Question

The equation for the state of real gas in terms of reduced form is

• A. \( (P_r + \frac{3}{V_r^2})(3V_r - 1) = 8T_r \)
• B. \( (P_r + \frac{3}{V_r})(3V_r - 1) = 8T_r \)
• C. \( (P_r - \frac{3}{V_r^2})(3V_r + 1) = 8T_r \)
• D. \( (P_r - \frac{3}{V_r})(3V_r - 1) = 8T_r \)

Hint:

The reduced form of the van der Waals equation of state is a universal equation for real gases. It is derived by expressing pressure, volume, and temperature as fractions of their critical properties. The key is to remember the relations between the van der Waals constants (\( a \) and \( b \)) and the critical properties (\( P_c, V_c, T_c \)).

Answer 1

The Correct Option is A

Step 1: Understand the concept of the reduced equation of state for real gases.
Real gases do not perfectly follow the ideal gas law \( PV=nRT \), especially at high pressures and low temperatures. Various equations of state have been developed to model the behavior of real gases. One such equation is the van der Waals equation. The "reduced form" of an equation of state expresses pressure, volume, and temperature as ratios to their respective critical values (critical pressure \( P_c \), critical volume \( V_c \), and critical temperature \( T_c \)). These ratios are called reduced properties: \[ P_r = \frac{P}{P_c}, \quad V_r = \frac{V}{V_c}, \quad T_r = \frac{T}{T_c} \]
Step 2: Recall the van der Waals equation of state.
The van der Waals equation for a real gas is: \[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \] where \( a \) and \( b \) are van der Waals constants specific to each gas, and \( V \) is the molar volume.
Step 3: Express van der Waals constants in terms of critical properties.
The van der Waals constants \( a \) and \( b \) can be related to the critical properties as follows: \[ a = \frac{27 R^2 T_c^2}{64 P_c} \] \[ b = \frac{RT_c}{8P_c} = \frac{V_c}{3} \] From \( b = \frac{V_c}{3} \), we have \( V_c = 3b \).
Step 4: Substitute the expressions for \( P, V, T \) in terms of reduced properties and van der Waals constants.
Substitute \( P = P_r P_c \), \( V = V_r V_c \), and \( T = T_r T_c \) into the van der Waals equation: \[ \left(P_r P_c + \frac{a}{(V_r V_c)^2}\right)(V_r V_c - b) = R T_r T_c \] Now, substitute \( P_c = \frac{a}{27b^2} \), \( V_c = 3b \), and \( T_c = \frac{8a}{27Rb} \): \[ \left(P_r \frac{a}{27b^2} + \frac{a}{(V_r 3b)^2}\right)(V_r 3b - b) = R T_r \frac{8a}{27Rb} \] \[ \left(P_r \frac{a}{27b^2} + \frac{a}{9V_r^2 b^2}\right)(b(3V_r - 1)) = T_r \frac{8a}{27b} \] Factor out \( \frac{a}{9b^2} \) from the first parenthesis: \[ \frac{a}{9b^2} \left(P_r \frac{9}{27} + \frac{1}{V_r^2}\right) b(3V_r - 1) = T_r \frac{8a}{27b} \] \[ \frac{a}{9b} \left(\frac{P_r}{3} + \frac{1}{V_r^2}\right) (3V_r - 1) = T_r \frac{8a}{27b} \] Multiply both sides by \( \frac{27b}{a} \): \[ \frac{27b}{a} \frac{a}{9b} \left(\frac{P_r}{3} + \frac{1}{V_r^2}\right) (3V_r - 1) = T_r \frac{8a}{27b} \frac{27b}{a} \] \[ 3 \left(\frac{P_r}{3} + \frac{1}{V_r^2}\right) (3V_r - 1) = 8T_r \] \[ \left(P_r + \frac{3}{V_r^2}\right) (3V_r - 1) = 8T_r \] This is the generalized compressibility chart equation for real gases, also known as the van der Waals equation in reduced form. The final answer is $\boxed{\text{1}}$.