A sample of gas at 47 C and 1.03 atm occupies a volume of 2.20 L. What volume would this gas occupy at 107 C and 0.789 atm?
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 = initial pressure of the gas (in atm)
V1 = initial volume of the gas (in liters)
T1 = initial temperature of the gas (in Kelvin)
P2 = final pressure of the gas (in atm)
V2 = final volume of the gas (unknown)
T2 = final temperature of the gas (in Kelvin)
First, let's convert the given temperatures from Celsius to Kelvin by adding 273.15:
Initial temperature (T1) = 47 + 273.15 Kelvin = 320.15 K
Final temperature (T2) = 107 + 273.15 Kelvin = 380.15 K
Now, let's plug in the values into the combined gas law equation and solve for V2:
(1.03 atm * 2.20 L) / 320.15 K = (0.789 atm * V2) / 380.15 K
Now, let's cross-multiply and solve for V2:
(1.03 atm * 2.20 L) * 380.15 K = (0.789 atm * V2) * 320.15 K
(1.03 * 2.20 * 380.15) / (0.789 * 320.15) = V2
V2 ≈ 5.148 L
Therefore, the gas would occupy approximately 5.148 liters at 107°C and 0.789 atm.
To solve this problem, we can use the combined gas law equation:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
Now, let's plug in the given values:
P₁ = 1.03 atm
V₁ = 2.20 L
T₁ = 47°C = 47 + 273.15 = 320.15 K
P₂ = 0.789 atm
T₂ = 107°C = 107 + 273.15 = 380.15 K
Using the formula, we can solve for V₂:
(1.03 atm * 2.20 L) / (320.15 K) = (0.789 atm * V₂) / (380.15 K)
Simplifying the equation:
(2.266) / (320.15) = (0.789) / (380.15) * V₂
V₂ = (2.266 / 320.15) * (380.15 / 0.789)
V₂ = 2.266 * 480.77 / 0.789
V₂ = 1383.62 / 0.789
V₂ ≈ 1754.45
Therefore, the gas would occupy approximately 1754.45 L at 107°C and 0.789 atm.