Imagine that you have a 7.00 gas tank and a 3.50 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 115 , 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.
You don't have any units listed for the 7.00 and 3.50.
Also, no units on the 115 ??pressure.
How many acetylene molecules do you need for each O2 molecule?
2C2H2 + 5O2 --> 4CO2 + 2H2O
5 O2 molecules for every 2 2C2H2 molecules
larger tank has O2
let's say it has 5x moles of O2
p v = 5x r t
115 * 7 = 5 x r t
x r t = 115 * 7/(5 )
now the other tank has 2x moles of C2H2
p * 3.5 = 2 x r t
p = (2/3.5) x r t
p = (2/3.5)(115*7)/5
I assume that whatever the pressure is, it is absolute and not gage.
I assumed 115 is atm and 7.00 and 3.50 are L and came out with 92 also.
124 atm
It's 84 atm
i just sneezed
To solve this problem, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to the number of moles of gas and its temperature, while inversely proportional to its volume.
First, let's determine the volume of each gas tank. We have a 7.00 gas tank and a 3.50 gas tank.
Next, let's assume that the initial pressures in both tanks are the same (P_initial) and the final pressures in both tanks are the same as well (P_final).
Since we want both tanks to run out of gas at the same time, the changes in moles of each gas should be equal. Therefore, the ratio of moles of oxygen to moles of acetylene should be equal to the ratio of their volumes.
Let's assume that the number of moles of oxygen is n_oxygen and the number of moles of acetylene is n_acetylene.
The ratio of their volumes is given by:
n_oxygen / n_acetylene = V_oxygen / V_acetylene
The ideal gas law states:
P_initial * V_initial / n = R * T
P_final * V_final / n = R * T
Since P_initial = P_final and T is constant, we can write:
V_initial / n = V_final / n
Substituting the volumes, we have:
7.00 / n_oxygen = 3.50 / n_acetylene
Simplifying further, we can cross multiply:
7.00 * n_acetylene = 3.50 * n_oxygen
Since we want both tanks to run out of gas at the same time, the moles of oxygen and acetylene should be equal. Therefore:
n_oxygen = n_acetylene
Dividing both sides by n_acetylene, we can find the ratio of moles:
7.00 = 3.50
Therefore, the pressure at which you should fill the acetylene tank to ensure that you run out of each gas at the same time is 3.50 atm.