Is the Use of C_V in Adiabatic Expansion of Carnot Cycle Incorrect?

[Pushoam]

Homework Statement

The second law of thermodynamics says that we cannot have an engine which converts all the heat into the work.This is because an engine has to work in cyclic process and we cannot have cyclic process which converts all the heat into the work. Right?
For example, in Carnot cycle, in isothermal expansion process, all the heat could be converted into the work if the system is ideal gas. Adiabatic process doesn't exchange heat.
Again, in the isothermal compression, all the work done on the system gets converted into the heat.
So, how is that, we come to the conclusion that in Carnot cycle,all heat could not be converted into work?

Homework Equations

The Attempt at a Solution

 

[Chestermiller]

Homework Statement

The second law of thermodynamics says that we cannot have an engine which converts all the heat into the work.This is because an engine has to work in cyclic process and we cannot have cyclic process which converts all the heat into the work. Right?
For example, in Carnot cycle, in isothermal expansion process, all the heat could be converted into the work if the system is ideal gas. Adiabatic process doesn't exchange heat.
Again, in the isothermal compression, all the work done on the system gets converted into the heat.
So, how is that, we come to the conclusion that in Carnot cycle,all heat could not be converted into work?

Homework Equations

The Attempt at a Solution

All the heat taken in by the system cannot be converted to work in a cyclic process. In the isothermal compression phase, some of the heat taken in must be rejected to the cold reservoir.

 

[Pushoam]

In the isothermal compression phase, some of the heat taken in must be rejected to the cold reservoir.

Aren't isothermal expansion and compression phases are independent of each other?
In isothermal expansion phase, all the heat absorbed by the system gets converted into the work done by the system on the surroundings.
In isothermal compression phase, all the work done on the system gets converted into the heat given to the surroundings.

 

[Chestermiller]

The key is that, over the entire cycle, you can't convert the intake heat entirely to work. Read the statement of the 2nd law carefully.

 

[CWatters]

Perhaps see..
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm

 

[Pushoam]

Perhaps see..
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/CarnotEngine.htm

I read that website. Thank you for posting it.
In that website work done by the gas during adiabatic expansion is given as W= nC$_V (T_H -T_L)$.
But $C_V$ is calculated at constant volume and in this process, the volume is changing. So, isn't it wrong?
Or, do we take the change in the volume during the adiabatic process negligible?

Carnot cycle:
Isothermal process
The system is: gas contained in a cylindrical container ( the shape of the container could be anything. Right? ) with a frictionless ( idealization) piston.
The system is kept in contact with reservoir at tem. $T_H$. The heat goes into the system. So, the tem. of the system raises to $T_H$. Till this time does the volume of the gas increase or not?

The system with the reservoir is an isolated system.

Assuming that the increase in the volume is negligible, when the tem. of the system i.e. gas becomes equal to that of the reservoir, why should heat flow from the reservoir to the gas?
Is it that the gas has a nature to expand so the gas expands while coming into the thermal eqbm, the expansion leads to a decrease into the tem. of the gas and so the heat flow from the reservoir to the gas and this process goes on?
Does it take infnite time to reach in thermal eqbm ideally?
But we start with the assumption that the system is in thermal eqbm. with the reservoir. Under this assumption, heat should not flow from the reservoir to the gas.
I think the correct assumption is the system is in thermal eqbm. at tem. $T'_H$ which is infinitesimally small than $T_H$ and so the heat goes into the system form the reservoir in very small amount per unit time per unit area.
Is this correct?

Adiabatic process

When the gas has expanded isothermally to a sufficient volume i.e. all the heat coming to the gas has got converted into the work done by the gas,we need to compress it so that it would reach its original volume. If we do this at this tem.i.e. $T'_H$ isothermally, then the work done on the gas for compression will be equal to the work done by the gas during expansion.(Even if we try to compress isothermally, then, first the tem. of the system will become slightly more than that of the reservoir, then the heat will flow into the reservoir from the gas. Right?)
To make the amount of work done on the gas less than that done by the gas, we compress it isothermally at lower tem $T'_L$.
The lower tem. is acheived by letting the gas expand adiabatically i.e. removing the reservoir from the contact of the gas.
This way, too, the work is done by the gas (at the cost of its internal energy).

Isothermal compression

Now, the gas is compressed isothermally at tem. $T'_C$ which is slightly more than the tem. of the cold reservoir $T_C$ so that the heat could flow from the gas to the cold reservoir.
During isothermal process,is the expansion in the volume equal to the compression in the volume?
If yes, then it's clear that the work done by the gas is greater than the work done on the gas.

Adiabatic compression

To rise the tem. of the gas to $T'_H$, we do the adiabatic compression.
I think, during the adiabatic process ( assuming that the volume expansion is equal to the volume compression), the amount of work done by the gas should be equal to that of the work done on the gas.

Anyway for calculating efficiency, we don't need to calculate the work done on\by during the adiabatic process.

Is this correct so far?

 

[Pushoam]

Why do we want each process of the Carnot cycle reversible ?
Is it because we can do the calculation only for the reversible process, by definition?

 

[CWatters]

But CVCVC_V is calculated at constant volume and in this process, the volume is changing. So, isn't it wrong?
Or, do we take the change in the volume during the adiabatic process negligible?

See replies..
https://physics.stackexchange.com/q...pansion-of-carnot-cycle-to-calculate-internal