Imagine that you have a 7.00 l gas tank and a 3.00 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 145 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.

The tank filled with acetylene should have a pressure of 135.3 atm

Explanation:

Step 1: Data given

Tank 1 = 7.00 L → filled with oxygen to a pressure of 145 atm

Tank 2 = 3.00 L → filled with acetylene

Step 2: The balanced equation

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

Step 2: Calculate moles of oxygen

p*V = n*R*T

⇒ with p = the pressure = 145 atm

⇒ with V = the volume of the gas = 7.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = unknown, so we will just use T

n = (p*V)/(R*T)

n( 145*7.00)/(0.08206*T)

n = 12369 T

Step 3: Calculate moles of acetylene

For 2 moles of acetylene we need 5 moles of oxygen to produce 4 moles of CO2 and 2 moles of H2O

For 12369* T moles of oxygen, we have 4947.6*T moles of acetylene

Step 4: Calculate pressure of acetylene

p = (nRT)/V

p = (4947.6* T*0.08206) / 3.00 L

p = 135.3 atm

The tank filled with acetylene should have a pressure of 135.3 atm

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