Welcome to Lecture 101 of our comprehensive JEE Physics Master Series. In this lecture, we examine the physics of Real Gases, formulate the van der Waals Equation of State from microscopic intermolecular interactions, analyze the continuity of state across Andrews’ experimental $P$-$V$ isotherms, and derive the Critical Constants ($P_c, V_c, T_c$), Boyle temperature, and inversion temperature from first principles.
1. Microscopic Deviations from Ideal Behavior
The ideal gas equation of state ($P V = n R T$) assumes two fundamental postulates that break down under high pressure and low temperature:
- Zero Molecular Volume: Real gas molecules possess finite, non-zero hard-sphere volumes. Consequently, the actual volume accessible for free thermal motion is less than the container volume $V$.
- Zero Intermolecular Forces: Real molecules experience attractive van der Waals dispersion forces at intermediate separations ($r \sim 1\text{ to }2\text{ nm}$) and strong repulsive Pauli exclusion forces at close contact ($r \le \sigma$).
2. Derivation of the van der Waals Equation
Johannes Diderik van der Waals (1873) introduced two deterministic empirical corrections to the ideal gas law:
1. Volume Correction (Co-volume $b$):
For two spherical molecules of diameter $d$, the center-to-center distance cannot be less than $d$. The excluded volume per pair is $\frac{4}{3}\pi d^3 = 8 v_{\text{mol}}$ (where $v_{\text{mol}} = \frac{4}{3}\pi (d/2)^3$). The excluded volume per molecule is therefore $4 v_{\text{mol}}$. For $n$ moles ($N = n N_A$ molecules):
$$b = 4 N_A \left(\frac{4}{3}\pi r^3\right)$$
The effective volume available for molecular motion becomes:
$$V_{\text{ideal}} = V – n b$$
2. Pressure Correction (Internal Cohesive Pressure $P_{\text{int}}$):
A molecule inside the bulk fluid experiences isotropic attraction with zero net force. However, as a molecule strikes the container wall, it experiences an inward net cohesive pull from the interior molecules, reducing its momentum transfer. The inward force is proportional to both the concentration of impacting molecules and the concentration of attracting interior molecules:
$$P_{\text{int}} \propto \left(\frac{n}{V}\right) \times \left(\frac{n}{V}\right) = \frac{a n^2}{V^2}$$
where $a$ characterizes the strength of attractive intermolecular forces. The measured wall pressure $P$ is thus lower than the ideal pressure $P_{\text{ideal}} = P + \frac{a n^2}{V^2}$.
Combining both corrections yields the van der Waals equation of state:
$$\left(P + \frac{a n^2}{V^2}\right)(V – n b) = n R T$$
For one mole of gas ($n = 1$, molar volume $V_m = V/n$):
$$\left(P + \frac{a}{V_m^2}\right)(V_m – b) = R T$$
3. Compressibility Factor ($Z$) and Asymptotic Regimes
The compressibility factor is defined as $Z = \frac{P V_m}{R T}$. For an ideal gas, $Z \equiv 1$ under all conditions.
1. Low Pressure Regime ($P \to 0$, $V_m \gg b$):
The volume correction $b$ is negligible compared to $V_m$, so $(V_m – b) \approx V_m$:
$$\left(P + \frac{a}{V_m^2}\right)V_m \approx R T \implies P V_m + \frac{a}{V_m} = R T \implies Z = 1 – \frac{a}{R T V_m} \approx 1 – \frac{a P}{(R T)^2}$$
Here $Z < 1$, indicating that attractive forces dominate and the gas is more compressible than an ideal gas.
2. High Pressure Regime ($P \to \infty$):
The pressure term $P \gg \frac{a}{V_m^2}$, so internal attraction is negligible compared to repulsive exclusion:
$$P (V_m – b) = R T \implies P V_m – P b = R T \implies Z = 1 + \frac{P b}{R T}$$
Here $Z > 1$, showing that repulsive forces dominate and the gas is less compressible than an ideal gas.
3. Extremely High Temperatures ($T \gg T_B$):
Thermal kinetic energy dominates attractive potential wells ($R T \gg a/V_m$), making $Z \ge 1$ across all pressures (typical for $H_2$ and $He$ at room temperature).
