consider a sample of gas in a container on a comfortable spring day. the Celsius temperature suddenly doubles, and you transfer the gas to a container with twice the volume of the first container. If the original pressure was 12 atm, what is a good estimate for the new pressure?

6.6 atm

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

Let P₂ be the new pressure

So, P₂ = P₁V₁T₂/V₂T₁

Since V₂ = 2V₁, P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius

T₂ = 273 + 2t (since its Celsius temperature doubles).

Substituting these values into the equation for P₂, we have

P₂ = P₁V₁ (273 + 2t) / 2V₁ (273 + t)

P₂ = 12 (273 + 2t) / [2 (273 + t) ]

P₂ = 6 (273 + 2t) / (273 + t) ]

assume t = 30 °C on a comfortable spring day

P₂ = 6 (273 + 2 (30)) / (273 + 30) ]

P₂ = 6 (273 + 60)) / (273 + 30) ]

P₂ = 6 (333)) / (303) ]

P₂ = 6.6 atm