Isobaric transformation. Isobaric transformation of gases

An isobaric transformation occurs when the gas is at a constant pressure. For example, if done in an open environment, the transformation will be isobaric as the pressure will be atmospheric pressure that will not change.

In this case, temperature and volume vary. Two leading scientists have studied how this variation in isobaric transformations occurs. The first to relate the volume and temperature of gases was Jacques Charles (1746-1823), in 1787, and then, in the year 1802, Joseph Gay-Lussac (1778-1850) quantified this relationship.

Thus, a law emerged that explains the isobaric transformations of gases, which became known as the Charles/Gay-Lussac Law. It is stated as follows:

"In a constant pressure system, the volume of a fixed mass of a gas is directly proportional to the temperature."

This means that if we double the temperature, the volume occupied by the gas will also double. On the other hand, if we decrease the temperature, the gas volume will also decrease in the same proportion.

This can be seen in a very simple experiment. If we place a balloon in the neck of a bottle, a fixed mass of air will be trapped. If we dip this bottle into a bowl of ice water, the balloon will deflate. Now, if we put it in a bowl of hot water, the balloon will fill.

This is because as the temperature increases, the kinetic energy of the gas molecules increases and the speed at which they move also increases. Thus, the gas expands, increasing the volume it occupies, and the balloon inflates. The opposite occurs when we lower the temperature, putting it in cold water.

This relationship between temperature and volume in isobaric transformations is given by the following relationship:

V = k
T

"k" is a constant, as can be seen in the following graph:

Note that the V/T ratio always gives a constant:

_V_ =_2V_ = _4V_
100 200 400

Thus, we can establish the following relationship for isobaric transformations:

V_initial = V_Final
T_initial  T_Final

This means that when there is any change in the temperature of the gas at constant pressure, we can find out its volume through this mathematical expression. The opposite is also true, knowing the volume of the gas, we find out what temperature it is at. See an example:

"A gaseous mass occupies a volume of 800 cm³ at -23°C, at a given pressure. What is the temperature recorded when the gaseous mass, at the same pressure, occupies a volume of 1.6 L?”

Resolution:

Data:

V_initial = 800 cm³
T_initial = -23 ºC, adding to 273 we have 250 K (Kelvin)
V_Final = 1.6 L
T_Final = ?

* First we have to leave the volume on the same unit. It is known that 1 dm³ equals 1 liter. like a 1 dm³ is the same as 1000 cm³, it appears that 1 liter = 1 000 cm³:

1 L 1000 cm³
x 800 cm³
x = 0.8 L

* Now we replace the formula values ​​and find the final temperature value:

V_initial = V_Final
T_initial  T_Final
0,8_ = 1,6
250 T_Final
0.8 T_Final = 250. 1,6
T_Final = 400
0,8
T_Final = 500K

* Moving to the Celsius scale, we have:

T (K) = T (°C) + 273
500 = T (°C) + 273
T (°C) = 500 - 273

T (°C) = 227°C

By Jennifer Fogaça
Graduated in Chemistry

Source: Brazil School - https://brasilescola.uol.com.br/quimica/transformacao-isobarica.htm