Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2 C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g) Imagine that you have a 5 L gas tank and a 3.5 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 127 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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Tringa0 Tringa0 · Answer 1 · 2025-03-28T16:44:15-04:00

Answer:

72.6 atm should be the pressure at which acetylene tank.

Explanation:

[tex]2C_2H_2 (g)+5O_2 (g)\rightarrow 4CO_2 (g)+2H_2O[/tex]

Let the temperature of the both tanks be same as T.

Volume of the tank in which oxygen is filled = [tex]V_1=5L[/tex]

Pressure of the oxygen in tank =[tex]P_1[/tex]= 127 atm

According to reaction 5 moles of oxygen reacts with 2 moles of acetylene.

[tex]n_1= 5 moles[/tex]

[tex]P_1V_1=n_1RT[/tex]

[tex]T=\frac{P_1V_1}{n_1}[/tex]..[1]

[tex]n_2=2 mol[/tex]

Volume of the tank in which acetylene is filled = [tex]V_2=3.5L[/tex]

Pressure of the acetylene in tank =[tex]P_2[/tex]= ?

[tex]T=\frac{P_2V_2}{n_2}[/tex]..[2]

[1] = [2]

[tex]\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}[/tex]

[tex]P_2=\frac{P_1\times V_1\times n_2}{n_1\times V_2}[/tex]

[tex]=\frac{127 atm\times 5 L \times 2 mol}{5 mol\times 3.5 L}[/tex]

[tex]P_2=72.6 atm[/tex]

72.6 atm should be the pressure at which acetylene tank.