Question

Imagine that you have a 7.00 L gas tank and a 3.00 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 145 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.

Answer 1

Assume both are operating at 20oC which = 273 + 20 = 293 oK

2C2H2 + 5O2 ===> 4CO2 + 2H2O
What the equation tells you is that you need 5/2 = 2.5 times as much oxygen as you do acetylene.

I was hoping that you would not have to use PV = nRT. No such luck.

P = 145
V = 7.0 L
T = 293^oK

R = 0.08205746
n = ???
n = P*V(R*T)
n = 145 * 7.0 /(0.0821*293)
n = 42.1944 moles of oxygen.

Step Two
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Now get the acylene moles.
For every 5 moles of oxygen you need 2 moles of acetylene
for every 42.1944 moles of oxygen you need x moles of acetylene

5/42.1944 = 2 / x
x = 16.877 moles of acetylene.

n = 16.877
V = 3.00
T = 293
R = 0.0821
P = ???

P = 16.877*0.0821*293/2
P = 203 atmospheres.