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

this a false

explanation: