Question

A commercial cylinder contains $ 6.01{\text{ }}{{\text{m}}^{\text{3}}} $ of $ {O_2} $ at $ 15.18{\text{MPa}} $ and $ 21^\circ C $ . The critical constants of $ {O_2} $ are $ {{\text{T}}_{\text{c}}} = - 118.4^\circ {\text{C,}}{{\text{P}}_{\text{c}}} = 50.1{\text{atm}} $ . Determine the reduced pressure and reduced temperature for $ {O_2} $ under these conditions.
(A) $ {P_r} = 2.99,{T_r} = 1.90 $
(B) $ {P_r} = 1.9,{T_r} = 2.99 $
(C) $ {P_r} = 5.98,{T_r} = 3.80 $
(D) None of these

Answer 1

Hint: In thermodynamics, reduced properties of a substance (preferably a fluid) are a set of properties that determine the state of a system at a particular time that are measured by the fluid’s state properties at its critical point. The reduced state of these variables is the base for the van der Waals theorem of corresponding states that all the state variables involved in describing the nature and state of a fluid have the same compressibility factor $ Z $ . And the compressibility factor describes the deviation of these fluids from ideal gas behaviour.

Formulas used: We will be using the formula to determine the reduced pressure that is given by, $ {P_r} = \dfrac{P}{{{P_c}}} $ where $ {P_r} $ is the reduced pressure of the system, $ P $ is the current pressure of the fluid system and $ {P_c} $ is the critical pressure experienced by the fluid system. We will also be using the formula to determine the reduced temperature behaviour that is given by, $ {T_r} = \dfrac{T}{{{T_c}}} $ where $ {T_r} $ is the reduced temperature behaviour of the fluid system, $ T $ is the current temperature of the system and $ {T_c} $ is the critical temperature experienced by the system.

Complete Step by Step answer
The Van Der Waals’ theorem of corresponding states says that there will be a minor deviation in the ideal gas behaviour exhibited by fluids in some conditions. This deviation can be computed by a factor called compressibility factor, $ Z $ .
The reduced form of van der Waals equation contains the reduced state variables like $ {P_r},{T_r},{V_r} $ . In the given problem you have the current state features of the fluid and the critical state variables of the fluid. To find the reduced pressure and reduced temperature we can use the formulas,
 $ {P_r} = \dfrac{P}{{{P_c}}} $ for reduced pressure and $ {T_r} = \dfrac{T}{{{T_c}}} $ for reduced pressure.
We know that $ {P_c} = (50.1 \times 0.1){\text{MPa}} $ and $ P = 15.18{\text{MPa}} $ . Thus, reduced pressure $ {P_r} $ is given by
 $ {P_r} = \dfrac{{15.18}}{{\left( {50.1 \times 0.1} \right)}} = \dfrac{{15.18}}{{5.01}} $
 $ \Rightarrow {P_r} = 2.99 $
We also know that $ {T_c} = 273 + ( - 118.4) = 154.6K $ and $ T = 21 + 273 = 294K $ .Thus reduced temperature, $ {T_r} $ is given by,
 $ {T_r} = \dfrac{{294}}{{154.6}} $
 $ \Rightarrow {T_r} = 1.90 $
Thus, reduced pressure is $ {P_r} = 2.99 $ and the reduced temperature is $ {T_r} = 1.90 $ .
Hence option A is correct.

Note
Although the reduced pressure is a measure of deviation of the pressure it is a ratio of the actual pressure to the pressure at critical conditions, hence the reduced pressure has no units. Similarly reduced temperature also has no units.