Equation of State for Gases (Clapeyron Equation)

THE clapeyron equation, also known as perfect gas state equation or yet general gas equation, created by the Parisian scientist Benoit Paul Emile Clapeyron (1799-1864), is shown below:

for. V = n. A. T

Being that:

p = gas pressure;

V = gas volume;

n = amount of matter in the gas (in moles);

T = gas temperature, measured on the kelvin scale;

R = universal constant of perfect gases.

But how did you arrive at this equation?

well in the text General gas equation, it is shown that when a fixed mass of a gas undergoes transformation into its three fundamental quantities, which are pressure, volume and temperature, the relationship below remains constant:

for_initial. V_initial = for_Final. V_Final
T_initial T_Final

or

for. V = constant
T

This constant, however, is proportional to the amount of matter in the gas, so we have:

for. V = n .constant
T

Passing the temperature to the other member, we have:

for. V = n. constant. T

This is the equation of state for perfect gases proposed by Clapeyron.

Italian chemist Amedeo Avogadro (1776-1856) proved that

equal volumes of any gases, which are at the same temperature and pressure conditions, have the same number of molecules. Thus, 1 mole of any gas has always the same amount of molecules, which is 6,0. 10²³ (number of Avogadro). This means that 1 mole of any gas also always occupies the same volume, which, in the Normal Conditions of Temperature and Pressure (CNTP), in which the pressure is equal to 1 atm and the temperature is 273 K (0°C), is equal to 22.4L.

With this data in hand, we can figure out the value of the constant in the equation above:

for. V = n. constant. T
constant = for. V
n. T

constant = 1 atm. 22.4 L
1 mol. 273K

constant = 0.082 atm. L. mol^-1. K^-1

Thus, this value is now defined as the universal gas constant and it also came to be symbolized by the letter R.

Under different conditions, we have:

R = PV = 760 mmHg. 22.4 L = 62.3 mmHg. L/mol. K
nT 1 mol. 273.15K

R = PV = 760 mmHg. 22 400 ml = 62 300 mmHg. mL/mol. K
nT 1 mol. 273.15K

R = PV = 101 325 Pa. 0.0224 m³ = 8,309 Pa.m³/mol. K
nT 1 mol. 273.15K

R = PV = 100,000 Pa. 0.02271 m³ = 8,314 Pa.m³/mol. K
nT 1 mol. 273.15K

We can then solve problems involving gases under ideal conditions using the Clapeyron equation, as it holds for any type of situation. However, it is important to emphasize that careful attention should be paid to the units that are used to apply the correct value for the universal gas constant, R.

Furthermore, since the amount of matter can be determined by the formula:

n = pasta → n = m
molar mass M

we can substitute “n” in the Clapeyron equation and obtain a new equation that can be used in cases where the value of the number of moles of the gas is not directly provided:

for. V = m . A. TM

By Jennifer Fogaça
Graduated in Chemistry

Source: Brazil School - https://brasilescola.uol.com.br/quimica/equacao-estado-dos-gases-equacao-clapeyron.htm