Question

A commercial cylinder contains 6.91\[{{m}^{3}}\] oxygen at 15 Pa and 21℃. The critical constants for \[{{O}_{2}}\] are Tc = -126 ℃, Pc = 50 bar. Determine the reduced pressure and reduced temperature for \[{{O}_{2}}\] under these conditions:
a.) \[{{P}_{r}}\]= 2, \[{{T}_{r}}\]= 3
b.) \[{{P}_{r}}\]= 3, \[{{T}_{r}}\]= 2
c.) \[{{P}_{r}}\]= 6, \[{{T}_{r}}\]= 4
d.) \[{{P}_{r}}\]= 4, \[{{T}_{r}}\]= 6

Answer 1

Hint: The physical constants that express the property of a substance in its critical state are called critical constants. The critical constants of gas are temperature, pressure and volume.

Complete step by step solution:

Gases cannot be liquified unless its temperature is below a certain value depending upon the nature of the gas. This temperature is called critical temperature.
Critical temperature (\[{{T}_{c}}\]) is the maximum temperature at which a gas can be liquified, above which liquid does not exist.
Critical pressure (\[{{P}_{c}}\]) is maximum pressure required to cause liquefaction.
Critical volume (\[{{V}_{c}}\]) is the volume occupied by one mole of gas at critical temperature.
So, in this case the critical constants for critical temperature and pressure are -126 ℃, and 50 bar.
The reduced pressure \[{{P}_{r}}=\dfrac{Given\,pressure}{Critical\,pressure}\]
\[\dfrac{P}{{{P}_{C}}}\]= \[\dfrac{15Mpa}{50\,\times 0.1Mpa}\] = 3
The reduced temperature \[{{T}_{r}}=\,\dfrac{Given\,temperature}{Critical\,temperature}\]
\[\dfrac{T}{{{T}_{c}}}\]= \[\dfrac{21\,+\,273}{273-126}\]= 2
So, from the above solution we can conclude that option (b) is the correct answer.

Note: The constant of the Vander Waals equation provides a correction for the intermolecular forces whereas constant b adjusts for the volume occupied by the gas particles. It is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules. Always remember to change the units in their SI form.