consider a mixture of air and gasoline vapor in a piston. original V is 30cm^3. if the combustion of misture releases 985J energy, to what volume will the gases expand against a constant pressure of 655 torr if all energy of combustion is converted into work ro push back the piston?

in L

Mel--This answer doesn't look right to me so check it out.

work = -p(V2-V2)= -985
change p to atmospheres.
change V to liters.
I get something like 1,000 liters and that sounds far to large. Perhaps Bob Pursley will look at this.

That should be -P(V2-V1) and solve for V2.

Also, last sentence should read "far too large."

type this in the Google search window:

985joules/655torr in liters

That gives you delta Volume.

I found my problem. I didn't convert J to L-atm.

985 J/101.325 = 9.72
Then -9.72 = -(655/760)(V2-V1)
9.72 = 0.862V2 - 0.862*0.03
V2 = 11.3 liters.

To determine the final volume of the gases, we can use the ideal gas law to relate the initial and final volumes with the pressure and temperature of the system.

The ideal gas law is given by:
PV = nRT

Where:
P is the pressure in atmospheres (atm)
V is the volume in liters (L)
n is the number of moles of gas
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K)
T is the temperature in Kelvin (K)

In this case, the pressure is given as 655 torr, which we need to convert to atm:
1 atm = 760 torr
Therefore, 655 torr is approximately 0.862 atm.

The initial volume is V = 30 cm^3, which we also need to convert to liters:
1 L = 1000 cm^3
Therefore, the initial volume is 30 cm^3 / 1000 cm^3/L = 0.03 L.

We are given that all the energy of combustion is converted into work to push back the piston, which means that the released energy is equal to the work done by the gases expanding. We can relate the energy released to the work done by the gases using the equation: work = P * ΔV.

In this case, the work done (ΔW) is given as 985 J, and the pressure is 0.862 atm. Solving for ΔV, we get: ΔV = ΔW / P.

ΔV = 985 J / (0.862 atm)

Now we can substitute the values of ΔV, P, and the initial volume V into the ideal gas law to find the final volume of the gases (Vf):

(P * V) = (n * R * T)
(0.862 atm * 0.03 L) = (n * 0.0821 L·atm/(mol·K) * T)

Since we're considering a mixture of air and gasoline vapor, we assume air behaves as an ideal gas. Therefore, we can use the ideal gas equation to find the number of moles (n). Rearranging the equation, we have:

n = (P * V) / (R * T)

Substituting the values of P, V, and R, we get:

n = (0.862 atm * 0.03 L) / (0.0821 L·atm/(mol·K) * T)

Now, we can substitute the value of n back into the ideal gas law equation to find the final volume (Vf):

Vf = (n * R * T) / P

Substituting the values of n, R, and T, we have:

Vf = [(0.862 atm * 0.03 L / (0.0821 L·atm/(mol·K) * T)] * 0.0821 L·atm/(mol·K) * T / 0.862 atm

Simplifying the equation, we find:

Vf = 0.03 L

Therefore, the final volume of the gases will be 0.03 L.