Answer to Question #258650 in Chemistry for mymy

Question

Initially, there is a gas at a pressure of 12.0 atm, a volume of 23.0 liters, and a

temperature of 200. K. Then the pressure is raised to 16.0 atm and the temperature is

increased to 250. K. What is the new volume of the gas?

Answer 1

P₁ = 12.0 atm

V₁ = 23.0 L

T₁ = 200. K

P₂ = 16.0 atm

V₂ = unknown

T₂ = 250. K

Solution:

The Combined Gas law can be used:

P₁V₁/T₁ = P₂V₂/T₂

Cross-multiply to clear the fractions:

P₁V₁T₂ = P₂V₂T₁

Divide to isolate V₂:

V₂ = (P₁V₁T₂) / (P₂T₁)

Plug in the numbers and solve for V₂:

V₂ = (12.0 atm × 23.0 L × 250. K) / (16.0 atm × 200. K) = 21.5625 L = 21.6 L

V₂ = 21.6 L

Answer: The new volume of the gas is 21.6 L