4. Critical State and Analytical Derivation of Critical Constants
Expanding the molar van der Waals equation into a cubic polynomial in $V_m$:
$$P V_m^3 – (P b + R T) V_m^2 + a V_m – a b = 0 \implies V_m^3 – \left(b + \frac{R T}{P}\right) V_m^2 + \frac{a}{P} V_m – \frac{a b}{P} = 0$$
At temperatures below the critical temperature ($T < T_c$), this cubic equation yields three real roots for $V_m$ within the coexistence vapor-liquid dome. As temperature rises to $T = T_c$, the three roots merge into a single threefold degenerate critical point $(P_c, V_c, T_c)$ where the isotherm exhibits a horizontal point of inflection:
$$\left(\frac{\partial P}{\partial V_m}\right)_{T_c} = 0 \quad \text{and} \quad \left(\frac{\partial^2 P}{\partial V_m^2}\right)_{T_c} = 0$$
At this critical point, the polynomial can be expressed as $(V_m – V_c)^3 = 0$:
$$V_m^3 – 3 V_c V_m^2 + 3 V_c^2 V_m – V_c^3 = 0$$
Equating coefficients with the van der Waals cubic equation at $(P_c, T_c)$:
$$3 V_c = b + \frac{R T_c}{P_c} \quad \text{— (1)}$$
$$3 V_c^2 = \frac{a}{P_c} \quad \text{— (2)}$$
$$V_c^3 = \frac{a b}{P_c} \quad \text{— (3)}$$
Dividing Eq. (3) by Eq. (2) gives the Critical Molar Volume:
$$V_c = 3 b$$
Substituting $V_c = 3b$ into Eq. (2) gives the Critical Pressure:
$$P_c = \frac{a}{3 V_c^2} = \frac{a}{3(3b)^2} = \frac{a}{27 b^2}$$
Substituting $V_c$ and $P_c$ into Eq. (1) yields the Critical Temperature:
$$3(3b) = b + \frac{R T_c}{a / (27 b^2)} \implies 8 b = \frac{27 b^2 R T_c}{a} \implies T_c = \frac{8 a}{27 R b}$$
Universal Critical Compressibility Factor:
$$Z_c = \frac{P_c V_c}{R T_c} = \frac{\left(\frac{a}{27 b^2}\right)(3b)}{R \left(\frac{8 a}{27 R b}\right)} = \frac{3}{8} = 0.375$$
This dimensionless number is independent of the chemical identity of the gas, demonstrating the universal scaling of the van der Waals framework.
5. Characteristic Thermodynamic Temperatures
1. Boyle Temperature ($T_B$):
The temperature at which a real gas obeys the ideal gas equation over a substantial range of pressures ($\lim_{P \to 0} \frac{\partial Z}{\partial P} = 0$):
$$T_B = \frac{a}{R b}$$
2. Inversion Temperature ($T_i$):
The temperature at which the Joule-Thomson coefficient $\mu_{\text{JT}} = \left(\frac{\partial T}{\partial P}\right)_H$ changes sign from cooling ($\mu_{\text{JT}} > 0$) to heating ($\mu_{\text{JT}} < 0$):
$$T_i = \frac{2 a}{R b} = 2 T_B = \frac{27}{4} T_c = 6.75 \, T_c$$
6. Solved Examples
Example 1 (Foundational – JEE Main):
The van der Waals constant $b$ for a gas is $4.42 \times 10^{-2}\text{ L/mol}$. Assuming gas molecules are spherical, estimate the radius of a single molecule of this gas. (Take Avogadro’s constant $N_A = 6.023 \times 10^{23}\text{ mol}^{-1}$).
Solution:
By definition, $b = 4 N_A v_{\text{mol}} = 4 N_A \left(\frac{4}{3}\pi r^3\right) = \frac{16}{3}\pi N_A r^3$.
Converting $b$ to SI units ($\text{m}^3/\text{mol}$):
$$b = 4.42 \times 10^{-2}\text{ L/mol} = 4.42 \times 10^{-5}\text{ m}^3/\text{mol}$$
$$r^3 = \frac{3 b}{16 \pi N_A} = \frac{3 \times 4.42 \times 10^{-5}}{16 \times 3.1416 \times 6.023 \times 10^{23}} = \frac{1.326 \times 10^{-4}}{3.027 \times 10^{25}} \approx 4.38 \times 10^{-30}\text{ m}^3$$
Taking cube roots:
$$r = (4.38 \times 10^{-30})^{1/3} \approx 1.636 \times 10^{-10}\text{ m} = \mathbf{1.64\text{ \AA}}$$
Example 2 (Critical Constants Determination – JEE Main/Advanced):
The critical temperature and critical pressure of carbon dioxide are $T_c = 304.2\text{ K}$ and $P_c = 72.8\text{ atm}$, respectively. Calculate the van der Waals constants $a$ (in $\text{atm}\cdot\text{L}^2/\text{mol}^2$) and $b$ (in $\text{L/mol}$). Take $R = 0.0821\text{ L}\cdot\text{atm/(mol}\cdot\text{K)}$.
