I really don't get this. The calculations shown don't make sense.

Demonstrate that work is not a state function by calculating the work involved in expanding a gas from an initial state of 1.00: and 10.0atm of pressure to (a) 10,0 L and 1.0 atm pressure. (b) 5.00L and 2.00 atm and then to 10,0L and 1.00 atm pressure.

Solution
(a) w=-PV = -(1.00atm(10.0L-5.0L)
= -9.0L atm

(b) w=-PV = -(2.00atm)(5.0L-1.00L)-(0.0L- 5.0L)
=-13.0L atm

I don't see how the calculations correlate with what was given in the question.

Can someone please explain to me? Thanks!

Referring to Part (b)

••Assume [PV] is a state function:
PV = (5 L)(2 atm) = 10 L.atm
PV = (10L)(1 atm) = 10 L.atm
∆(PV) = [PV]2 - [PV]1 = 10L.atm - 10L.atm = 0 work (not true)

••Actual work = -P∆V which could not be equal to 0 since P is not 0, and ∆V = is not 0. P varies from 2 atm to 1 atm., and ∆V = 5L.

For more in depth information on this check:
http://www.chem.arizona.edu/~salzmanr/480a/480ants/pvwork/pvwork.html

To understand the calculations, we need to first understand the concept of work in thermodynamics and why it is not a state function.

In thermodynamics, work (denoted as w) is defined as the energy transfer that occurs due to the force acting through a distance. Work can be done in various ways, such as by expanding or compressing a gas, or by moving an object against a force.

Now, the reason work is not a state function is because it depends on the path taken to reach a particular state, not just the initial and final states themselves. This means that the work done during a process can vary depending on how the process occurs.

In the question, we are given two different paths for expanding a gas from an initial state to a final state. Let's analyze each path step by step:

(a) Initial state: 1.00 atm, 10.0 L
Final state: 1.00 atm, 10.0 L

To calculate the work for this path, we use the formula w = -PΔV, where P is the pressure and ΔV is the change in volume. Here, the change in volume is (10.0 L - 5.0 L) = 5.0 L. Plugging the values into the formula, we get:
w = -(1.00 atm)(5.0 L) = -5.0 L atm

(b) Initial state: 1.00 atm, 10.0 L
Intermediate state: 2.00 atm, 5.00 L
Final state: 1.00 atm, 10.0 L

To calculate the work for this path, we need to consider two steps:
(i) From the initial state to the intermediate state:
Here, the change in volume is (5.00 L - 1.00 L) = 4.00 L. Plugging the values into the formula, we get:
w1 = -(2.00 atm)(4.00 L) = -8.0 L atm

(ii) From the intermediate state to the final state:
Here, the change in volume is (10.0 L - 5.0 L) = 5.0 L. Plugging the values into the formula, we get:
w2 = -(1.00 atm)(5.0 L) = -5.0 L atm

Adding w1 and w2, we get the total work for this path:
w = w1 + w2 = -8.0 L atm - 5.0 L atm = -13.0 L atm

So, as you can see, the calculations are correctly performed by using the equation w = -PV, where P is the pressure and ΔV is the change in volume. The difference in the calculations arises from the fact that work is path-dependent, and the different paths taken from the initial to final states result in different amounts of work.

I hope this explanation helps you understand the calculations involved. Let me know if you have any further questions!