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.

Question

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.

Answer 1

find the number of moles of O2

n=PV/RT

now balance the equation, and look at the mole ratio. How many moles of acetylene do you need for n moles of O2?

now, find the pressure

P=molesAcy*RT/V

Answer 2

To determine the pressure to fill the acetylene tank, we need to consider the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we are assuming ideal behavior for all gases, so we can ignore the effect of temperature. Therefore, we can rewrite the ideal gas law equation as P₁V₁ = P₂V₂, where P₁ and V₁ are the initial pressure and volume of the oxygen tank, and P₂ and V₂ are the pressure and volume of the acetylene tank.

Given that the oxygen tank has a volume of 5.50 L and is filled to a pressure of 155 atm, we have:
P₁ = 155 atm
V₁ = 5.50 L

We want to determine the pressure (P₂) to fill the acetylene tank so that we run out of each gas at the same time. Considering that we have two tanks, the total volume of gas used will be the sum of their volumes (V₁ + V₂). In this case, we want the tanks to run out of gas simultaneously, so the gas used from each tank will be equal, which means that the volume of gas used from each tank will also be equal.

Therefore, the volume of gas used from the oxygen tank (V₁) will also be equal to the volume of gas used from the acetylene tank (V₂). Now, we can set up the equation using the given information:
P₁V₁ = P₂V₂

Substituting the values we have:
155 atm * 5.50 L = P₂ * 3.50 L

Now we can solve for P₂:
P₂ = (155 atm * 5.50 L) / 3.50 L

Calculating the result:
P₂ ≈ 243.57 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 243.57 atm.