T is constant so we don't need to worry about that so let's just use 300 K for it. The reaction is
2C2H2 + 5O2 ==> 4CO2 + 2H2O
For the larger 7 L tank filled with O2 at a pressure of 145, what is n?
Use PV = nRT and calculate n = number of moles O2.
Now use the coefficients in the balanced equation to convert mols O2 to mols C2H2. That's mols C2H2 = mols O2 x 2/5 = ? mols C2H2.
Now use PV = nRT again. Substitute mols C2H2 and the other values and solve for P in atm for the smaller tank..
Imagine that you have a 7.00L gas tank and a 4.00L 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 145atm , 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.
5 answers
I have no idea if this is correct, but I'll take a stab at it.
Use the ideal gas law to start, and I will have to do some assuming.
I'm going to first assume that the temperature of both gases are the same, and I will also assume that I have equal quantities of each gas.
First step, is to use the ideal gas law and solve for n:
PV=nRT
T=PV/nR
Since both gases are in equal molar quantities, and they need to diffuse at the same rate, then their velocities must be the same.
Using the equation for Kinetic Energy:
1/2mv^2=3/2RT
Solve for v:
mv^2=3RT
v=Sqrt*[3RT/m]
Let m1= mass of acteylene
and
Let m2=mass of oxygen
Since v's are the same, my new equation is the following:
Sqrt*[3RT/m1]=Sqrt*[3RT/m2]
Use the rearranged gas law equation and substitute it into the equation above:
Sqrt*[3R(PV/nR)/m1]=Sqrt*[3R(PV/nR)/m2]
Simplification gives me the following:
[(PV/n)/m1]=[(PV/n)/m2]
Since moles are in equal quantities, moles cancel out of the equation, and simplification gives me the following:
[(P1V1)/m1]=[(P2V2)/m2]
m1=26.04 g
V1=4.00L
P1=??
m2=32.00 g
V2=7.00L
and
P2=145atm
Solve for P1:
(P1V1)/m1=(145atm*7.00L)/32.00g
(P1V1)/m1=31.719 atm*L*g^-1
[(P1V1)/m1]=31.719 atm*L*g^-1
[(P1*4.00L)/26.04g]=31.719 atm*L*g^-1
P1*4.00L=26.04g*31.719 atm*L*g^-1
P1=(26.04g*31.719 atm*L*g^-1)/4.00L
P1=206atm
Use the ideal gas law to start, and I will have to do some assuming.
I'm going to first assume that the temperature of both gases are the same, and I will also assume that I have equal quantities of each gas.
First step, is to use the ideal gas law and solve for n:
PV=nRT
T=PV/nR
Since both gases are in equal molar quantities, and they need to diffuse at the same rate, then their velocities must be the same.
Using the equation for Kinetic Energy:
1/2mv^2=3/2RT
Solve for v:
mv^2=3RT
v=Sqrt*[3RT/m]
Let m1= mass of acteylene
and
Let m2=mass of oxygen
Since v's are the same, my new equation is the following:
Sqrt*[3RT/m1]=Sqrt*[3RT/m2]
Use the rearranged gas law equation and substitute it into the equation above:
Sqrt*[3R(PV/nR)/m1]=Sqrt*[3R(PV/nR)/m2]
Simplification gives me the following:
[(PV/n)/m1]=[(PV/n)/m2]
Since moles are in equal quantities, moles cancel out of the equation, and simplification gives me the following:
[(P1V1)/m1]=[(P2V2)/m2]
m1=26.04 g
V1=4.00L
P1=??
m2=32.00 g
V2=7.00L
and
P2=145atm
Solve for P1:
(P1V1)/m1=(145atm*7.00L)/32.00g
(P1V1)/m1=31.719 atm*L*g^-1
[(P1V1)/m1]=31.719 atm*L*g^-1
[(P1*4.00L)/26.04g]=31.719 atm*L*g^-1
P1*4.00L=26.04g*31.719 atm*L*g^-1
P1=(26.04g*31.719 atm*L*g^-1)/4.00L
P1=206atm
The way that Dr. Bob22 went about doing it could be correct, and the author of the question is assuming that the gases would combine in a combustion reaction, which is something that I did not think of when approaching this problem.
Devron is correct; this is a combustion reaction/problem.
Thought about it some more, and this is a combustion problem. I could have saved myself some work. Dr. Bob222 is correct.
1 answer