁾/80 cm³ = 45.6 kPa.

Make sure the answer makes sense. Because the volume is decreasing, the pressure is expected to be higher.