A sample of ammonia gas at 65.5°C and 0.921 atm has a volume of 15.31 L. What is its volume when the temperature is -15.8°C and its pressure is 0.921 atm?
Use the combined gas law
P2V2/T2=P1V1/T1
temps in K
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample.
The combined gas law is expressed as follows:
(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
Given values:
P1 = 0.921 atm
V1 = 15.31 L
T1 = 65.5°C = 65.5 + 273.15 = 338.65 K
P2 = 0.921 atm
T2 = -15.8°C = -15.8 + 273.15 = 257.35 K
Now, we can substitute these values into the combined gas law and solve for V2.
(0.921 atm * 15.31 L) / (338.65 K) = (0.921 atm * V2) / (257.35 K)
To find V2, we can rearrange the equation:
V2 = (0.921 atm * 15.31 L * 257.35 K) / (338.65 K * 0.921 atm)
Simplifying the equation:
V2 = (3944.24265) / (310.03615)
V2 ≈ 12.714 L
Therefore, the volume of the ammonia gas at -15.8°C and 0.921 atm is approximately 12.714 L.