Imagine that you have a 6.00 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 135 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 your answer with the appropriate units.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Since we want the two tanks to run out of gas at the same time, the number of moles of oxygen gas (n1) should be equal to the number of moles of acetylene gas (n2) when each tank is at its respective pressure.
For the oxygen tank:
P1V1 = n1RT
(135 atm)(6.00 L) = n1(0.0821 L atm/mol K)(298 K)
n1 = 33.276 moles
For the acetylene tank:
P2V2 = n2RT
(P2)(4.50 L) = n2(0.0821 L atm/mol K)(298 K)
n2 = P2(4.50 L) / (0.0821 L atm/mol K)(298 K)
Since we want n1 = n2:
33.276 = P2(4.50) / (0.0821)(298)
P2 = (33.276*0.0821*298) / 4.50
P2 = 44.99 atm
Therefore, you should fill the acetylene tank to a pressure of approximately 45 atm to ensure that you run out of oxygen and acetylene at the same time.