Consider an ideal gas encloesd in a 1.00 L container at an internal pressure of 10.0 atm. Calculate the work, w, if the gas expands against a constant external pressure of 1.00 atm to a final volume of 20.0 L.
w = ____ J
now calculate the work done if this process is carried out in two steps.
1] first let, the gas expand against a constant external pressure of 5.00 atm to a volume of 4.00 L.
2] From there, let the gas expand to 20.0 L against a constant external pressure of 1.00 atm.
w = _____ J
i know that the formula to solve this is w=-(p)(Delta V) but when they give you 2 pressures does that mean the formula turns to w=-(delta p)(delta v)?
1)
-Pdelta V = -1 atm x (v2-v1) = -1(20-1) = -19 L*atm work.
2)
The first one is -5 x (4-1) = -15 L*atm
Second stage is -1 x (20-4) = -16 L*atm
Total = -31 L*atm
Here is a good site you can read; in fact there is an example problem that is ALMOST the same as this one.
(Broken Link Removed)
Thank you Dr. Bob! That was very helpful
2=2
wow ilovedrbob69 that was so not helpful :)
thanks SO much dr bob this was SO helpful ;);;:;:);0));)
I love ihateilovedrbob69 >w<
The formula to calculate the work done by a gas during an expansion or compression is indeed given by:
w = -PΔV
where w represents the work done, P is the external pressure, and ΔV is the change in volume of the gas.
In the given scenario, we are initially given the internal pressure, but we need to consider the external pressure as well. Let's solve each part step by step:
1) In the first step, the gas expands against a constant external pressure of 5.00 atm to a volume of 4.00 L. Using the formula mentioned above, we have:
w1 = -P1ΔV1 = -(5.00 atm - 1.00 atm) * (4.00 L - 1.00 L)
Here, ΔV1 is the change in volume, which is 3.00 L (4.00 L - 1.00 L), and P1 is the external pressure during this step. Evaluating the equation:
w1 = -(4.00 atm) * (3.00 L) = -12.00 atm * L
2) In the second step, the gas expands from 4.00 L to 20.0 L against a constant external pressure of 1.00 atm. Again, using the same formula:
w2 = -P2ΔV2 = -(1.00 atm - 1.00 atm) * (20.0 L - 4.00 L)
Here, ΔV2 is the change in volume, which is 16.0 L (20.0 L - 4.00 L), and P2 is the external pressure during this step. Notice that since the pressure remains constant at 1.00 atm, the difference in pressure becomes zero, and the work done in this step is zero.
w2 = -(0 atm) * (16.0 L) = 0 J
Now, let's calculate the total work done in the process by adding the work done in each step:
w_total = w1 + w2 = -12.00 atm * L + 0 J
Therefore:
w_total = -12.00 atm * L = -12.00 L·atm
Note: The unit for work (Joules) is not applicable here since one of the steps has zero work done.
Finally, the total work done in the process is -12.00 L·atm.