Thermodynamics-work, heat, & internal energy

[mbrmbrg]

Thermodynamics--work, heat, & internal energy

Homework Statement

When a system is taken from state i to state f along path iaf in Figure 18-41, Q = 60 cal and W = 20 cal. Along path ibf, Q = 51 cal.

Fig. 18-41 (see attatched)

(a) What is W along path ibf?
(b) If W = -13 cal for the return path fi, what is Q for this path?
(c) If E_int,i = 7 cal, what is Eint,f?
(d) If E_int,b = 18 cal what is Q for path ib?
(e) For the same value of E_int,b, what is Q for path bf?

Homework Equations

[tex]\Delta E = Q - W[/tex] and is path independant

The Attempt at a Solution

(a) [tex]\Delta E_{iaf} = \Delta E_{ibf}[/tex]
hereon, iaf will be abbreviated a, and ibf will be abbreviated b.
[tex] Q_a - W_a = Q_b - W_b[/tex]
[tex] Q_a - W_b + Q_b = -W_b [/tex]
[tex] 60-20+51=-W_b=91[/tex]
So W_b = -91 cal. Wrong. And yes the problem wants the answer in calories.
On the chance that they're giving work performed on the system rather than work performed by the system, I changed Q-W to Q+W, and got that W_b=29 cal, which is still wrong.

I used the same exact reasoning for part b (and got it wrong).

I got part C correct [47 cal] using the fact that DeltaE is path independent, and Delta E = final energy - initial energy.

Parts d and e I just have no idea where to begin. So I guessed 18 cal for both. Surprise! Both wrong...

 

[Andrew Mason]

Homework Statement

When a system is taken from state i to state f along path iaf in Figure 18-41, Q = 60 cal and W = 20 cal. Along path ibf, Q = 51 cal.

Fig. 18-41 (see attatched)

(a) What is W along path ibf?
(b) If W = -13 cal for the return path fi, what is Q for this path?
(c) If E_int,i = 7 cal, what is Eint,f?
(d) If E_int,b = 18 cal what is Q for path ib?
(e) For the same value of E_int,b, what is Q for path bf?

Homework Equations

[tex]\Delta E = Q - W[/tex] and is path independant

The Attempt at a Solution

(a) [tex]\Delta E_{iaf} = \Delta E_{ibf}[/tex]
hereon, iaf will be abbreviated a, and ibf will be abbreviated b.
[tex] Q_a - W_a = Q_b - W_b[/tex]
[tex] Q_a - W_b + Q_b = -W_b [/tex]
[tex] 60-20+51=-W_b=91[/tex]
So W_b = -91 cal. Wrong. And yes the problem wants the answer in calories.
On the chance that they're giving work performed on the system rather than work performed by the system, I changed Q-W to Q+W, and got that W_b=29 cal, which is still wrong.

I used the same exact reasoning for part b (and got it wrong).

I got part C correct [47 cal] using the fact that DeltaE is path independent, and Delta E = final energy - initial energy.

Parts d and e I just have no idea where to begin. So I guessed 18 cal for both. Surprise! Both wrong...

a) First of all, I would recommend that you use U instead of E. W and U are both forms of energy, as is heat flow: Q). Singling out internal energy as E suggests that Q and W are something other than energy, which is not true.

dQ = dU + dW (where dW is the work done by the gas: PdV)

We know from the data given that the area under the path from a to f (dW, the work done by the gas in following the iaf path) is 20. We also know that this required a heat flow, dQ, into the gas of 60. So the change in internal energy, dU from i to f is 40, as you appear to have found.

Since, as you correctly stated, internal energy path independent, we know that following ibf, the change in U is the same: 40. So work done is just dQ - dU, where dQ is now the heat flow into the gas following ibf (51): dW = dQ - dU = 51-40 = 11 Cal.

I think you will see that in your algebra you just got a sign wrong.

b) the same reasoning as above applies to b).

c) just add the change in U to Ui

d) what is the change in U from i to b? You already know the work done (area under the graph from i to b). So you can get dQ.

e) Hint: Is there any work done in going from b to f (what is the area under the graph from b to f)?

AM

 

[m2003]

Hello,

Thank you for your question. I am a scientist and I would be happy to provide a response to this content.

Firstly, it is important to note that thermodynamics is the branch of physics that deals with the relationships between heat, work, and internal energy in a system. It is governed by the laws of thermodynamics, which are fundamental principles that describe the behavior of energy in a system.

In this problem, we are dealing with a system that has undergone changes along two different paths, iaf and ibf. Along path iaf, the system has gained 60 cal of heat (Q = 60 cal) and has done 20 cal of work (W = 20 cal). Along path ibf, the system has gained 51 cal of heat (Q = 51 cal).

(a) To find the work along path ibf, we can use the fact that the change in internal energy (\Delta E) is the same for both paths. This is because \Delta E is a path-independent quantity. Therefore, we can write the following equation:

\Delta E_{iaf} = \Delta E_{ibf}

Since we know that \Delta E = Q - W, we can substitute in the values we know for path iaf:

\Delta E_{iaf} = Q_{iaf} - W_{iaf} = 60 cal - 20 cal = 40 cal

Now, we can solve for the work along path ibf:

\Delta E_{ibf} = Q_{ibf} - W_{ibf} = 40 cal

Therefore, W_{ibf} = -40 cal. This means that 40 cal of work is performed on the system along path ibf.

(b) To find Q for the return path fi, we can use the fact that work is path-independent, so the value of work remains the same regardless of the path taken. Therefore, we can write the following equation:

\Delta E_{iaf} = \Delta E_{fi}

Substituting in the known values for \Delta E_{iaf} and W_{iaf}, we get:

40 cal = Q_{fi} + 20 cal

Solving for Q_{fi}, we get Q_{fi} = 20 cal. This means that the system gains 20 cal of heat along path fi.

(c) To find the final internal energy, we can use the same