To find total work done from multiple reversible processes

In summary, In State 1, the pressure, temperature, and volume were all the same. In State 2, the pressure was increased but the temperature and volume remained the same. In State 3, the pressure was increased but the temperature remained the same and the volume was decreased. In State 4, the pressure was increased and the temperature and volume were both increased.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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
 

Related to To find total work done from multiple reversible processes

1. What is the definition of work in the context of reversible processes?

In thermodynamics, work is defined as the energy transfer that occurs when a force is applied to an object and causes it to move a certain distance. In the context of reversible processes, work refers to the total energy transfer that occurs during a series of reversible processes.

2. How is work calculated in a reversible process?

The total work done in a reversible process is calculated by taking the integral of the pressure with respect to the volume, or W = ∫PdV. This integral represents the area under the curve on a pressure-volume graph and is equal to the energy transferred during the process.

3. What is the significance of reversible processes in thermodynamics?

Reversible processes are important in thermodynamics because they represent the idealized way in which energy can be transferred and work can be done. They allow us to calculate the maximum possible work that can be obtained from a given system, and serve as a benchmark for real-world processes.

4. Can the total work done in a series of reversible processes be negative?

Yes, the total work done in a series of reversible processes can be negative. This occurs when the system does work on its surroundings, such as in the case of an expansion process where the volume increases and the pressure decreases, resulting in a negative work value.

5. How does the efficiency of a reversible process compare to that of an irreversible process?

The efficiency of a reversible process is always greater than or equal to that of an irreversible process. This is because reversible processes are idealized and do not involve any energy losses, while irreversible processes always result in some energy being lost as heat. In other words, the efficiency of a reversible process represents the maximum possible efficiency for a given system.

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