Imagine that you have a 7.00L gas tank and a 4.00L 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 145atm , 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.
2C2H2 + 5O2 --> 4CO2 + 2H2O
I assume the T is constant.
Use PV = nRT. Plug in P = 145 atm
V = 7.00 L
N = ?
R = 0.08206
T = not given but I would use any convenient number, say 300.
Solve for n of O2.
Use n and the equation to convert to mols C2H2, plug that n back into PV = nRT and the same T and R, along with v = 4.00L and solve for P C2H2.
Post your work if you get stuck.
To determine the pressure at which you should fill the acetylene tank, we need to use the concept of the ideal gas law. The ideal gas law is given by:
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.
In this case, we have two tanks, one filled with oxygen and the other with acetylene. We want to find the pressure in the acetylene tank that will ensure we run out of both gases at the same time.
Since the amount of gas consumed is directly proportional to the number of moles (n) and the pressure (P) of the gas in each tank, we can set up the following equation:
(Volume of oxygen tank) × (Pressure of oxygen) = (Volume of acetylene tank) × (Pressure of acetylene)
To make the equation easier to solve, we can use the ratio between the volumes of the two tanks. The ratio of the oxygen tank volume to the acetylene tank volume is 7.00L / 4.00L = 1.75.
Now, let's substitute the values into the equation:
(7.00L) × (145atm) = (4.00L) × (Pressure of acetylene)
Now, solve for the pressure of acetylene by rearranging the equation:
Pressure of acetylene = (7.00L) × (145atm) / (4.00L)
Pressure of acetylene = 253.75 atm
Therefore, you should fill the acetylene tank to a pressure of 253.75 atm to ensure that you run out of both gases at the same time.