Problem 20

Question

The equation of state of a gas is given by the Berthelot equation $\left(P+a / T v^{2}\right)(v-b)=R T$. (a) Find values of the critical temperature $T_{c}$, the critical molar volume $v_{c}$, and the critical pressure $P_{c}$, in terms of $a, b$, and $R$. (b) Does the Berthelot equation satisfy the law of corresponding states? (c) Find the critical exponents $\beta, \delta$, and $\gamma$ from the Berthelot equation.

Step-by-Step Solution

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Answer

The critical temperature is $ T_c = \frac{a}{27bR} $, the critical molar volume is $ v_c = 3b $, and the critical pressure is $ P_c = \frac{a}{27b^2} $. The Berthelot equation does not strictly satisfy the law of corresponding states. Critical exponents are $ \beta = \frac{1}{2} $, $ \delta = 3 $, and $ \gamma = 1 $.

1Step 1: Expressing the Critical Point Conditions

At the critical point, the first and second partial derivatives of pressure with respect to volume are zero. Hence, we write $ \left(\frac{\partial P}{\partial v}\right)_{T_c,v_c} = 0$ and $ \left(\frac{\partial^2 P}{\partial v^2}\right)_{T_c,v_c} = 0 $.

2Step 2: Rewrite Berthelot Equation

Rewrite the Berthelot equation in terms of pressure: \[ P = \frac{R T}{v - b} - \frac{a}{T v^2} \].

3Step 3: Partial Derivative with Respect to Volume

Calculate the first partial derivative of \[ P = \frac{R T}{v - b} - \frac{a}{T v^2} \] with respect to volume, set it to zero and solve for $ v_c $: $ \left(\frac{\partial P}{\partial v}\right)_{T_c, v_c} = \ - \frac{RT}{(v - b)^2} + \frac{2a}{T v^3} = 0 $.

4Step 4: Solve for Critical Molar Volume

Solve the equation from Step 3 for $ v_c $, which gives $ v_c = 3b $.

5Step 5: Second Partial Derivative with Respect to Volume

Calculate the second partial derivative and set it to zero: $ \left(\frac{\partial^2 P}{\partial v^2}\right)_{T_c, v_c} = \frac{2RT}{(v - b)^3} - \frac{6a}{T v^4} = 0 $.

6Step 6: Solve for Critical Temperature

Use the value $ v_c = 3b $ in the second partial derivative equation to find $ T_c $, resulting in $ T_c = \frac{a}{27bR} $.

7Step 7: Calculate Critical Pressure

Substitute $ T_c = \frac{a}{27Rb} $ and $ v_c = 3b $ into the original equation to get $ P_c = \frac{a}{27b^2} $.

8Step 8: Check Law of Corresponding States

Express the equation of state in terms of reduced variables: $ P_r = \frac{P}{P_c} $, $ v_r = \frac{v}{v_c} $, and $ T_r = \frac{T}{T_c} $. Show that it reduces to a universal form if it satisfies the law of corresponding states. The Berthelot equation becomes $ \bigg(\frac{P_r}{T_r v_r^2} + \frac{1}{T_r} \bigg)(v_r - 1) = 1 $, which shows that it does not satisfy the law strictly.

9Step 9: Calculate Critical Exponents

Determine exponents $ \beta $, $ \delta $, and $ \gamma $ as: $ \beta = \frac{1}{2} $ from $ P \sim (T_c - T)^\beta $, $ \delta = 3 $ from $ P \sim (v-v_c)^\delta $, and $ \gamma = 1 $ from $ \chi \sim (T - T_c)^\gamma $.

Key Concepts

Berthelot equationCritical temperatureCritical molar volumeCritical pressurePartial derivativesLaw of corresponding statesCritical exponents

Berthelot equation

The Berthelot equation is a modification of the ideal gas law that includes parameters to account for intermolecular forces and molecular size. This can be written as \rand is expressed as: $ (P + \frac{a}{T v^2})(v - b) = R T $\rtIt modifies the pressure and volume terms by introducing constants 'a' and 'b'. Here:

'a' accounts for intermolecular forces.
'b' represents the volume occupied by the gas molecules.

This equation is more accurate for real gases than the ideal gas law, especially under high pressure or low temperature conditions.

Critical temperature

Critical temperature $T_{c}$ is the temperature above which a gas cannot be liquefied, no matter how much pressure is applied. For the Berthelot equation, it is determined by setting the first and second partial derivatives of pressure with respect to volume to zero at the critical point. From the earlier solution, we have:\rand $T_{c} = \frac{a}{27 b R}$\rtCritical Temperature\rat a temperature where the distinctions between liquid and gas phases disappear, marking the end of a liquid-vapor boundary for pure substances.

Critical molar volume

Critical molar volume, $v_{c}$, is the volume at which the second partial derivative of pressure with respect to volume is zero at the critical temperature. For the Berthelot equation, it involves solving the partial derivative equation and yields:$ v_c = 3b $\rThe critical molar volume indicates the molar volume of a substance at its critical point. In essence, it shows us the volume occupied by one mole of gas at the critical point where the liquid and gas phases coexist.

Critical pressure

Critical pressure, $P_{c}$, is the pressure required to liquefy a gas at its critical temperature. For the Berthelot equation, substituting $T_{c}$ and $v_{c}$ into the original equation gives us: $ P_c = \frac{a}{27b^2} $\rThis critical pressure represents the moment when gas turns to liquid under critical conditions. It is a critical point which aids in understanding the conditions under which a gas condenses.

Partial derivatives

Partial derivatives are used to determine how a function changes as one variable changes while keeping other variables constant. In our context, partial derivatives of pressure with respect to volume at constant temperature help locate the critical point. We employed:$ \left( \frac{ \partial P }{ \partial v } \right)_{T_c,v_c} = 0 $ and $ \left( \frac{ \partial^2 P }{ \partial v^2 } \right)_{T_c,v_c} = 0 $\rThese equations ensure that we've found the point where the gas transitions into either a supercritical fluid or back into a liquid from a gaseous state. Understanding partial derivatives in this context allows us to determine how pressure changes as volume undergoes infinitesimal change at the critical point.

Law of corresponding states

The law of corresponding states suggests that gases behave similarly under corresponding conditions of reduced pressure, volume, and temperature. It states that all gases, when in corresponding states (same reduced temperature, pressure, volume), exhibit similar behavior. The critical values ($P_r = \frac{P}{P_c} $, $v_r = \frac{v}{v_c} $, $T_r = \frac{T}{T_c} $) are used to form the reduced variables.\rFor the Berthelot equation, when substituted in terms of these reduced variables, it doesn't strictly conform to the law of corresponding states. This tells us that Berthelot's equation isn't universally applicable across all gases under all conditions.

Critical exponents

Critical exponents characterize the behavior of physical quantities near critical points. For the Berthelot equation, the key exponents we derive are:

$ \beta $: Describes how the order parameter (like density) changes with temperature. Here, $ \beta = \frac{1}{2} $.
$ \delta $: Describes how pressure relates to volume changes. Here, $ \delta = 3 $.
$ \gamma $: Describes how the isothermal compressibility diverges near the critical point. Here, $ \gamma = 1 $.

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