- #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..

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..