Question

An ideal gas is sealed inside a container with an initial volume of 1 m3 ; the gas has an initial temperature of 300 K and a pressure of 1 atmosphere. If the temperature of the gas is increased to 450 K and the volume of the container decreases to 0.75 m3 , what is the new pressure of the ideal gas?

(a) 2.0 atm
(b) 0.89 atm
(c) 1.1 atm
(d) 0.50 atm
(e) 2.5 atm

Answer 1

The new pressure of the ideal gas is approximately 2.0 atm.

Step-by-step explanation:

To find the new pressure of the ideal gas, we can use the combined gas law, which states that P₁V₁/T₁ = P₂V₂/T₂. In this case, the initial pressure (P₁) is 1 atmosphere, the initial volume (V₁) is 1 m³, the initial temperature (T₁) is 300 K, the final temperature (T₂) is 450 K, and the final volume (V₂) is 0.75 m³. Plugging in these values, we can solve for the final pressure (P₂).

Substituting the known values:

P₁V₁/T₁ = P₂V₂/T₂

1 * 1 / 300 = P₂ * 0.75 / 450

0.00333 ≈ P₂/600

P₂ ≈ 0.00333 * 600

P₂ ≈ 2.0 atm

Therefore, the new pressure of the ideal gas is approximately 2.0 atm.