Solution:
Using the critical equations:
$$T_c = \frac{8a}{27 R b} \quad \text{and} \quad P_c = \frac{a}{27 b^2}$$
Dividing $T_c$ by $P_c$:
$$\frac{T_c}{P_c} = \frac{8a / (27 R b)}{a / (27 b^2)} = \frac{8b}{R} \implies b = \frac{R T_c}{8 P_c}$$
Substituting values:
$$b = \frac{0.0821 \times 304.2}{8 \times 72.8} = \frac{24.975}{582.4} \approx \mathbf{0.0429\text{ L/mol}}$$
Now calculate $a$ from $P_c = \frac{a}{27 b^2}$:
$$a = 27 P_c b^2 = 27 \times 72.8 \times (0.04288)^2 = 1965.6 \times 0.001839 \approx \mathbf{3.61\text{ atm}\cdot\text{L}^2/\text{mol}^2}$$
Example 3 (Law of Corresponding States – JEE Advanced):
Define reduced variables $P_r = \frac{P}{P_c}$, $V_r = \frac{V_m}{V_c}$, and $T_r = \frac{T}{T_c}$. Derive the Reduced Equation of State from the molar van der Waals equation and explain its physical significance.
Solution:
Express state variables in terms of reduced coordinates:
$$P = P_r P_c = P_r \left(\frac{a}{27 b^2}\right), \quad V_m = V_r V_c = V_r (3b), \quad T = T_r T_c = T_r \left(\frac{8a}{27 R b}\right)$$
Substitute these into $\left(P + \frac{a}{V_m^2}\right)(V_m – b) = R T$:
$$\left(P_r \frac{a}{27 b^2} + \frac{a}{(3 b V_r)^2}\right)(3 b V_r – b) = R T_r \left(\frac{8a}{27 R b}\right)$$
$$\frac{a}{27 b^2} \left(P_r + \frac{3}{V_r^2}\right) \cdot b (3 V_r – 1) = \frac{8a}{27 b} T_r$$
Cancelling $\frac{a}{27 b}$ from both sides:
$$\mathbf{\left(P_r + \frac{3}{V_r^2}\right)(3 V_r – 1) = 8 T_r}$$
Significance (Law of Corresponding States): This equation contains no parameters ($a, b, R$) specific to any individual substance. If two different gases have the same reduced pressure and reduced temperature, they must occupy the identical reduced volume.
Example 4 (Virial Form & Boyle Temperature – JEE Advanced):
Express the molar van der Waals equation in the Virial expansion form $Z = 1 + \frac{B(T)}{V_m} + \frac{C(T)}{V_m^2} + \dots$ and explicitly determine the second virial coefficient $B(T)$. Hence, prove that $T_B = \frac{a}{R b}$.
