At a deep-sea station that iss 200. m below the surface of the Pacific Ocean, workeres live in a highly pressurized enviornmetn. how many liters of gas at STP must be compressed on the surface to fill the underwater enviorment wiht 2.00*10^7 L of gast a 20.0 atm? Assume that temperature reamins constant. I did this problem using v1p1=v2p2. and got 400,000,000L. I know that is totally wrong, so can someone please help me with this problem! Thanks:)
well, in round terms, every ten meters adds one atmospher, so 200 meters is about 21 atmosphers, so volume should be increased by a factor of 21 (about).
P1V1=P2V2 is correct, but you need barometric pressure in meters of water. That would be .76*13.2 meters (the 13.2 is the density of mercury) or 10.03m
The pressure at 200 m is 200mwater+atmospheric, or 210mwater
Your problem states that the pressure is 20.0 atm, but that does not include atmospheric pressure above the ocean, it is pressing down also.
10.03*V=210*2E7liter
solve for V . Just looking, your answer is not far off.
Oops, the density of mercury is 13.6, so atmospheric pressure is 10.34 meters water
This means pressure at 200 m is 200m+10.3m, which changes the final answer slightly.
To solve this problem, we can use the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. We can assume that the temperature remains constant for this problem.
First, let's convert the given pressure of 20.0 atm to the appropriate units of pressure. 1 atm is equal to 101.325 kPa, so 20.0 atm is equal to 20.0 × 101.325 kPa = 2026.5 kPa.
Now, we need to find the number of moles of gas required to fill the underwater environment. We know the volume at the deep-sea station is 2.00 * 10^7 L. However, this volume is given at the pressure of 20.0 atm.
To account for the pressure difference, we can use the combined gas law equation: P1V1 / T1 = P2V2 / T2. Rearranging this equation, we have V1 = (P2 / P1) * (V2 * T1) / T2. P1 and T1 can be assumed to be at standard temperature and pressure (STP), which is 1 atm and 273.15 K, respectively.
Substituting the values into the equation:
V1 = (2026.5 kPa / 101.325 kPa) * (2.00 * 10^7 L * 273.15 K) / 273.15 K
= 20 * (2.00 * 10^7 L)
= 4.00 * 10^8 L
Therefore, the correct answer is 4.00 * 10^8 L of gas at STP must be compressed on the surface to fill the underwater environment with 2.00 * 10^7 L of gas at 20.0 atm.