Question

The compression and expansion of gases from the basis of how air is cooled by air conditioners. Suppose \[1.55L\] of an ideal gas under \[6.38\] atm of pressure at \[20.5^\circ C\] is expanded to \[6.95L\] at \[1.00\] atm. What is the new temperature?

Answer 1

Hint:Gas is portrayed as a homogeneous liquid with low thickness and low viscosity, and it is accepted that the volume of the gas is equivalent to the vessel volume. The two kinds of gases are there: 1) Ideal gas and 2) Non-ideal gas.

Complete solution:
The pressure, temperature, and volume of the gas will change while going from its underlying state to its current state, which discloses to us that we should utilize the consolidated gas law to locate the new temperature of the gas.
As we know that the number of moles are constant in this process so from combined gas law the relationship between pressure, volume and temperature are as follows:
\[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
\[{P_1}\]\[ = \]Pressure of the gas at initial state
\[{V_1}\]\[ = \]Volume of the gas at initial state
\[{T_1}\]\[ = \]Temperature of the gas at initial temperature
\[{P_2}\]\[ = \]Pressure of the gas at final state
\[{V_2} = \]Volume of the gas at final state
\[{T_2} = \]Temperature of the gas at final state
The most important point we must remember during the course of using this reaction would be we need to convert the temperature from Celsius to kelvin.
From the above form of the reaction, by rearranging it we can write:
\[{T_2} = \dfrac{{{P_2}}}{{{P_1}}}.\dfrac{{{V_2}}}{{{V_1}}}.{T_1}\]
Now we need to put all the given values above in the equation in order to get the final temperature.
\[{T_2} = \dfrac{{1.00atm}}{{6.38atm}}.\dfrac{{6.95L}}{{1.55L}}(273.15 + 20.5)K\]
\[{T_2} = 206.38K\]
\[{T_2}\]\[[^\circ C] = 206 - 273.15 = - 67.2^\circ C\]

Note: In an ideal gas, all the crashes between particles or atoms are totally flexible and no intermolecular power of attraction exists in ideal gas on the grounds that the atoms of an ideal gas move so quick, and they are so distant from one another that they don't interact by any means. On account of real gas, they have irrelevant intermolecular attractive forces. Ideal gas doesn't exist normally.