Solution:
Rearrange the molar van der Waals equation for pressure:
$$P = \frac{R T}{V_m – b} – \frac{a}{V_m^2}$$
Multiply by $\frac{V_m}{R T}$ to obtain $Z$:
$$Z = \frac{P V_m}{R T} = \frac{V_m}{V_m – b} – \frac{a}{R T V_m} = \left(1 – \frac{b}{V_m}\right)^{-1} – \frac{a}{R T V_m}$$
For $V_m \gg b$, perform a binomial expansion $(1 – x)^{-1} = 1 + x + x^2 + \dots$ on $\left(1 – \frac{b}{V_m}\right)^{-1}$:
$$Z = \left(1 + \frac{b}{V_m} + \frac{b^2}{V_m^2} + \dots\right) – \frac{a}{R T V_m} = 1 + \left(b – \frac{a}{R T}\right)\frac{1}{V_m} + \left(\frac{b}{V_m}\right)^2 + \dots$$
Comparing with $Z = 1 + \frac{B(T)}{V_m} + \mathcal{O}\left(\frac{1}{V_m^2}\right)$, the second virial coefficient is:
$$B(T) = b – \frac{a}{R T}$$
At the Boyle temperature $T = T_B$, the second virial coefficient identically vanishes ($B(T_B) = 0$):
$$b – \frac{a}{R T_B} = 0 \implies \mathbf{T_B = \frac{a}{R b}}$$
7. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): The dimensions of the van der Waals parameter $a$ are equivalent to:
(A) $[\text{M}\text{L}^5\text{T}^{-2}\text{mol}^{-2}]$
(B) $[\text{M}\text{L}^2\text{T}^{-2}\text{mol}^{-1}]$
(C) $[\text{M}\text{L}^{-1}\text{T}^{-2}\text{mol}^{-1}]$
(D) $[\text{L}^3\text{mol}^{-1}]$
Q2 (Single Choice): At moderately low pressure, the compressibility factor $Z$ of one mole of a van der Waals gas is best approximated by:
(A) $1 + \frac{P b}{R T}$
(B) $1 – \frac{a}{R T V_m}$
(C) $1 + \frac{a}{R T V_m}$
(D) $\frac{R T}{P b}$
Q3 (Single Choice): For a van der Waals gas, the critical molar volume $V_c$ is related to the co-volume constant $b$ by:
(A) $V_c = b$
(B) $V_c = 2b$
(C) $V_c = 3b$
(D) $V_c = \frac{b}{3}$
Q4 (Single Choice): If the critical temperature $T_c$ of a gas is $160\text{ K}$, its inversion temperature $T_i$ is:
(A) $320\text{ K}$
(B) $540\text{ K}$
(C) $1080\text{ K}$
(D) $80\text{ K}$
Q5 (Single Choice): For hydrogen and helium gases at room temperature ($300\text{ K}$), the value of the compressibility factor $Z$ at all positive pressures is:
(A) Strictly less than 1
(B) Always greater than 1
(C) Equal to 1 up to $100\text{ atm}$
(D) Zero
Q6 (Single Choice): The ratio of Boyle temperature $T_B$ to critical temperature $T_c$ for any van der Waals gas is:
(A) $\frac{8}{27}$
(B) $\frac{27}{8}$
(C) $2$
(D) $\frac{3}{8}$
Q7 (One or More than One Correct): Which of the following statements regarding a van der Waals gas are strictly TRUE?
(A) The critical compressibility factor is a universal constant equal to $0.375$.
(B) Above the critical temperature $T_c$, a gas cannot be liquefied, no matter how high the applied pressure.
(C) At the inversion temperature $T_i$, the Joule-Thomson expansion produces neither cooling nor heating.
(D) The constant $b$ is equal to the actual volume of the gas molecules.
Q8 (One or More than One Correct): At the critical point of a fluid, which of the following mathematical conditions must be satisfied?
(A) $\left(\frac{\partial P}{\partial V_m}\right)_{T} = 0$
(B) $\left(\frac{\partial^2 P}{\partial V_m^2}\right)_{T} = 0$
(C) $\left(\frac{\partial^2 P}{\partial T^2}\right)_{V_m} = 0$
(D) $\left(\frac{\partial P}{\partial T}\right)_{V_m} = \frac{R}{2b}$
Q9 (Numerical Value): The critical temperature of a real gas is $T_c = 40\text{ K}$ and its critical pressure is $P_c = 20\text{ atm}$. The Boyle temperature of the gas in Kelvin is ________.
Q10 (Numerical Value): One mole of a real gas obeys the equation $P(V_m – b) = R T$. A plot of compressibility factor $Z$ versus $P$ at $T = 300\text{ K}$ gives a straight line with a slope equal to $2.0 \times 10^{-3}\text{ atm}^{-1}$. If $R = 0.0821\text{ L}\cdot\text{atm/(mol}\cdot\text{K)}$, the value of the van der Waals constant $b$ in $\text{L/mol}$ (rounded to two decimal places) is ________.
8. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (A) | Q2: (B) | Q3: (C) | Q4: (C) | Q5: (B) | Q6: (B) | Q7: (A, B, C) | Q8: (A, B, D) | Q9: 135 | Q10: 0.05
Detailed Solutions
Solution 1:
In $\left(P + \frac{a n^2}{V^2}\right)$, the term $\frac{a n^2}{V^2}$ has dimensions of pressure $[P] = [\text{M}\text{L}^{-1}\text{T}^{-2}]$.
$$[a] = \frac{[P][V^2]}{[n^2]} = \frac{[\text{M}\text{L}^{-1}\text{T}^{-2}][\text{L}^6]}{[\text{mol}^2]} = [\text{M}\text{L}^5\text{T}^{-2}\text{mol}^{-2}]$$
Hence, (A) is correct.
