# Doubt regarding proof of Clausius Inequality.

• I
• weezy
In summary, the conversation discusses the difficulty in converting all heat to work in a thermodynamic cycle due to the violation of the 2nd law and the use of the Clausius inequality in thermodynamics. It also questions the concept of ideal reservoirs returning to their original state and the transfer of heat in a reversible case. The conversation also explains the work done in a Carnot cycle and the relationship between work done and gas expansion/contraction. It concludes with a clarification on the work done in a combined system, showing that the sum of the individual works must be less than or equal to zero.

#### weezy

I have attached two images from my textbook one of which is a diagram and the other a paragraph with which I am having problems. The last sentence mentions that due to violation of 2nd law we cannot convert all the heat to work in this thermodynamic cycle. However what is preventing the carnot cycles from doing work dW_i ? Shouldn't ΔW be the only work that is impossible to achieve by the machine? Why take ∑dW_i + ΔW <=0 and not just ΔW<=0?

I would also like to know that since at the end of one cycle reservoir the system returns to it's initial state and T loses some heat and if total work performed by that system is zero(for reversible case) then where does the heat go?

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Firstly I am not understanding your difficulty. But here I raise two points of views:
1. There cannot be any proof of Clausius inequality as it is the fundamental axiom in the theory of thermodynamics. One can only relate it to other popular statements of laws of thermodynamics.

2.The ideal reservoirs returning to its original state is also not true in the strictest possible sense because we can go on depositing heat or extracting heat from it and it remains in the same state, one does not know what does it mean totally but only one aspect we know its temperature does not change because of assumed infinite thermal capacity. That is why these heat transfers are considered reversible.

weezy said:
Shouldn't ΔW be the only work that is impossible to achieve by the machine? Why take ∑dW_i + ΔW <=0 and not just ΔW<=0?

No. The system undergoing a thermodynamic cycle is the Carnot cycle plus the other cycle. This is the system that appears to take heat from one reservoir and get work out without rejecting heat to another reservoir. Since for any thermodynamic cycle that is connected to only one reservoir the cyclical work must be less than or equal to zero, it must be so with this cycle too. The cyclical work involved for this cycle is the sum of the individual works outputs from anywhere in the system.

weezy said:
I would also like to know that since at the end of one cycle reservoir the system returns to it's initial state and T loses some heat and if total work performed by that system is zero(for reversible case) then where does the heat go?

If the net work performed by that system is zero, then what is the net heat transfer? Think about it in terms of the first law.

mfig said:
No. The system undergoing a thermodynamic cycle is the Carnot cycle plus the other cycle. This is the system that appears to take heat from one reservoir and get work out without rejecting heat to another reservoir. Since for any thermodynamic cycle that is connected to only one reservoir the cyclical work must be less than or equal to zero, it must be so with this cycle too. The cyclical work involved for this cycle is the sum of the individual works outputs from anywhere in the system.
If the net work performed by that system is zero, then what is the net heat transfer? Think about it in terms of the first law.
But the carnot engine which produces dW_i work is connected to 2 reservoirs. How would this work be negative or zero?

weezy said:
But the carnot engine which produces dW_i work is connected to 2 reservoirs. How would this work be negative or zero?

Was there more text defining what the quantities $\Delta W$ and $dW_i$ mean?

In the simplest ideal Carnot cycle, there are 4 steps:
1. Isothermal expansion: Put a quantity of gas into a hot reservoir (temperature $T_h$) and let it expand.
2. Isentropic expansion: Take it out of the reservoir and let it expand some more, extracting work from the gas. During this expansion, the temperature drops to $T_c$.
3. Isothermal contraction: Put the gas into a cold reservoir (temperature $T_c$) and let it contract.
4. Isentropic contraction: Take it out of the reservoir and force it to contract some more, putting work into the gas.
In steps 1 and 2, the gas is doing work, and in steps 3 and 4, work is being done on the gas (which is considered to be negative work done by the gas).

The work done by a gas is $P dV$. If the gas is expanding, then it is doing work, and this is a positive number. If the gas is contracting, then work is being done on the gas, and this quantity is a negative number.

weezy said:
But the carnot engine which produces dW_i work is connected to 2 reservoirs. How would this work be negative or zero?

It is not the Carnot engine that is considered in the equation you are asking about. It is the combined system, shown in red in the image below. For this system, would you agree that ##\Delta W+\Sigma dW_i \leq 0##? That is all that they are saying. Once you see this, you should be able to see that ##\Delta W \leq -\Sigma dW_i##.

## 1. What is Clausius Inequality and why is it important?

Clausius Inequality is a fundamental principle in thermodynamics that states that the total change in entropy within a closed system must be greater than or equal to zero. It is important because it helps us understand the direction of heat flow and the efficiency of thermodynamic processes.

## 2. How is Clausius Inequality derived?

Clausius Inequality is derived from the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time. It can also be derived from the Kelvin-Planck statement of the second law, which states that it is impossible to construct a perfectly efficient heat engine.

## 3. What evidence supports Clausius Inequality?

There is a wealth of experimental and observational evidence that supports Clausius Inequality. For example, the efficiency of real-world thermodynamic processes is always less than 100%, which is consistent with the principle that total entropy must increase over time.

## 4. Can Clausius Inequality be violated?

No, Clausius Inequality is a fundamental law of thermodynamics and cannot be violated. It has been extensively tested and has been shown to hold true for all known systems and processes.

## 5. How is Clausius Inequality used in practical applications?

Clausius Inequality is used in various practical applications, such as in the design and analysis of heat engines and refrigeration systems. It also helps us understand the limitations of energy conversion and the direction of heat transfer in various processes.