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.