Imagine that you have a 5.00 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 135 , 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.
Equation is 2C2H2 + 5O2 ==> 4CO2 + 2H2O for burning acetylene.
Use PV = nRT
Calculate n, number of moles of oxygen in the 5.00 L tank. Use any T that's convenient for both calculations (this one and the last one).
Next, convert moles oxygen to moles acetylene. Use the equation for burning acetylene in oxygen.
Now use PV = nRT again, this time for the 3.00 L tank with the new n calculated for acetylene. Post your work if you get stuck.
5.8
To determine the pressure to which you should fill the acetylene tank, you need to consider the ratio of the volumes of the two gas tanks.
Let's denote the pressure to which you fill the acetylene tank as P_acetylene.
First, we find the volume ratio of the two tanks:
Volume ratio = Volume of oxygen tank / Volume of acetylene tank
Volume ratio = 5.00 L / 3.00 L
Volume ratio = 1.67
Next, we need to find the pressure ratio of the two gases at the same temperature:
Pressure ratio = P_acetylene / 135
According to the ideal gas law, PV = nRT, the pressure and volume of an ideal gas are directly proportional, given n (number of moles), R (universal gas constant), and T (temperature) are constant.
Assuming the temperature is constant, we can set up the following equation based on the ratios:
Volume ratio = Pressure ratio
1.67 = P_acetylene / 135
Now, solve for P_acetylene:
P_acetylene = 1.67 * 135
P_acetylene ≈ 225.45
Therefore, to ensure that you run out of oxygen and acetylene at the same time, you should fill the acetylene tank to a pressure of approximately 225.45 .