Question

2. A sample of helium gas has a volume of 200.0 mL at 0.960 atm. What pressure, in atmospheres, is needed to reduce the volume at constant tem-perature to 50.0 mL?

Answer 1

Answer:

3.840 atm

Explanation:

We use Boyle's law, which relates the pressure (P) and the volume (V) of a gas at a constant temperature. The change from initial P and V (P₁ and V₁) to final P and V (P₂ and V₂) is expressed as:

P₁V₁ = P₂V₂

We have the following data:

P₁= 0.960 atm

V₁= 200.0 mL

V₂ = 50.0 mL

Thus, we introduce the data in the equation and calculate the final pressure P₂:

P₂ = P₁V₁/V₂= (0.960 atm x 200.0 mL)/50.0 mL = 3.840 atm

Therefore, a pressure of 3.840 atm is needed to reduce the volume of the gas from 200.0 mL to 50.0 mL.