A 3.2 liter sample of gas is at 40.0oC and 1.0 atmosphere of pressure. If the temperature decreases to 20oC and the pressure decreases to 0.60 atmospheres, what is the new volume in liters?
(P1V1/T1)=(P2V2/T2)
T must be in Kelvin.
To solve this problem, we can use the combined gas law, which is expressed as:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
where:
P₁ = initial pressure (1.0 atm)
V₁ = initial volume (3.2 L)
T₁ = initial temperature (40.0°C + 273.15 = 313.15 K)
P₂ = final pressure (0.60 atm)
V₂ = final volume (unknown)
T₂ = final temperature (20.0°C + 273.15 = 293.15 K)
Plugging in the known values, we can rearrange the equation to solve for V₂:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
(1.0 atm * 3.2 L) / (313.15 K) = (0.60 atm * V₂) / (293.15 K)
Now, we can solve for V₂:
V₂ = [(1.0 atm * 3.2 L) / (313.15 K)] * (293.15 K / 0.60 atm)
V₂ ≈ 3.0 liters (rounded to one decimal place)
Therefore, the new volume of the gas is approximately 3.0 liters.