Imagine that you have a 5.50 L gas tank and a 3.50 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 155 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.
2C2H2 + 5O2 ==> 4CO2 + 2H2O
Use PV = nRT for O2. Plug in 155 atm for P, V = 5.5L, Assume any value for T(in kelvin) and use your chosen value for T throughout. Solve for n = number of mols O2.
Knowing you need that many mols O2, use the coefficients in the balanced equation to convert mols O2 to mols C2H2.
Then use PV = nRT for the C2H2 tank and solve for P. Use the same value of T you used in the first calculation.
To find the pressure at which you should fill the acetylene tank, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
In this case, we want to determine the pressure at which the two tanks run out of gas at the same time. Since they have different volumes, we can assume that they have different amounts of gas initially. Let's assign the number of moles of oxygen in the larger tank as n1 and the number of moles of acetylene in the smaller tank as n2.
Since we want both tanks to run out of gas at the same time, we can equate the number of moles in each tank:
n1 = n2
Using the ideal gas law, we can rewrite this equation for each gas:
(P1 * V1) / (n1 * R * T) = (P2 * V2) / (n2 * R * T)
Simplifying the equation, we can cancel out the R and T terms:
(P1 * V1) / n1 = (P2 * V2) / n2
Plugging in the given values:
P1 = 155 atm (pressure of oxygen tank)
V1 = 5.50 L (volume of oxygen tank)
n1 = ?
V2 = 3.50 L (volume of acetylene tank)
n2 = ?
We still need to determine the values of n1 and n2. To do that, we can use the ideal gas law and rearrange it to solve for the number of moles:
n = (P * V) / (R * T)
Given that the temperature and gas constant are the same for both tanks, we can substitute the values and find the number of moles for each gas:
For oxygen:
n1 = (P1 * V1) / (R * T)
For acetylene:
n2 = (P2 * V2) / (R * T)
Now we can substitute these expressions back into the equation we derived earlier:
(P1 * V1) / ((P1 * V1) / (R * T)) = (P2 * V2) / ((P2 * V2) / (R * T))
Simplifying, we can cancel out some terms:
V1 = V2
This means that the volumes of the two tanks should be equal. Therefore, the acetylene tank should also be filled to a pressure of 155 atm to ensure that you run out of each gas at the same time.
In summary, to ensure that the two tanks run out of gas at the same time, you should fill the acetylene tank to a pressure of 155 atm, which is the same pressure as the oxygen tank.