Imagine that you have a 6.00L gas tank and a 3.50L gas tank. You need to fill on
ID: 788343 • Letter: I
Question
Imagine that you have a 6.00L gas tank and a 3.50L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115atm to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
2C 2 H 2 (g)+5O 2 (g)?4CO 2 (g)+2H 2 O(g)2C2H2(g)+5O2(g) -> 4CO2(g)+2H2O(g)2C 2 H 2 (g)+5O 2 (g)?4CO 2 (g)+2H 2 O(g)
Imagine that you have a 6.00L gas tank and a 3.50L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115atm to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases. Acetylene torches are used for welding. These torches use a mixture of acetylene gas,C2H2 and oxygen gas O2 to produce the following combustion reaction: 2C 2 H 2 (g)+5O 2 (g)?4CO 2 (g)+2H 2 O(g)2C2H2(g)+5O2(g) -> 4CO2(g)+2H2O(g)2C 2 H 2 (g)+5O 2 (g)?4CO 2 (g)+2H 2 O(g)Explanation / Answer
The balanced equation is C2H2 + 5/2O2 = 2CO2 + H2O
Therefore, you need 2.5 times as much oxygen as acetylene to fully react both the two gases.
Looking at the ideal gas equation PV=nRT has n = PV/RT
Assuming the temperature is constant for the two gases, the product of the pressure and volume is directly related to the number of moles.
The amount of oxygen you have is 6 L * 115 atm = 690Latm
For the acetylene as said above, you need only 1 / 2.5 times as much or 690 L atm / 2.5 = 276 L atm.
The volume of the smaller tank is 3.5 L so the pressure you need = L atm / 3.5 L = 78.857 atm.