Imagine that you have a 6.00L gas tank and a 3.00L 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 125atm , 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.
Note: Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
You have 5 O2 molecules for every 2 acetylene molecules
P V = n R T
where n (mols)is proportional to the number of molecules
so for O2
125 (6) = 5 k R T for O2
P(3) = 2 k R T for C2H2
k R T are the same for both
so
125(6/5) = P(3/2)
5142
To find out the pressure at which you should fill the acetylene tank, you can use the ideal gas law which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature (which remains constant in this case)
First, let's calculate the number of moles of oxygen present in the larger tank. Since the amount of oxygen in moles will be the same as the amount of acetylene in moles, we can use the balanced chemical equation:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
This equation tells us that the ratio between the number of moles of C2H2 and O2 is 2:5. So if we have x moles of acetylene, we will have (5/2)x moles of oxygen.
Next, we can use the ideal gas law for the oxygen tank:
(P_oxygen)(V_oxygen) = (n_oxygen)(R)(T)
Similarly, we can use the ideal gas law for the acetylene tank:
(P_acetylene)(V_acetylene) = (n_acetylene)(R)(T)
Since we want both tanks to run out of gas at the same time, we can set the number of moles of oxygen equal to the number of moles of acetylene:
(n_oxygen)(R)(T) = (n_acetylene)(R)(T)
Canceling out the common terms, we get:
(P_oxygen)(V_oxygen) = (P_acetylene)(V_acetylene)
Substituting the given values, we have:
(125 atm)(6.00L) = (P_acetylene)(3.00L)
Now, we can solve for the pressure of acetylene:
P_acetylene = (125 atm)(6.00L) / (3.00L)
P_acetylene = 250 atm
Therefore, you should fill the acetylene tank to a pressure of 250 atm to ensure that both tanks run out of gas at the same time.
To determine the pressure at which you should fill the acetylene tank, we can use the ideal gas law and the stoichiometry of the combustion reaction.
1. Start by identifying the relevant information:
- Oxygen tank: 6.00 L, filled to a pressure of 125 atm.
- Acetylene tank: 3.00 L, unknown pressure.
2. We can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant (0.08206 L⋅atm/(mol⋅K))
T = temperature (assumed constant)
3. Rearrange the equation to solve for the number of moles, n:
n = PV / RT
4. Calculate the number of moles of oxygen gas in the oxygen tank:
n_oxygen = (125 atm) * (6.00 L) / (0.08206 L⋅atm/(mol⋅K) * T)
Note: The temperature is assumed to be constant and cancels out in this calculation.
5. Determine the number of moles of acetylene gas needed for combustion:
According to the balanced equation, 2 moles of acetylene (C2H2) react with 5 moles of oxygen (O2) to produce 4 moles of carbon dioxide (CO2) and 2 moles of water (H2O). Therefore, the stoichiometric ratio between oxygen and acetylene is 5:2.
So, the number of moles of acetylene needed = 2*(n_oxygen * 2/5)
6. Substitute the calculated number of moles of acetylene into the ideal gas law and solve for the pressure in the acetylene tank.
n_acetylene = 2*(n_oxygen * 2/5)
P_acetylene = n_acetylene * RT / V_acetylene
P_acetylene = (2*(n_oxygen * 2/5)) * (0.08206 L⋅atm/(mol⋅K)) / (3.00 L)
7. Calculate the pressure at which you should fill the acetylene tank to ensure both gases run out at the same time.
Note: The assumption of ideal behavior for all gases allows us to use the ideal gas law and stoichiometry with confidence in this calculation.
Please note that the specific temperature is required to obtain a precise value for the pressure in the acetylene tank.