Answer:
[tex]P_2=139atm[/tex]
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:
[tex]P_1V_1=P_2V_2[/tex]
Hence, we solve for the P2, representing the pressure of the acetylene as shown below:
[tex]P_2=\frac{P_1V_1}{V_2} =\frac{5.00L*125atm}{4.50L} \\\\P_2=139atm[/tex]
Best regards.
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
