Understanding System Properties and Exact Differentials

In thermodynamics, a property of a system is a characteristic that describes its state independently of how that state was reached. These are often called point functions or state functions. Key examples include temperature (T), pressure (P), volume (V), and internal energy (U).

Mathematically, a quantity is a property if its differential is exact. The integral of an exact differential between two points (states) depends only on the points themselves, not the path taken between them. This is why changes in properties are path-independent.

Analyzing Thermodynamic Property Options

We are given four integral expressions. We need to determine which one represents a property of the system. For an integral $\smallint dF$ to represent a property, the differential $dF$ must be exact.

1. $\smallint \left( {\frac{{dT}}{T} - \frac{{vdp}}{T}} \right)\;$
2. $\smallint \left( {\frac{{dT}}{T} - \frac{{pdV}}{V}} \right)\;$
3. $\smallint vdp$
4. $\smallint pdV$

Options 3 ($\smallint vdp$) and 4 ($\smallint pdV$) represent types of work done during processes. Work is a form of energy transfer that is path-dependent. Therefore, these integrals do not represent properties of the system.

Justifying Option 2 as a System Property

Option 2 is the integral $\smallint \left( {\frac{{dT}}{T} - \frac{{pdV}}{V}} \right)$. For this integral to represent a property, the differential form $dF = \frac{dT}{T} - \frac{PdV}{V}$ must be exact.

A differential $M(T, V) dT + N(T, V) dV$ is exact if $\frac{{\partial M}}{{\partial V}} = \frac{{\partial N}}{{\partial T}}$.

In our case, $M = \frac{1}{T}$ and $N = -\frac{P}{V}$. The variables are T and V.

We check the exactness condition:

$\frac{{\partial M}}{{\partial V}} = \frac{{\partial }}{{\partial V}}\left( {\frac{1}{T}} \right) = 0$ (treating T and V as independent variables).

$\frac{{\partial N}}{{\partial T}} = \frac{{\partial }}{{\partial T}}\left( { - \frac{P}{V}} \right)$. For the differential to be exact, this must also be 0.

Thus, the condition for option 2 to represent a property is $\frac{{\partial }}{{\partial T}}\left( { - \frac{P}{V}} \right) = 0$, which simplifies to $\frac{{\partial }}{{\partial T}}\left( {\frac{P}{V}} \right) = 0$. This means the ratio $P/V$ must be independent of temperature T.

While this condition ($P/V$ being independent of T) is not true for all substances under all conditions, it is a specific requirement for the differential $\frac{dT}{T} - \frac{PdV}{V}$ to be exact. If a system behaves such that $P/V$ is a function of V only (i.e., $P/V = f(V)$), then the differential is exact and its integral represents a property.

Comparing with Option 1 in Thermodynamics

Option 1 is the integral $\smallint \left( {\frac{{dT}}{T} - \frac{{vdp}}{T}} \right)$. The differential form is $dG = \frac{dT}{T} - \frac{v}{T}dp$. For this to be exact, where variables are T and p, the condition is $\frac{{\partial (1/T)}}{{\partial p}} = \frac{{\partial (-v/T)}}{{\partial T}}$.

$\frac{{\partial (1/T)}}{{\partial p}} = 0$.

So, exactness requires $\frac{{\partial (-v/T)}}{{\partial T}} = 0$, meaning $v/T$ must be independent of T. This condition is met for an ideal gas, where $v/T = R/p$. In this specific case (ideal gas), option 1 represents the integral of an exact differential $\frac{dT}{T} - \frac{R}{p}dp$, which integrates to $\ln T - R \ln p$, a property.

Considering the possible conditions under which options 1 and 2 could represent properties based on the exactness of their differentials, option 2 is the expression that represents a property under the specific condition that its differential is exact.