Question

Imagine that you have a 5.00 LL gas tank and a 4.50 LL 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 125 atmatm , 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

Answer:

$P_2=139atm$

Explanation:

Hello,

In this case, even we are given two different gases, as it says "ensure that you run out of each gas at the same time", they shall behave as only one, for that reason, no matter the amounts, they are going to be modeled by means of the Boyle's law, as temperature remains unchanged, which allows us to understand the pressure-volume behavior as an inversely proportional relationship:

$P_1V_1=P_2V_2$

Hence, we solve for the P2, representing the pressure of the acetylene as shown below:

$P_2=\frac{P_1V_1}{V_2} =\frac{5.00L*125atm}{4.50L} \\\\P_2=139atm$

Best regards.

Answer 2

Answer:

The pressure of the acetylene tanks is 55.56 atm

Explanation:

Step 1: Data given

Volume of oxygen = 5.00 L

Volume of acetylene = 4.50 L

Pressure of oxygen tank = 125 atm

Step 2: The balanced equation

2C2H2 +5O2 ⇒ 4CO2+2H20

For 2 moles acetylene we need 5 moles O2 to produce 4 moles CO2 and 2 moles H2O

Step 3: Calculate the pressure of the acetylene tanks

n = p*V/RT

⇒ since R and T is constatn

n1/n2 = P1*V1 / P2*V2

⇒with n1 = the number of moles acetylene = 2 moles

⇒with n2 = the number of moles of oxygen = 5 moles

⇒with P1 = the pressure of acetylene = TO BE DETERMINED

⇒with P2 = the pressure of oxygen = 125 atm

⇒with V2 = the volume of oxygen = 5.00L

⇒with V1 = the volume of acetylene = 4.50 L

 2/5 =  P1 * 4.50 L / 125atm * 5.00

0.4 = P1*4.50  /  625

250 = P1*4.50

P1 = 55.56 atm

The pressure of the acetylene tanks is 55.56 atm