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 105 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.
60.0 atm
Use PV = nRT for the 5.50 L tank of oxygen. Calculate n = # mols O2 in the tank. There is no T listed so pick any T and use it throughout the problem. I would pick 27 C because that's 300 K and that is very close to room T. (Don't forget to use Kelvin for T in PV = nRT).
Using n mols oxygen, and the equation you wrote earlier, calculate mols acetylene needed to exactly react with the oxygen in the 5.50 L tank. Then use PV = nRT again and solve for P of the acetylene in the 3.5 L tank.
To determine the pressure at which you should fill the acetylene tank, we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Since we are assuming ideal behavior for all gases, we can use the same equation for both tanks.
First, let's calculate the number of moles of oxygen in the larger tank. We know the volume (5.50 L) and the pressure (105 atm), and the gas constant is approximately 0.0821 L·atm/(mol·K) at standard conditions. Assuming standard conditions, we can convert the pressure to units of Pascals (Pa) by multiplying by 101325 Pa/atm.
(105 atm) × (101325 Pa/atm) = 1.062 × 10^7 Pa
Now, we can rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
n = (1.062 × 10^7 Pa) × (5.50 L) / (0.0821 L·atm/(mol·K) × 298 K)
n ≈ 250.4 moles of oxygen
Next, we need to calculate the number of moles of acetylene in the smaller tank. We know the volume (3.50 L) and the pressure that we're trying to find. Let's call it P_acetylene.
n_acetylene = P_acetylene × V_acetylene / (R × T)
Now, we want to ensure that we run out of each gas at the same time, which means that the number of moles in each tank should be equal. Therefore, we can set up an equation:
n_oxygen = n_acetylene
250.4 moles = P_acetylene × (3.50 L) / (0.0821 L·atm/(mol·K) × 298 K)
Simplifying this equation, we can solve for P_acetylene:
P_acetylene = (250.4 moles) × (0.0821 L·atm/(mol·K) × 298 K) / (3.50 L)
P_acetylene ≈ 174.1 atm
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 174.1 atm.