A sample of gas at 47°C and 1.03 atm occupies a volume of 2.20L. 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
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Let's plug in the values given in the question:
P1 = 1.03 atm
V1 = 2.20 L
T1 = 47°C (which needs to be converted to Kelvin)
P2 = 0.789 atm
T2 = 107°C (which also needs to be converted to Kelvin)
Converting temperatures to Kelvin:
T1(K) = 47 + 273.15 = 320.15 K
T2(K) = 107 + 273.15 = 380.15 K
Now we can substitute these values into the combined gas law equation:
(1.03 atm * 2.20 L) / (320.15 K) = (0.789 atm * V2) / (380.15 K)
Simplifying the equation:
(1.03 * 2.20) / (320.15) = (0.789 * V2) / (380.15)
Solving for V2:
V2 = (1.03 * 2.20 * 380.15) / (0.789 * 320.15)
V2 ≈ 2.689 L
Therefore, the gas would occupy a volume of approximately 2.689 L at 107°C and 0.789 atm.
(P1V1/T1) = (P2V2/T2)
Post your work if you get stuck.