Oxygen Tank Problem/Ideal Gas Law - solving for pressure

In summary, the new pressure in the holding tank at 300 K after transferring the contents of the compressed gas cylinder is 223 atm.
  • #1
~anne~
1
0
Problem: A cylinder of compressed Oxygen is carried on a spacecraft headed for Mars. The
compressed gas cylinder has a volume of 17000 L and is filled to a pressure of 157
atm at 234 K. The maximum pressure the cylinder can hold is 1000 atm.

Question: The contents of the cylinder are then entirely transferred to a partially filled holding
tank of volume 10,000 L, originally at pressure 22 atm and 300 K. What is the new
pressure in the holding tank at 300 K?

Relevant equation: PV=nRT

Attempt at Solution (hopefully it's quite clear enough to tell for where the error is):

I first solved for number of moles in the tank
--> n=PV/RT= (2228977 Pa)(10 m^3) / (8.31 J/molK)(300 K) = 8940.942639 moles

I then solved for the previous number of moles from the cylinder with 17000 L
---> n= PV/RT= (101325 Pa)(0.01764 m^3) / (8.31 J/molK)(294 K) = 0.7315884477 moles

I added both of these to get the total number of moles in the new tank and then solved for the new pressure
-----> P=nRT/V = (8941.674228 moles)(8.31 J/molK)(300K) / (10 m^3) = 2229159.385 Pa

This is the same as before in atm - 22 atm. I don't understand how I'm getting the same answer. Is there an error in how I calculated the initial number of moles before the transfer? Please specify. Thanks in advance!

SORRY! Seems there actually was a mistake in my initial calculation (typing it out here helped me reflect). I was confusing the calculations from two different problems. I redid the problem correctly and did it right this time.
 
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  • #2
I first solved for number of moles in the tank --> n=PV/RT= (101325 Pa)(10 m^3) / (8.31 J/molK)(300 K) = 3739.108531 molesI then solved for the additional number of moles from the cylinder with 17000 L---> n= PV/RT= (157 atm)(0.01764 m^3) / (8.31 J/molK)(294 K) = 11.71808721 molesI added both of these to get the total number of moles in the new tank and then solved for the new pressure-----> P=nRT/V = (3751.82644 moles)(8.31 J/molK)(300K) / (10 m^3) = 2230354.328 PaThis is the same as before in atm - 223 atm.
 

Related to Oxygen Tank Problem/Ideal Gas Law - solving for pressure

1. What is the Oxygen Tank Problem?

The Oxygen Tank Problem is a physics problem that involves using the Ideal Gas Law to calculate the pressure of a gas inside a container. It is commonly used in introductory chemistry and physics courses.

2. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and amount of gas in a closed system. It is written as PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the universal gas constant, and T is temperature in Kelvin.

3. How do you solve for pressure in the Oxygen Tank Problem?

To solve for pressure in the Oxygen Tank Problem, you need to have values for the other variables in the Ideal Gas Law equation. Once you have those values, you can rearrange the equation to solve for pressure by dividing both sides by volume (P = nRT/V).

4. What are the units for pressure, volume, temperature, and moles in the Ideal Gas Law equation?

The units for pressure are typically in kilopascals (kPa) or atmospheres (atm). Volume is measured in liters (L) or cubic meters (m^3). Temperature is measured in Kelvin (K). Moles are a unitless quantity, but they must be consistent with the units used for pressure, volume, and temperature.

5. How does the Ideal Gas Law relate to real gases?

The Ideal Gas Law is an idealized equation that assumes certain conditions, such as low pressure and high temperature. In real gases, there are deviations from this ideal behavior, but the equation is still a useful approximation in many situations. Factors such as intermolecular forces and molecular size can affect the behavior of real gases.

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