Atmospheric pressure on the peak of Mount Everest can be as low as 0.20 atm. If the volume of an oxygen tank is 10.0 L, at what pressure must the tank be filled so that the gas inside would occupy a volume of 1.2x10^3L at this pressure?
Tried using the P1V1 = P2V2 formula and subbed all the values in, but it did not work. I don`t know what I did wrong.
To solve this problem, you can use the combined gas law equation which takes into account the initial and final conditions of pressure, volume, and temperature.
The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
In this case, the volume and temperature are held constant, so you can simplify the equation to:
P1 / P2 = V1 / V2
Let's plug in the given values:
P1 = 1 atm (standard pressure)
V1 = 10.0 L (initial volume)
V2 = 1.2 * 10^3 L (final volume)
Now, let's solve for P2, which is the pressure we want to find:
P2 = (P1 * V1) / V2
Substituting the values into the equation:
P2 = (1 atm * 10.0 L) / (1.2 * 10^3 L)
P2 = 0.0083 atm
Therefore, the tank must be filled with gas at a pressure of approximately 0.0083 atm to occupy a volume of 1.2 * 10^3 L at the given pressure on the peak of Mount Everest.
To solve this problem using the ideal gas law equation (PV = nRT), you need to convert the given volumes to liters and the pressure to atm.
Given:
Initial volume, V1 = 10.0 L
Final volume, V2 = 1.2x10^3 L
Final pressure, P2 = 0.20 atm
First, let's find the initial pressure, P1.
Using the ideal gas law equation:
(P1)(V1) = (P2)(V2)
Plug in the known values:
(P1)(10.0 L) = (0.20 atm)(1.2x10^3 L)
Now, divide both sides of the equation by V1 to solve for P1:
P1 = (0.2 atm)(1.2x10^3 L) / 10.0 L
P1 = 24 atm
So, the oxygen tank must be filled with a pressure of 24 atm so that the gas inside would occupy a volume of 1.2x10^3 L at the atmospheric pressure on the peak of Mount Everest.