Imagine that you have a 5.00 L gas tank and a 4.00 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 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

2C2H2 + 5O2 ==> 4CO2 + 2H2O

The long way is to use PV = nRT and solve for n= number of mols O2 in the 5L tank. Convert that to mols C2H2 used and resubmit to PV = nRT and solve for P.(Note that you don't have a T listed but you can make up one and use it for both sets of data.)
The shorter way is to see that
PV= nRT can be modified to
PVk = n
(V1/V2) x (2 mol C2H2/5 mols O2) = ?

0.4444

To determine the pressure at which you should fill the acetylene tank, you can use the ideal gas law. The ideal gas law equation is PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T represents temperature.

Since we want both tanks to run out of gas at the same time, the number of moles of oxygen and acetylene should be equal.

First, we need to calculate the number of moles of oxygen in the 5.00 L tank. We know the pressure of the oxygen tank is 105 atm, and assuming ideal behavior, we can use the ideal gas law to calculate the number of moles:

PV = nRT

n = PV / RT
= (105 atm) * (5.00 L) / (0.0821 L*atm/(mol*K)) * T

Now, we can calculate the number of moles of acetylene needed based on the number of moles of oxygen.

n(acetylene) = n(oxygen)

Now we can use the ideal gas law again, this time with the acetylene tank's volume and the moles of acetylene:

PV = nRT

P(acetylene) = n(acetylene) * RT / V(acetylene)

Let's substitute the values we know:

P(acetylene) = (105 atm) * [(105 atm) * (5.00 L) / (0.0821 L*atm/(mol*K)) * T] / (4.00 L)

Simplifying the equation, the L units cancel out, as well as the atm units:

P(acetylene) = (105/4) * (105/0.0821) * T

Now, we can calculate the value of T to get the pressure at which we should fill the acetylene tank:

T = (4 / 105/0.0821) * P(acetylene)

Finally, substituting the known values, we can solve for T:

T = (4 / 105/0.0821) * P(acetylene)
=> T ≈ 3.57 * P(acetylene)

Therefore, the pressure at which you should fill the acetylene tank to ensure the oxygen and acetylene run out at the same time is approximately 3.57 times the desired pressure in the oxygen tank.

To determine the pressure at which the acetylene tank should be filled to ensure that both gases run out at the same time, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

Since we are assuming ideal behavior for the gases and the given information does not include the temperature, we can assume the temperature remains constant for both tanks. Therefore, the ideal gas law can be simplified to:

PV = constant

Now, let's analyze the situation step by step:

1. Determine the initial amount of gas in each tank.
- The larger tank has a volume of 5.00 L and will be filled with oxygen to a pressure of 105 atm.
- The smaller tank has a volume of 4.00 L and will be filled with acetylene.
- Since we want both tanks to run out of gas at the same time, we can assume that the initial amount of oxygen and acetylene in each tank is the same.

2. Find the number of moles of oxygen in the larger tank using the ideal gas law.
- Rearrange the equation PV = nRT to solve for n (number of moles): n = PV/RT.

Given:
P = 105 atm (pressure of oxygen)
V = 5.00 L (volume of the larger tank)
R = ideal gas constant (0.0821 L atm/(mol K))
T = constant (assumed the same for both tanks)

Solve for n:
n = (105 atm * 5.00 L) / (0.0821 L atm/(mol K) * T)

3. Assume the initial amount of acetylene in the smaller tank is also 'n' moles.

4. Determine the pressure at which the acetylene tank should be filled to run out at the same time as the oxygen tank.
- The volume of the acetylene tank is 4.00 L.
- Set up an equation using the ideal gas law for acetylene: PV = nRT.
- We want the final pressure of acetylene tank to be such that it will use up all 'n' moles of acetylene before it runs out.
- Since the volume is given, we can rearrange the equation to solve for pressure:

P_acetylene = (n * R * T) / V_acetylene

5. Substitute the calculated value of 'n' from step 2 into the pressure equation from step 4 to find the pressure at which the acetylene tank should be filled to run out simultaneously with the oxygen tank.