Deriving $P{V}^{n}=\text{constant}$ for polytropic processes?

Taniya Burns
2022-07-22
Answered

Deriving $P{V}^{n}=\text{constant}$ for polytropic processes?

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bardalhg

Answered 2022-07-23
Author has **15** answers

The dU here is an exact differential:

$\mathrm{d}U(T,\phantom{\rule{thinmathspace}{0ex}}V)={\left(\frac{\mathrm{\partial}U}{\mathrm{\partial}T}\right)}_{V}\mathrm{d}T+{\left(\frac{\mathrm{\partial}U}{\mathrm{\partial}V}\right)}_{T}\mathrm{d}V$

rather than the derivatives you have. Hence, you cannot cancel the dT's and dV's as you've done.

$\mathrm{d}U(T,\phantom{\rule{thinmathspace}{0ex}}V)={\left(\frac{\mathrm{\partial}U}{\mathrm{\partial}T}\right)}_{V}\mathrm{d}T+{\left(\frac{\mathrm{\partial}U}{\mathrm{\partial}V}\right)}_{T}\mathrm{d}V$

rather than the derivatives you have. Hence, you cannot cancel the dT's and dV's as you've done.

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Entropy change in a calorimetry problem

A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat ${c}_{a}$ and initial temperature ${T}_{A}$ and object $B$ has specific heat ${c}_{b}$ with initial temperature ${T}_{B}$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.

The standard solution is to use

$S=\int \frac{dQ}{T}$

where $dQ=mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.

The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.

My question is - what exactly is the reversible path we are using when we use $dQ=mcdT$?

A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat ${c}_{a}$ and initial temperature ${T}_{A}$ and object $B$ has specific heat ${c}_{b}$ with initial temperature ${T}_{B}$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.

The standard solution is to use

$S=\int \frac{dQ}{T}$

where $dQ=mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.

The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.

My question is - what exactly is the reversible path we are using when we use $dQ=mcdT$?

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