Please justify the mechanism

Question

Does adiabatic process actually happen in real life?

I think everyone is familiar with the equation $pv^{\gamma}=constant$ and how it was derived.

But does everyone actually know $\frac{du}{dv}=-p$ which the p here actually mean instantaneous pressure exert by the gas system to the surrounding?

This can be prove by consider the mechanism change in internal energy which is $\frac{1}{\gamma-1}pv$ is equal to $-p$

consider a gas system that occupied volume v and pressure p, $p(v_0)=p_0$ undergo expansion of volume from $v_0$ to $v_0+h$ where h is very small increase and obtain pressure $p_1$ so

$\frac{1}{\gamma-1}p_0v_0-p_0(h)=\frac{1}{\gamma-1}p_1(v_0+h)$

By using binomial approximation, one can obtain

$p_1\approx p_0-h\gamma\frac{p}{v}$

While from $p_0v_0^{\gamma}=constant$

$p_0v_0^{\gamma}=p_1(v_0+h)^{\gamma}$

$p_1=p_0(\frac{1}{1+\frac{h}{v_0}})^{\gamma}=p_0(1+\frac{h}{v_0})^{-\gamma}\approx p_0-h\gamma\frac{p}{v}$

This hold even for -h

$\frac{1}{\gamma-1}p_0v_0-p_0(-h)=\frac{1}{\gamma-1}p_1(v_0-h)$

By using binomial approximation, one can obtain

$p_1\approx p_0+h\gamma\frac{p}{v}$

While from $p_0v_0^{\gamma}=constant$

$p_0v_0^{\gamma}=p_1(v_0-h)^{\gamma}$

$p_1=p_0(\frac{1}{1-\frac{h}{v_0}})^{\gamma}=p_0(1-\frac{h}{v_0})^{-\gamma}\approx p_0+h\gamma\frac{p}{v}$

In adiabatic free expansion,$\frac{du}{dv}=0$

Hence for +h

$\frac{1}{\gamma-1}p_0v_0-0(+h)=\frac{1}{\gamma-1}p_1(v_0+h)$

while for -h

$\frac{1}{\gamma-1}p_0v_0-0(-h)=\frac{1}{\gamma-1}p_1(v_0-h)$

which implies pv=constant

I think there a physics question in physics book long ago with the question an insulated cylinder contain gas system on left and right which is isolated by a partition, the pressure on the left side is higher than pressure on right, need to find the pressure when both side come to the same pressure. The book actually solve it using conservation of internal energy and taking adiabatic expansion on left side and compression on right side, I think this is not true? First for left partition to expand adiabatically, it must do a work which is same as its instantaneous pressure p but on the right side is different pressure than $p$, so it is not possible right? Also if both side happen adiabatically, the change of internal energy at the left partition will not be the same as right partition?

Answer 1

I think everyone is familiar with the equation $pv^{\gamma}=constant$ and how it was derived.

Just to make sure you understand, this equation only applies to an ideal gas and only to a reversible adiabatic process.

But does everyone actually know $\frac{du}{dv}=-p$ which the p here actually mean instantaneous pressure exert by the gas system to the surrounding?

This relationship comes from combining the first law and an adiabatic process.

$$du=\delta q-\delta w$$

$$\delta q=0$$

$$\delta w=pdv$$

$$du=-pdv$$

So for an adiabatic process the change in internal energy equals the work done on or by the system. Note, however, that this need not be a reversible adiabatic process described that only applies for your first equation. It could be an irreversible adiabatic process. For an irreversible adiabatic process the pressure $p$ that performs work is the external pressure, not the gas pressure which is not defined for the irreversible process. So you can't say it is the pressure exerted by the gas unless you stipulate it is a reversible adiabatic process.

This can be prove by consider the mechanism change in internal energy which is $\frac{1}{\gamma-1}pv$ is equal to $-p$

CAUTION. WARNING. Again, this only applies to a reversible adiabatic process and an ideal gas.

I will not comment on the mathematical gymnastics that follows this because it, again, is limited to a reversible adiabatic process involving an ideal gas.

In adiabatic free expansion,$\frac{du}{dv}=0$.. etc., etc., ...which implies pv=constant

The conclusion that for an adiabatic free expansion, which is an irreversible process, $pv=constant$, is incorrect. For an ideal gas $p_{f}v_{f}=p_{i}v_{i}$ only means the product of the initial equilibrium pressure and volume equals the product of the final equilibrium pressure and volume, not that the product is constant throughout the process. It also only means that the final and initial temperatures are the same ($T_{f}=T_{i}$) for an ideal gas. It does not mean that the temperature is constant throughout the expansion.

The error you are making is your conclusion was derived by using the initial equation for a reversible adiabatic process and applying it to an adiabatic free expansion, which is an irreversible process.

I think there a physics question in physics book long ago with the question an insulated cylinder contain gas system on left and right which is isolated by a partition, the pressure on the left side is higher than pressure on right, need to find the pressure when both side come to the same pressure.

What is being described is not a "free adiabatic expansion". For it to be a free adiabatic expansion the right side would need to be a vacuum, not simply at a lower pressure than the pressure on the left.

....Also if both side happen adiabatically, the change of internal energy at the left partition will not be the same as right partition?

Any loss/gain of internal energy on one side of the partition must equal the gain/loss of internal energy on the other side of the partition, since the total change in internal energy is necessarily zero. That is because you have insulated cylinder, for which there can be no heat transfer to or from the system (gas). And because, assuming the cylinder is rigid, there can be no boundary work done on or by the system (gas). From these the first law dictates that the total change in internal energy of the system (gas) has to be zero. Any work done by (or on) one part of the gas on (or by) another part of the gas is internal work and does not alter the internal energy of the gas as a whole.

Yeah thank for correcting the error for adiabatic free expansion, I made an error as demonstrate above using "-h" which is not true. I see do you mean 𝑑𝑢/𝑑𝑣=0 is not true along the process? Is more sensible to consider initial internal energy=final internal energy of gas at equilibrium ... (and) $p_{i}v_{i}=p_{f}v_{f}$?

The internal energy is the same throughout the process since it is the sum of the molecular kinetic energies of all the particles, which does not change during the process, even though the distribution of the kinetic energies of the particles within the gas changes. $pv=$constant is not true throughout the process as that implies the temperature of the gas is constant throughout the process which is not true. There is no one temperature during the expansion. There are temperature (and pressure) gradients occurring during the process. However, the initial and final equilibrium temperatures are the same for the ideal gas which, together with the ideal gas law, means $p_{i}v_{i}=p_{f}v_{f}$.

I think I still have some confusion regarding reversible and irreversible adiabatic process so give me sometime to understand....

OK. Take the time.

Hope this helps.