# Is this against Kelvin-Planck statement?

• pitbull
In summary, the conversation discusses an ideal gas expanding isothermally in contact with a heat source and the resulting change in internal energy (∆U). It is determined that ∆U is zero in this case because T=constant. This leads to a discussion about the validity of the Kelvin-Planck statement, which states that there is no process where the sole result is the full transformation of heat into work. After considering the equations pv=nRT and dW=-pdV, it is concluded that all the heat absorbed by the gas in this process is converted into work, which goes against the Kelvin-Planck statement. However, there is a need for clarification in the wording of the statement, which states that any device used in
pitbull
Gold Member

## Homework Statement

An ideal gas expands isothermally in contact with a heat source. ∆U is zero in this case because it is an ideal gas and T=constant. Is this against Kelvin-Planck statement?

## Homework Equations

pv=nRT
dW=-pdV
Kelvin-Planck statement: There is no process whose only result is the full transformation of heat into work.

## The Attempt at a Solution

My guess is, ∆U=Q+W=0, therefore
Q=integral(pdv)=nRTlog(Vfinal/Vinitial)>0
W=-nRTlog(Vfinal/Vinitial)
So all the heat that the gas absorbs turns into work. Therefore, this is against K-P statement, but I have a feeling that I am wrong. Can you guys help me?

pitbull

## 1. Is it possible to violate the Kelvin-Planck statement?

No, the Kelvin-Planck statement is a fundamental law of thermodynamics and cannot be violated.

## 2. What does the Kelvin-Planck statement state?

The Kelvin-Planck statement states that it is impossible for any system to operate in a thermodynamic cycle and produce a net amount of work while only exchanging energy with a single reservoir.

## 3. Why is the Kelvin-Planck statement important?

The Kelvin-Planck statement is important because it sets limits on the efficiency of heat engines and helps us understand the transfer and conversion of energy in thermodynamic systems.

## 4. Can the Kelvin-Planck statement be applied to all thermodynamic systems?

Yes, the Kelvin-Planck statement applies to all thermodynamic systems, including both closed and open systems.

## 5. Are there any exceptions to the Kelvin-Planck statement?

No, there are no exceptions to the Kelvin-Planck statement. It has been experimentally verified and has never been observed to be violated.

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