Question

Show that \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T}=0\) for an ideal and for a van der Waals gas.

Short answer

In summary, for both ideal and van der Waals gases, the internal energy depends only on temperature and not on volume. Therefore, the heat capacity at constant volume (\(C_V\)) is given by: \(C_V = \left(\frac{\partial U}{\partial T}\right)_V\). Differentiating \(C_V\) with respect to \(V\) while keeping the temperature constant, we find that \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = 0\) for both ideal and van der Waals gases.

Step-by-step solution

Recall the heat capacity at constant volume definition

For a given gas, the heat capacity at constant volume is defined as: \[C_{V} = \left(\frac{\delta Q}{dT}\right)_{V}\] where \(Q\) is the heat supplied to the gas and \(T\) is the temperature of the gas.

Ideal gas description

An ideal gas is described by the ideal gas law, which states: \[PV = nRT\] where \(P\) is the pressure of the gas, \(V\) is the volume of the gas, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature of the gas. This law describes the relationship between pressure and volume for an ideal gas when energy is transferred as heat at constant temperature.

van der Waals gas description

A van der Waals gas is described by the van der Waals equation, which takes into account the finite size of gas molecules and intermolecular attractions: \[P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}\] where \(a\) and \(b\) are van der Waals constants.

Calculate the partial derivative for ideal gas

For an ideal gas, the internal energy (\(U\)) depends only on the temperature. Therefore, the heat capacity at constant volume for an ideal gas is given by: \[C_V = \left(\frac{\partial U}{\partial T}\right)_V\] Now let's differentiate \(C_V\) with respect to \(V\) while keeping the temperature constant: \[\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = \left(\frac{\partial^2 U}{\partial T \partial V}\right)_{T}\] Since \(U\) depends only on temperature and not on volume for an ideal gas, this partial derivative is equal to zero: \[\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = 0\]

Calculate the partial derivative for van der Waals gas

Similarly, for a van der Waals gas, the internal energy depends only on the temperature. Thus, the heat capacity at constant volume is given by: \[C_V = \left(\frac{\partial U}{\partial T}\right)_V\] Differentiating \(C_V\) with respect to \(V\) while keeping the temperature constant, we get: \[\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = \left(\frac{\partial^2 U}{\partial T \partial V}\right)_{T}\] Again, since \(U\) depends only on temperature and not on volume for a van der Waals gas, this partial derivative is equal to zero: \[\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = 0\] In conclusion, we have shown that the partial derivative \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T} = 0\) for both ideal and van der Waals gases, as required by the exercise.

Key concepts

Heat Capacity

When we talk about heat capacity, we're delving into how much energy it takes to raise the temperature of a substance. Specifically, heat capacity at constant volume, denoted as \(C_{V}\), tells us how much heat (\(Q\)) is needed to raise the temperature (\(T\)) of a gas when the volume is held constant. The formula is \(C_{V} = \left(\frac{\delta Q}{dT}\right)_{V}\).
This concept is crucial in thermodynamics as it helps us understand how substances absorb heat differently. For gases, determining \(C_{V}\) helps us grasp how energy changes affect their temperature under constant volume conditions.
In exercises like the one provided, we're interested in differentiating heat capacity concerning volume, while holding the temperature constant, to explore deeper properties of gases.

Ideal Gas Law

The Ideal Gas Law is foundational in understanding gas behaviors and is described by the equation \(PV = nRT\). Here, \(P\) stands for pressure, \(V\) for volume, \(n\) for moles of gas, \(R\) for the ideal gas constant, and \(T\) for temperature.
This law provides a simplified model of a gas by assuming no interactions between gas molecules and that they occupy no volume themselves. It serves as a good approximation under many conditions, especially when dealing with gases at higher temperatures and lower pressures.
For an ideal gas, we can conclude that the internal energy \(U\) depends solely on temperature, and changes in volume, at constant temperature, do not influence internal energy. This result lays the groundwork for further explorations concerning heat capacity changes with volume.

Van der Waals Equation

The van der Waals equation provides a more realistic description of gas behaviors than the Ideal Gas Law, especially under conditions of high pressure and low temperature. Given by \(P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}\), this equation accounts for the finite volume of gas molecules (with \(b\) being the correction for volume) and the attraction between them (with \(a\) being the correction for attraction).
This model recognizes that gas molecules have volume and experience inter-molecular forces, addressing some limitations of the ideal gas approximation.
Even with these complexities, for a van der Waals gas, the internal energy still depends only on temperature, leading to the same conclusion that \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T}\) equals zero.

Internal Energy

Internal energy, denoted as \(U\), is the total energy contained within a system, encompassing both kinetic and potential energies of molecules. In the context of gases, internal energy primarily depends on temperature.
For ideal gases, internal energy is a function of temperature only, meaning changes in volume do not affect internal energy, as no energy is stored in molecular interactions or volume changes. This simplicity aids the differentiation in our heat capacity exercise, showing no dependence on volume.
Similarly, in van der Waals gases, despite acknowledging molecular interactions via corrections \(a\) and \(b\), internal energy remains temperature-bound. Hence, for both gas models, the volume does not factor into internal energy's role, solidifying the zero result for the partial derivative of heat capacity.