Solution 2:
At low pressure, $V_m \gg b \implies (V_m – b) \approx V_m$.
$$\left(P + \frac{a}{V_m^2}\right)V_m = R T \implies P V_m + \frac{a}{V_m} = R T \implies Z = \frac{P V_m}{R T} = 1 – \frac{a}{R T V_m}$$
Hence, (B) is correct.
Solution 3:
From the derivation of the cubic van der Waals equation roots, $V_c = 3b$. Hence, (C) is correct.
Solution 4:
We established $T_i = \frac{2a}{Rb}$ and $T_c = \frac{8a}{27Rb}$.
$$\frac{T_i}{T_c} = \frac{2a / (Rb)}{8a / (27Rb)} = \frac{27}{4} = 6.75$$
$$T_i = 6.75 \times T_c = 6.75 \times 160\text{ K} = \mathbf{1080\text{ K}}$$
Hence, (C) is correct.
Solution 5:
Because hydrogen and helium molecules have very small masses and very weak attractive forces, their $a$ constant is extremely tiny. At $300\text{ K}$, $R T \gg a/b$, so the pressure correction is completely negligible ($P + a/V_m^2 \approx P$), leaving $P(V_m – b) = R T \implies Z = 1 + \frac{P b}{R T} > 1$ at all pressures. Hence, (B) is correct.
Solution 6:
Boyle temperature $T_B = \frac{a}{R b}$ and critical temperature $T_c = \frac{8 a}{27 R b}$.
$$\frac{T_B}{T_c} = \frac{a / (R b)}{8 a / (27 R b)} = \frac{27}{8}$$
Hence, (B) is correct.
Solution 7:
(A) True: $Z_c = \frac{P_c V_c}{R T_c} = \frac{3}{8} = 0.375$ for all van der Waals gases.
(B) True: Above $T_c$, thermal kinetic energy prevents meniscus formation and condensation regardless of pressure.
(C) True: At $T_i$, the Joule-Thomson coefficient $\mu_{\text{JT}} = 0$, so no temperature change occurs on throttling.
(D) False: Constant $b = 4 \times (\text{actual molecular volume})$, not the actual volume itself.
Hence, (A, B, C) are correct.
Solution 8:
(A) and (B) are the fundamental definition of the critical point inflection point: $\left(\frac{\partial P}{\partial V_m}\right)_{T} = 0$ and $\left(\frac{\partial^2 P}{\partial V_m^2}\right)_{T} = 0$.
From $P = \frac{R T}{V_m – b} – \frac{a}{V_m^2}$, evaluating at $(T_c, V_c)$ where $V_c = 3b$:
$$\left(\frac{\partial P}{\partial T}\right)_{V_m} = \frac{R}{V_m – b} \implies \left(\frac{\partial P}{\partial T}\right)_{V_c} = \frac{R}{3b – b} = \frac{R}{2b}$$
Thus (D) is also true. Hence, (A, B, D) are correct.
Solution 9:
Given $T_c = 40\text{ K}$ and $P_c = 20\text{ atm}$.
Using the universal ratio $\frac{T_B}{T_c} = \frac{27}{8}$:
$$T_B = \frac{27}{8} \times T_c = \frac{27}{8} \times 40 = 27 \times 5 = \mathbf{135\text{ K}}$$
Solution 10:
For $P(V_m – b) = R T$, dividing by $R T$ gives:
$$\frac{P V_m}{R T} – \frac{P b}{R T} = 1 \implies Z = 1 + \left(\frac{b}{R T}\right) P$$
The plot of $Z$ vs $P$ is a straight line with slope $m = \frac{b}{R T}$.
$$m = \frac{b}{R T} \implies b = m \times R T$$
Substituting numerical values:
$$b = (2.0 \times 10^{-3}\text{ atm}^{-1}) \times (0.0821\text{ L}\cdot\text{atm/(mol}\cdot\text{K)}) \times (300\text{ K})$$
$$b = 2.0 \times 10^{-3} \times 24.63 = 0.04926\text{ L/mol} \approx \mathbf{0.05\text{ L/mol}}$$