
#1
Dec2206, 12:18 PM

P: 424

1. The problem statement, all variables and given/known data
It is said that the work done between 2 isothermals is independent of the adiabatic. 2. Relevant equations The work done during an adiabatic process is given by, W = C(v){T1 – T2) 3. The attempt at a solution From the above equation it is clear that W depends only on the initial and final temperatures. So W is the same along any adiabatic curve. But work is a path function. Then the work done along different adiabatic curves should be different. It is contradictory to each other. Could somebody please clear my doubt? Here C(v) is the specific heat of the gas at constant volume. 



#2
Dec2206, 06:25 PM

HW Helper
P: 1,391

Does this explain it? 



#3
Dec2206, 11:07 PM

Sci Advisor
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P: 6,561

For the reversible adiabatic path the temperature change is: [tex]\Delta T = \frac{K(V_f^{1\gamma}V_i^{1\gamma})}{nC_v (1\gamma)}[/tex] So you are right that Work is a path function. But so is the temperature change in an adiabatic process. AM 



#4
Dec2306, 01:33 PM

P: 424

work done in an adiabatic process
I didn't get you. Could you please explain it in a more simple way? I am herewith attaching a PV diagram in which I have drawn two adiabatics 1 & 2 between 2 isothermals at temperatures T1 & T2. As we move from A to B in adiabatic 1 and from D to C in adiabatic 2, will the change in temperature be the same i.e. T2  T1?




#5
Dec2306, 05:56 PM

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P: 6,561

If you take some path other than the adiabatic path from A to B the work done will be different. (which simply means that heat is flowing into or out of the system, so it would not be adiabatic). AM 



#6
Dec2406, 03:12 AM

P: 424





#7
Dec2406, 06:39 AM

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P: 6,561

[tex]PV^{\gamma} = K \text{and} PdV =  dU = nC_vdT[/tex], the work done in this process is: [tex]W = \int_{V_i}^{V_f}PdV = \int_{V_i}^{V_f}KV^{\gamma}dV = \frac{K(V_f^{1\gamma}V_i^{1\gamma})}{(1\gamma)} =  \int_{T_i}^{T_f} nC_vdT =  nC_v(T_f  T_i)[/tex] If you draw another reversible adiabatic path from A to T2 it will be AB because there is only one reversible adiabatic path from A to T2. If you draw a reversible adiabatic path between some other point on T1 to T2, the work done will be the same. But it is not the same path because the initial and final volumes and pressures are different. AM 



#8
Dec2406, 01:28 PM

P: 424

Thank you very much for your guidance.




#9
Oct1010, 02:09 AM

P: 7

In the relationship
[tex]W =  n C_{v} (T_{f}  T_{i})[/tex] If the volume changes in an adiabatic process why is [tex]C_{v}[/tex] used? 



#10
Oct1010, 02:15 AM

P: 3,015

The work done during an adiabatic process, according to the First Law of Thermodynamics is equal to the change of what? If this is not dependent on which adiabatic we go along, but only on the initial and final temperatures, then what can we say that the quantity in the first question only depends on? Do you know of any physical systems with this property?




#11
Oct1010, 02:37 AM

P: 7

In an adiabatic process dQ = 0 so that dU = dW.
So then you use, [tex] C_{v} = ( \frac{\Delta U}{\Delta T})_{v}[/tex] to make [tex] W =  n C_{v} (T_{f}  T_{i}) [/tex] Is that right? 


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