To find total work done from multiple reversible processes

AI Thread Summary
The discussion focuses on calculating the total work done from multiple reversible processes involving an ideal gas. In the first part, the isochoric process results in zero work done, with the final pressure being double the initial pressure due to a temperature increase. The second part involves an isothermal process where the final volume is half of the initial volume, leading to a negative work calculation, which raises questions about the gas's expansion. The final part returns to the original state with isobaric work calculated based on the change in volume. Participants suggest reviewing the calculations and ensuring clarity on whether the work is done on or by the gas, as well as accurately tracking pressure, temperature, and volume throughout the processes.
warhammer
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Homework Statement
One mole of an ideal gas is heated isochorically till its temperature is doubled. Then it is expanded isothermally till it reaches the original pressure. Finally it is cooled by an isobaric process and restored to the original state. By assuming all the processes to be reversible, show that the resultant work done is RT [2 log 2-1].
Relevant Equations
Isothermal W= nRT ln (V(2)/V(1))
Isobaric W= P (V(2)-V(1))
The question is given in 3 parts.

For first part, process is isochoric so Work done=0. We know here that at end of the process (a), T2=T1 while V remains constant (we can take it as V1) so P2=2P1.

For second part, process is isothermal so T is constant. At end of process we reach P1 again from 2P1, thus V2 for the end stage of this process=V1/2. Isothermal W= RT ln (1/2)

For the last part, original state is restored. Thus final volume for this particular state is V1 from V1/. P is kept constant. Thus Isobaric W = P1 (V1-V1/2)= (P1V1)/2.

Adding first and third results we will not get the value expressed in solution. I would be very grateful if someone looked over my solution and offered guidance as to where I am making the error..
 
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warhammer said:
Homework Statement:: One mole of an ideal gas is heated isochorically till its temperature is doubled. Then it is

warhammer said:
For first part, process is isochoric so Work done=0. We know here that at end of the process (a), T2=T1 while V remains constant (we can take it as V1) so P2=2P1.
If the temperarure is doubled, then T₂ = 2T₁ (not T₂ = T₁).

I haven't checked anything else in detail but also note that you need to be clear if you are being asked for the work done on the gas or the work done by the gas.
 
warhammer said:
For second part, process is isothermal so T is constant. At end of process we reach P1 again from 2P1, thus V2 for the end stage of this process=V1/2. Isothermal W= RT ln (1/2)
Note that your expression for ##W## is negative. Does this make sense if the gas is expanding?

Adding first and third results we will not get the value expressed in solution. I would be very grateful if someone looked over my solution and offered guidance as to where I am making the error..
You seem to have made a number of simple mistakes or you made a bunch of typos. I suggest you carefully determine the pressure, temperature, and volume at the beginning and end of each process.
 
State 1: T, V, P
State 2: 2T, V, 2P
State 3: 2T, 2V, P
State 4: T, V, P

Show these on a P-V diagram
 
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