Bernoulli's principle at molecular level

Question

I've seen many questions and answers about this topic, which keep saying that pressure decrease is the cause of acceleration of flow, and don't explain where that pressure drop comes from in the first place, while others say that pressure drop is just due to energy conservation as the flow speeds up.

My question is regarding ideal gases at low speeds. So, what is the actual reason that causes a drop in pressure (maybe explain it at molecular level), because if pressure is calculated from this relation: $PV=nRT$, and we don't observe a decrease in any of those factors, that would cause a drop in pressure. Also, given that pressure of a gas is due to collision of molecules with the walls and themselves, how does pressure decrease in terms of collision between surface and themselves?

Also, why don't we observe a drop in temperature if, in free gas expansion, the temperature drops momentarily as it gets turned to kinetic energy of flow, and then temperature returns back to normal as the kinetic energy of flow converts back. Why doesn't the same happen in this case?

Answer 1

if pressure is calculated from the this relation : $PV=nRT$ and we don't observe a decrease in any of those factors that would cause a drop in pressure

This statement is not true. $P$, $V$ do change, and $T$ may change as well. Let's see why:

The classical example of accelerated flow is when a fluid (make it an ideal gas for simplicity) flows adiabatically through a converging nozzle:

The energy balance between inlet (1) and outlet (2) is $$ \dot m h_1 + \dot m\frac{v_1^2}{2} + \dot m g z_1 + \dot Q = \dot m h_2 + \dot m\frac{v_2^2}{2} + \dot m g z_2, $$ where $\dot m$ is the constant mass flow rate through the entire nozzle and $\dot Q$ is the heat transferred from the surroundings, if any. With $z_1=z_2$ this becomes $$ \Delta h_{12} + \frac{\Delta v_{12}^2}{2} - \frac{\dot Q}{m} = 0 $$ with all deltas calculated as (outlet) $-$ (inlet). Since $\Delta v_{12}^2 > 0$ (the velocity is faster at the exit of the nozzle) we must have $$ \Delta h_{12} - \frac{\dot Q}{\dot m} < 0 . $$

• Reversible adiabatic flow In this case $\dot Q = 0$ and states 1 and 2 have the same entropy. Then $\Delta h_{12}<0$, which for an ideal gas means $T_2<T_1$ and since inlet and outlet have the same entropy we also have $$ \frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^{C_P/R} \Rightarrow \boxed{P_2 < P_1} $$

• Reversible isothermal flow If the flow is isothermal, i.e., by placing the nozzle inside a bath with temperature $T$, then $\Delta h_{12} = 0$ and $Q>0$. This means that the gas is being heated but for its temperature to stay constant its pressure must go down, so that the heat that enters the gas, instead of increasing the temperature, it is converted into the work to expand the gas to lower pressure. We must have then $$ \boxed{P_2 < P_1}, \quad V_2> V_1,\quad P_1 V_1 = P_2 V_2 $$

Summary In both cases pressure decreases. In the adiabatic case $P$, $T$ and $V$ all change, while in the isothermal case only $P$ and $V$ change, while $T$ is constant.

Answer 2

Suppose that an ideal gas is flowing through a pipe, and there is a reduction of section at some point. If the velocity of the gas is greater after this point, there is a region of transition where the gas accelerates. The acceleration corresponds to a net force in the direction of the acceleration, that is the direction of the flow, according to the second law of Newton.

The net force in the direction of the flow is the product of the difference of pressure on a fluid slice by the cross section of the slice. Without this difference, the gas can not accelerate.

According to the equation for ideal gases, a decrease of pressure corresponds to a decrease of temperature and/or density. In both cases it is intuitive to understand that the number of collision (pressure) upstream is greater than downstream.