Question

a gas that has a volume of 28 liters a temperature of 95 celsius and an unknown pressure has its volume increased go 36 liters and its temperature decreased to 35 celsius. if i measure the pressure after the change to be 8.5 atm what was the original pressure of the gas?

Answer 1

Using the combined gas laws (Boyle's law, Charles law and Gay-Lussac's law), we have the formula

(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂

where

P₁ = initial pressure = unknown pressure

V₁ = initial volume 28 L

T₂ = initial temperature = 95 ^oC = (95 + 273) K = 368 K

P₂ = final pressure = 8.5 atm

V₂ = final volume = 36 L

T₂ = final temperature = 35 ^oC = (35 + 273) K = 308 K

Substituting these values in above formula,

(P₁ * 28 L) / (368 K) = (8.5 atm * 36 L) / (308 K)

Rearranging,

P₁ = (8.5 atm) * (36 L / 28 L) * (368 K / 308 K)

P₁ = (8.5 atm) * (1.256) * (1.195)

P₁ = 13.06 atm

original pressure of gas is 13 atm (with correct significant figures)