A quantity of gas has a volume of 121 L when confined under a pressure of 2.50 atm at a temperature of 20.0 C. At what pressure will its volume be 30.0 L at 25.0 C?
(P1V1/T1) = (P2V2/T2)
At a constant pressure, the volume of a gas is increased from 150.0 L to 300.0 L by heating it. If the original temperature of the gas was 20.0 degrees C, what will its final temperature (in Kelvin) be?
To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
To solve for the unknown pressure, we need to keep the number of moles constant. Therefore, we can rewrite the equation as:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
P₁ = initial pressure = 2.50 atm
V₁ = initial volume = 121 L
T₁ = initial temperature = 20.0 C = 20.0 + 273.15 K (convert to Kelvin)
P₂ = final pressure (unknown)
V₂ = final volume = 30.0 L
T₂ = final temperature = 25.0 C = 25.0 + 273.15 K (convert to Kelvin)
Substituting the given values into the equation, we have:
(2.50 atm * 121 L) / (20.0 + 273.15 K) = (P₂ * 30.0 L) / (25.0 + 273.15 K)
Now, let's solve for P₂:
2.50 atm * 121 L * (25.0 + 273.15 K) = P₂ * 30.0 L * (20.0 + 273.15 K)
To isolate P₂, divide both sides of the equation by (30.0 L * (20.0 + 273.15 K)):
P₂ = (2.50 atm * 121 L * (25.0 + 273.15 K)) / (30.0 L * (20.0 + 273.15 K))
Calculating this expression will give you the value of the final pressure P₂.