Question

Assume that the equation of state for a gas can be written in the form \(P\left(V_{m}-b(T)\right)=R T .\) Derive an expression for \(\beta=1 / V(\partial V / \partial T)_{P}\) and \(\kappa=-1 / V(\partial V / \partial P)_{T}\) for such a gas in terms of \(b(T), d b(T) / d T, P,\) and \(V_{m^{-}}\).

Short answer

The expressions for the coefficients of thermal expansion \(\beta\) and isothermal compressibility \(\kappa\) are given by: \[\beta = \frac{1}{V_m}\left(\frac{R}{P} + \frac{db(T)}{dT}\right)\] and \[\kappa = \frac{RT}{V_m P^2}\]

Step-by-step solution

Rewrite the equation of state in terms of \(V_m

Start by rewriting the equation of state in terms of the molar volume \(V_m\) as follows: \[V_m = \frac{RT}{P} + b(T)\]

Differentiate the equation w.r.t. T under constant P

Now, differentiate the equation obtained in Step 1 with respect to temperature T while keeping pressure P constant: \[\frac{\partial V_m}{\partial T}\bigg|_P = \frac{R}{P}\cdot\frac{\partial T}{\partial T}\bigg|_{_P} + \frac{\partial b(T)}{\partial T}\bigg|_{_P}\] Since \(\frac{\partial T}{\partial T}\) is equal to 1, we have: \[\frac{\partial V_m}{\partial T}\bigg|_P =\frac{R}{P} + \frac{db(T)}{dT}\]

Calculate the coefficient of thermal expansion \(\beta\)

The coefficient of thermal expansion \(\beta\) is given by: \[\beta = \frac{1}{V_m} \cdot \frac{\partial V_m}{\partial T}\bigg|_P\] Substitute the expression for \(\frac{\partial V_m}{\partial T}\bigg|_P\) obtained in Step 2: \[\beta = \frac{1}{V_m}\left(\frac{R}{P} + \frac{db(T)}{dT}\right)\]

Differentiate the equation w.r.t. P under constant T

Next, differentiate the equation obtained in Step 1 with respect to pressure P while keeping temperature T constant: \[\frac{\partial V_m}{\partial P}\bigg|_T = -\frac{RT}{P^2}\cdot\frac{\partial P}{\partial P}\bigg|_{_T} + \frac{\partial b(T)}{\partial P}\bigg|_{_T}\] Since \(\frac{\partial P}{\partial P}\) is equal to 1, and \(\frac{\partial b(T)}{\partial P}\) is equal to 0, we have: \[\frac{\partial V_m}{\partial P}\bigg|_T =-\frac{RT}{P^2}\]

Calculate the isothermal compressibility \(\kappa\)

The isothermal compressibility \(\kappa\) is given by: \[\kappa = -\frac{1}{V_m}\cdot \frac{\partial V_m}{\partial P}\bigg|_T\] Substitute the expression for \(\frac{\partial V_m}{\partial P}\bigg|_T\) obtained in Step 4: \[\kappa = -\frac{1}{V_m}\left(-\frac{RT}{P^2}\right) = \frac{RT}{V_m P^2}\] Thus, we have the expression for \(\beta\) and \(\kappa\) as: \[\beta = \frac{1}{V_m}\left(\frac{R}{P} + \frac{db(T)}{dT}\right)\] \[\kappa = \frac{RT}{V_m P^2}\]

Key concepts

Thermal Expansion Coefficient (β)

The thermal expansion coefficient, denoted as \( \beta \), is a measure of how the volume of a substance changes in response to a change in temperature while the pressure remains constant. It is particularly important in understanding the behavior of gases and solids as they expand or contract with temperature variations.
The mathematical expression for \( \beta \) is given by:

• \( \beta = \frac{1}{V_m} \cdot \frac{\partial V_m}{\partial T}\bigg|_P \)

where \( V_m \) is the molar volume, and \( \frac{\partial V_m}{\partial T} \bigg|_P \) is the partial derivative of molar volume with respect to temperature at constant pressure.
In the context of the provided equation of state \( P(V_{m} - b(T)) = RT \), it means that if the temperature increases, one should expect the molar volume \( V_m \) to increase accordingly. To find \( \beta \), we use the derived formula:

• \( \beta = \frac{1}{V_m} \left( \frac{R}{P} + \frac{db(T)}{dT}\right) \)

This equation shows that the thermal expansion will depend on the rate at which \( b(T) \), a function of temperature in the equation of state, changes with temperature. Understanding \( \beta \) helps in predicting how substances behave under heating, which is crucial in various engineering applications.

Isothermal Compressibility (κ)

The isothermal compressibility \( \kappa \) provides insight into how a material's volume changes with pressure at constant temperature. It is a crucial parameter in thermodynamics and fluid mechanics.
The formula for \( \kappa \) is expressed as:

• \( \kappa = -\frac{1}{V_m} \cdot \frac{\partial V_m}{\partial P}\bigg|_T \)

where \( V_m \) stands for molar volume, and \( \frac{\partial V_m}{\partial P} \bigg|_T \) represents the rate of change of molar volume with respect to pressure at a constant temperature.
In our initial example, using the expression from our differentiated \ equation of state \( \frac{\partial V_m}{\partial P} \bigg|_T = -\frac{RT}{P^2} \), the formula for \( \kappa \) becomes:

• \( \kappa = \frac{RT}{V_m P^2} \)

\( \kappa \) being inversely proportional to pressure squared highlights that compressibility decreases as pressure increases. This means it's harder to compress a gas under higher pressures, a principle heavily utilized in designing pressurized systems and understanding natural phenomena.

Molar Volume (V_m)

Molar volume \( V_m \) is defined as the volume occupied by one mole of a substance. It is a significant concept in chemistry and physics, aiding in the understanding of the microscopic properties of materials.
Typically expressed in cubic meters per mole (\( \text{m}^3/\text{mol} \)), the molar volume offers insights when discussing gas behaviors under varying conditions of temperature and pressure. The given equation of state simplifies the relationship as:

• \( V_m = \frac{RT}{P} + b(T) \)

Here, \( R \) is the universal gas constant, \( T \) the temperature, \( P \) the pressure, and \( b(T) \) a temperature-dependent correction factor. This equation suggests that molar volume increases with temperature and decreases with pressure, a familiar effect seen in gas expansion and compression.
Understanding \( V_m \) helps in predicting how substances will perform under different thermodynamic conditions. It plays an essential role in everything from chemical reactions to material sciences and engineering applications, making it an indispensable tool in both academic and practical contexts.