Question

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:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

imagine that you have a 7.00 L gas tank and a 4.50 L 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 155 atm , 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.

Express answer with appropriate units.

Answer 1

First thing to understand is how you can manupulate the Ideal Gas Law to your needs. Since PV = nRT, at a constant temperature, PV/n = RT = constant. That means

P1V1/n1 = P2V2/n2!

If we say that subscript 1 = oxygen and subscript 2 = acetylene, then for acetylene, P2 = (P1V1n2) / (V2n1).

Next, we realize from the reaction equation given above that 2 moles of acetylene (n2) react with 5 moles of oxygen (n1). We can plug that right into our equation for P2, along with the rest of the conditions given in the question:

P2 = (P1V1n2) / (V2n1) = [(155 atm)*(7L)*(2 moles)] / [(4.50 L)*(5 moles)] = 96.44 atm