Imagine that you have a 5.00L gas tank and a 3.50L 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 145atm , 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

For the large tank, n = PV/RT
Solve for n = # moles O2

Convert mols O2 to moles C2H2 and use PV = nRT and solve for P. I get a little over 500 atm.

To solve this problem, we can use the ideal gas law, which states:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

Given that the temperature is constant and both gases have the same temperature, we can ignore it for this calculation.

Let's define the variables for each gas:
For oxygen:
POxygen = 145 atm
VOxygen = 5.00 L

For acetylene:
PAcetylene = ?
VAcetylene = 3.50 L

Since we want both tanks to run out of gas at the same time, the number of moles of oxygen consumed (nOxygen) should be equal to the number of moles of acetylene consumed (nAcetylene). Therefore, we can set up the following equation:

POxygen * VOxygen = PAcetylene * VAcetylene

Now we can solve for PAcetylene:

PAcetylene = (POxygen * VOxygen) / VAcetylene

Substituting the given values:
PAcetylene = (145 atm * 5.00 L) / 3.50 L

PAcetylene = 207.14 atm

Therefore, you should fill the acetylene tank to a pressure of approximately 207.14 atm to ensure that you run out of each gas at the same time.

To solve this problem, we need to use the concept of the ideal gas law, which states:

PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin

Since we want both gases to run out at the same time, we can set the number of moles of each gas equal to each other.

To find the pressure at which to fill the acetylene tank, we need to compare the number of moles of oxygen gas to the number of moles of acetylene gas.

The number of moles (n) can be calculated using the ideal gas law by rearranging the equation to solve for n:

n = PV / RT

The ideal gas constant, R, is the same for all gases, so it cancels out in the comparison between the two gases.

Therefore, we can set up the following equation:

(Poxygen)(V_oxygen) = (P_acetylene)(V_acetylene)

Substituting the given values:
(145 atm)(5.00 L) = (P_acetylene)(3.50 L)

Now we can solve for P_acetylene:

P_acetylene = (145 atm)(5.00 L) / (3.50 L)
P_acetylene = 208.57 atm

So, you should fill the acetylene tank to a pressure of approximately 208.57 atm to ensure that both gases run out at the same time.