A sample of oxygen gas has a volume of 2.1 L at 3.6 atm pressure and 200 K temperature.
What would the volume of this sample be at 1.0 atm and 400 K?
I need help.
Thanks for helping...
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure = 3.6 atm
V1 = initial volume = 2.1 L
T1 = initial temperature = 200 K
P2 = final pressure = 1.0 atm
T2 = final temperature = 400 K
V2 = final volume (unknown)
Let's plug in the known values into the equation:
(3.6 atm * 2.1 L) / (200 K) = (1.0 atm * V2) / (400 K)
Simplifying the equation, we get:
7.56 atm·L/K = V2 / 400
Now we can solve for V2:
V2 = 7.56 atm·L/K * 400 K
V2 = 3024 L
So, the volume of the sample at 1.0 atm and 400 K would be 3024 L.
To solve this problem, we can use the combined gas law, which gives the relationship between pressure, volume, and temperature of a gas. The equation for the combined gas law is:
(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 given values:
P1 = 3.6 atm
V1 = 2.1 L
T1 = 200 K
P2 = 1.0 atm
T2 = 400 K
Now we can solve for V2 (the final volume). Rearranging the equation, we get:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the values:
V2 = (3.6 atm * 2.1 L * 400 K) / (1.0 atm * 200 K)
Now we can solve it:
V2 = (3024) / (200)
V2 = 15.12 L
Therefore, the volume of the sample of oxygen gas at 1.0 atm and 400 K would be 15.12 L.
Let me know if you need any further assistance!
PV/T is constant, so you want V such that
1.0*V/400 = 3.6*2.1/200