# Bernoulli's principle at molecular level

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?

• Where are you seeing that we "don't observe a drop in temperature"? Commented Sep 18, 2023 at 23:33

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 $$T_2 and since inlet and outlet have the same entropy we also have $$\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.

• Thanks.Could you add another answer for ideal gases flowing in tubes instead of nozzles where the flow continues in the tube without an exit and explain pressure and temperature in the lower cross sectional area because that was what I wanted with my original question but I didn't mention it. Commented Sep 18, 2023 at 22:32
• Do you mean pressure drop due to frictional losses? This has nothing to do with acceleration that you mentioned in your original post. But I need to understand if you are talking about a pipe with (i) constant diameter or (ii) a pipe whose diameter changes. My answer above applies to (ii), just imagine that at the end of the nozzle the flow continues inside a pipe with constant diameter equal to the small diameter of the nozzle. Commented Sep 18, 2023 at 22:39
• It's in the math: We are adding heat to the gas, but we are also keeping its temperature constant. The only way to do that is if we use the heat to do work. And since this work cannot be done at the expense of temperature, it must be done at the expense of pressure. Add heat to a gas in a cylinder with a movable piston while keeping its $T$ fixed to see how this works. Commented Sep 18, 2023 at 22:53
• No I meant changing diameter. But you say the gas expands (in case of isothermal flow) while doing work so that its temperature doesn't rise but what is it doing work against and how does it expand if the volume is lower due to lower cross sectional area Commented Sep 18, 2023 at 22:56
• Sorry I actually didn't notice that but I already deleted the comment but you already viewed it Commented Sep 18, 2023 at 22:58

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.

• Okay but you haven't told me yet why would that pressure decrease in the first place to cause the pressure difference Commented Sep 18, 2023 at 22:24
• Well, that results from the Newton's second law. Net force and acceleration come always together. In this case, if the gas accelerates from left to right, $F_{net} = F_L - F_R = S(P_L - P_R)$ on an elementary slice. Commented Sep 18, 2023 at 22:41