2.1. Isothermal Process and Perfect Gas

Now, we are going to do perfect gas at constant temperature that is isothermal process, we start with the same old equation. And where gamma is a function of P. So, because in gas it is PV = nRT that is the thumb rule for isothermal. So, it can be written as n*M /V or in other terms we can write it as pM / RT, M is the molecular mass which we have seen in lecture number 1.

An isothermal process is a thermodynamic process in which the temperature of the system remains constant. For perfect gases, this can be expressed using the equation PV = nRT, where P is pressure, V is volume, n is the amount of gas in moles, R is the universal gas constant, and T is the absolute temperature. Here, M represents the molecular mass of the gas. This means that if you know the number of moles and the temperature, you can calculate the product of pressure and volume.

Think of a balloon filled with air. If you squeeze the balloon (decreasing the volume), the air inside gets compressed, increasing the pressure. However, if you then let the balloon return to its original shape without adding heat, the temperature of the air doesn’t change significantly, making the process isothermal.

So, dp will be , so this is. Correct? Because we already seen here, we just multiplied the density with G here. So, now, we start simply integrating from P 1 to P 2, and z1 to z2 simple integration, and if you integrate this, we are going to get because it is dp / p. So, that becomes ln (p2/p1) and dz is a simple integration yielding (Z 2 - Z 1).

In isothermal processes, the relationship between pressure and height can be derived through integration. By considering the equation for pressure as a function of height, you can see that as you integrate from an initial pressure P1 to a final pressure P2, the equation simplifies to involve logarithms. This gives a clear way to calculate how the pressure changes with height in an isothermal environment.

Imagine a sealed container of air. If you were to lift this container up a mountain, the pressure inside would change with the height due to the decrease in external atmospheric pressure. Using the integration we discussed, you could mathematically predict how much pressure drops as you ascend.

So, the pressure at point 2 in a constant temperature will be given as 𝑝 = 𝑝 e−[𝑀𝑔 𝑅𝑎 𝑇𝑠𝑔 (𝑍2−𝑍1)].

This equation shows how the pressure p at height Z2 is related to the pressure p1 at height Z1. The exponential part represents how gravity and the temperature of the gas influence the pressure. Essentially, it indicates that as you go higher (i.e., Z2 increases), the pressure decreases exponentially due to gravitational effects on the gas molecules.

Think of a hot air balloon. As the balloon rises, the pressure of the air inside decreases. The temperature needs to be maintained to keep the balloon floating. If the air inside cools too much, it will start to drop because the pressures equalize.

Pressure measurement devices, we have discussed one already barometers, there are manometers, a standard manometer or a differential manometer and the pressure transducers.

There are several instruments to measure pressure. A barometer is used to measure atmospheric pressure, while manometers can measure pressure relative to atmosphere or the pressure difference between two points. These devices employ different liquids to gauge pressure, such as water or mercury, each selected based on density and measurement accuracy.

Consider a barometer as a weather forecasting tool. When you check the barometer and see it drop, it's likely that rain is on the way because low atmospheric pressure typically signals weather changes.

For standard manometers, what is the pressure? So how it works is very simple example. So, what is if we are asked, what is the pressure at A given the height, more importantly we should mention what is the gauge pressure.

When measuring pressure in contexts like water distribution systems, gauge pressure is commonly used, which excludes atmospheric pressure. Calculating pressure involves knowing either the height of the liquid column in the manometer or using fluid density. This way, you can determine how high a column of liquid like water can rise based on the pressure applied.

Imagine a simple water fountain. The pressure driving the water up through the fountain changes based on the depth of the water source and any restrictions in the piping. Understanding these pressures allows you to design a fountain that works correctly without overflows or underwhelming spray heights.

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