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magine that you have a 7.00 L gas tank and a 2.00 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 115 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.

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jufsanabriasa

Answer:

The pressure you should fill the acetylene tank to ensure that you run out of each gas at the same time is 161 atm

Explanation:

To answer this question you need to know acethylene combustion:

C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O

The same P*V  in both oxygen and acetylene tanks will produce the same moles in each tank. As you need 5/2 of O₂ more than C₂H₂ moles:

115 atm* 7,00L = 805

805 * (1 mol Câ‚‚Hâ‚‚/5/2 mol Oâ‚‚) = 322

322 = PV

As volume of tank is 2,00L

P = 161 atm

I hope it helps!

I hope it helps!

dsdrajlin

Answer:

The pressure of acetylene should be 161 atm.

Explanation:

Let's consider the reaction between oxygen and acetylene. This is a combustion reaction that produces carbon dioxide and water.

C₂H₂ + 2.5 O₂  ⇄ 2 CO₂ + H₂O

For Oâ‚‚, we know it occupies 7.00 L and exerts a pressure of 115 atm. The welding with acetylene takes place at around 3373 K. In these conditions, we can use the ideal gas law to find out the amount of oxygen.

[tex]P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{115atm.7.00L}{(0.08206atm.l/mol.K). 3373K} =2.91mol[/tex]

Using the balanced equation, we can calculate how much acetylene we need for 2.91 moles of Oâ‚‚.

[tex]2.91molO_{2}.\frac{1molC_{2}H_{2}}{2.5molO_{2}} =1.16molC_{2}H_{2}[/tex]

So, 1.16 moles of Câ‚‚Hâ‚‚ occupy 2.00L at 3373 K. The pressure it exerts is:

[tex]P.V=n.R.T\\P=\frac{n.R.T}{V} =\frac{1.16mol.(0.08206atm.L/mol.K).3373K}{2.00L} =161atm[/tex]

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