Question

Because \(U\) is a state function, \(\left(\partial / \partial V(\partial U / \partial T)_{V}\right)_{T}=\) \(\left(\partial / \partial T(\partial U / \partial V)_{T}\right)_{V} .\) Using this relationship, show that \(\left(\partial C_{V} / \partial V\right)_{T}=0\) for an ideal gas.

Short answer

Using the given relationship \(\left(\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_{V}\right)_{T} = \left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_{T}\right)_{V}\), we first replace \(\frac{\partial U}{\partial T}\) with \(C_V\). Then, we make use of the fact that for an ideal gas, the internal energy U is independent of the volume, which implies \(\left(\frac{\partial U}{\partial V}\right)_{T} = 0\). Substituting this result into the modified equation, we find that \(\left(\frac{\partial C_V }{\partial V}\right)_{T} = 0\), so for an ideal gas, \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T}=0\).

Step-by-step solution

Recall the definition of heat capacity at constant volume

: Heat capacity at constant volume, \(C_V\), is defined as the amount of heat required to raise the temperature of a system by one degree while keeping the volume constant. Mathematically, it can be represented as: \[C_V = \left(\frac{\partial U}{\partial T}\right)_{V}\]

Use the relationship for the state function U

: Given, \[\left(\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_{V}\right)_{T} = \left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_{T}\right)_{V}\] Now, replacing \(\frac{\partial U}{\partial T}\) with \(C_V\), we have: \[\left(\frac{\partial C_V }{\partial V}\right)_{T} = \left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_{T}\right)_{V}\]

Use the properties of an ideal gas

: For an ideal gas, the internal energy U depends only on its temperature T and is independent of the volume. This means: \[\left(\frac{\partial U}{\partial V}\right)_{T} = 0\] We can substitute this into the previous equation from Step 2: \[\left(\frac{\partial C_V }{\partial V}\right)_{T} = \left(\frac{\partial}{\partial T}(0)\right)_{V}\]

Calculate the derivative

: For any constant, its derivative with respect to any variable is always zero. Hence, the derivative of 0 with respect to T (at constant V) will also be zero: \[\left(\frac{\partial C_V }{\partial V}\right)_{T} = 0\] So, we have shown that for an ideal gas, \(\left(\frac{\partial C_{V}}{\partial V}\right)_{T}=0\), as required.

Key concepts

State Function

State functions are properties of a system that depend only on the current state of the system, not on the path by which the system arrived at that state.

Examples of state functions include internal energy, enthalpy, and entropy. They are contrastingly different from path functions like heat and work, which depend on the specific transformations or processes the system undergoes.

In the exercise given, internal energy (\(U\)) is a state function, which implies that its change is the same no matter what path is taken between two states. This characteristic is used to derive that the change in heat capacity at constant volume (\(C_V\)) with respect to volume is zero for an ideal gas. Mathematically, this is expressed by the equality \(\left(\frac{\partial }{\partial V}\left(\frac{\partial U}{\partial T}\right)_{V}\right)_{T} = \left(\frac{\partial }{\partial T}\left(\frac{\partial U}{\partial V}\right)_{T}\right)_{V}\), demonstrating the path-independent nature of internal energy change.

Internal Energy

Internal energy (\(U\)) is a fundamental concept in thermodynamics, representing the total energy contained within a system. It includes the kinetic and potential energy of the molecules within the system.

For an ideal gas, the internal energy is a function of temperature alone, not volume or pressure. In other words, the internal energy change is independent of a gas's volume changes, which can be stated mathematically as \(\left(\frac{\partial U}{\partial V}\right)_{T} = 0\). Consequently, when we express the heat capacity at constant volume (\(C_V\)) with a derivative, \(\left(\frac{\partial C_V }{\partial V}\right)_{T}\), and considering the properties of an ideal gas, it's established that \(\left(\frac{\partial C_V }{\partial V}\right)_{T} = 0\). This follows from the fact that any derivative of a constant (in this case, the internal energy concerning volume) is zero.

Ideal Gas Law

The ideal gas law, expressed as \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature, describes the behavior of an ideal gas under various conditions.

The law connects four fundamental physical properties, assuming that the gas particles do not attract or repel each other and occupy no volume themselves. This simplification allows us to derive relationships between variables, such as in the exercise provided where the heat capacity at constant volume (\(C_V\)) is related to the internal energy of an ideal gas.

Because the internal energy of an ideal gas is directly related to its temperature, changes in volume at constant temperature do not affect the internal energy, which underpins the conclusion that the partial derivative of \(C_V\) with respect to volume at a constant temperature for an ideal gas is zero. The ideal gas law is fundamentally connected to this concept, simplifying the relationship between pressure, volume, temperature, and the amount of gas.