At a deep-sea station 200. m below the surface of the Pacific Ocean, workers live in a highly pressurized environment. How much gas at STP must be compressed on the surface to fill the underwater environment with 2 × 107 L of gas at 22 atm? Assume that temperature remains constant.
Answer in units of L.
I'm not understanding how to do the problem. Any and all help is appreciated!
Thanks in advance!
Isn't this just a P1V1 = P2V2 problem?
You have V1 liters gas at STP and you want it to go to 2E7 L at P of 22 atm. You want to solve for V1.
I probably should have stated that I know this is a P1V1=P2V2. However every time I try an answer I get it wrong. I don't know if I'm putting my numbers in wrong but so far nothing has been correct.
To solve this problem, we need to use the ideal gas law, which states that the pressure, volume, and temperature of a gas are related by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this case, the pressure and volume are given. We need to solve for the number of moles of gas, n. Since the temperature remains constant, we can leave it out of the equation.
First, we need to convert the pressure from atm to Pa (Pascals) so that the units are consistent with the ideal gas constant. The conversion factor is 1 atm = 101325 Pa.
Given:
Pressure = 22 atm
Volume = 2 × 10^7 L
Converting pressure to Pascals:
Pressure = 22 atm × 101325 Pa/atm = 2,227,150 Pa
Next, we need to rearrange the ideal gas law equation to solve for the number of moles of gas, n:
n = PV / RT
Since the temperature is constant, we can simplify the equation to:
n = (PV) / RT
Now we can substitute the given values into the equation:
n = (2,227,150 Pa) × (2 × 10^7 L) / (8.314 J/(mol·K) × T)
Since the temperature is not given, we can assume it to be room temperature, which is typically around 298 K.
n = (2,227,150 Pa) × (2 × 10^7 L) / (8.314 J/(mol·K) × 298 K)
Now we can solve for the number of moles of gas, n:
n = (2,227,150 × 2 × 10^7) / (8.314 × 298) mol
Finally, we need to convert the number of moles of gas to liters. This can be done by multiplying the number of moles by the molar volume at STP, which is approximately 22.4 L/mol.
Volume of gas at STP = n × 22.4 L/mol
Substituting the value of n we found:
Volume of gas at STP = [(2,227,150 × 2 × 10^7) / (8.314 × 298)] × 22.4 L
Now we can calculate the answer:
Volume of gas at STP = 1.923 × 10^9 L
Therefore, approximately 1.923 × 10^9 liters of gas at standard temperature and pressure (STP) must be compressed on the surface to fill the underwater environment with 2 × 10^7 L of gas at 22 atm.