Equation of state for ideal gases

The thermodynamic quantities we have defined, such as internal energy, enthalpy and heat capacities, are functions of temperature and pressure. We know that the variation of thermodynamic quantities with temperature is described by the heat capacity. This is the example of enthalpy variation with temperature. How about the variation of thermodynamic quantities with pressure P? For that, we need to know the relationship between pressure and temperature or other physical variables. This relationship is the equation of state. The equation of state is the relationship among the physical variables that describe the condition of a material. For a gas, the condition of a gas system can be described by pressure, volume, temperature and the number of moles n.

So the equation of states relates P, V, T and n. For ideal gas, the equation of states is PV equal to nRT. It is a result of combination of Boyle’s and Charles’s laws. Boyle’s law states that at constant temperature, pressure is inversely proportional to volume. In other words, PV product is constant. Charles’s laws is that at constant pressure, the volume is proportional to temperature. The combination gives the equation of states, PV equal to nRT. The proportionality constant R is called the gas constant and the value is 0.082 little atm/mole K or 8.314 J/mole K. We can apply this equation of states to induce other characteristics of ideal gas. These are some characteristics of ideal gas.

The internal energy of ideal gases does not change with volume at constant temperature. The enthalpy does not change with pressure at constant temperature. Let’s derive it. By definition, H is U + PV. To get the change of enthalpy with pressure at constant temperature, let’s take the partial differential with respect to pressure at constant temperature on both side. It is thus (dU over dP) + (P times (dV over dP)) + V. By chain rule, (dU over dP) is (dU over dV) times (dV over dP). Among them,(dU over dV) at constant temperature is zero. (dV over dP) is calculated as - RT over (P squared) from the equation of states for the ideal gas.

So (dH over dP) at constant temperature is (- RT/P + V), and again by applying equation of states for ideal gas, it is zero. From here, the Joule-Thompson coefficient defined like this is also zero for ideal gas. Another characteristic of ideal gas is the difference between Cp and Cv. It was the gas constant R before. Let’s derive this relationship here. Cp is (dH over dT) at constant P and Cv is (dU over dT) at constant v. Let’s express the (dH over dT) first. H is U + PV by definition. Thus at constant pressure, the partial differential on both side yields (dU over dT) + (P times (dV over dT)). Let’s turn back to Cp-Cv.

And insert the first part equation here. Then Cp-Cv is (dU over dT at constant P) + (P times dV over dT at constant P) - (dU over dT at constant V). Among this equation, let’s first look at (dU over dT at constant P). The total differential of dU as a function of temperature and volume is (dU over dT at constant volume) times dT + (dU over dV at constant temperature) times dV. Divide both side by (dT at constant pressure). It results in (dU over dT at constant V) + (dU over dV at constant T) times (dV over dT at constant P). Insert it into Cp-Cv. Then it becomes like this. Cancel out the same things.

Then finally Cp-Cv is P + (dU over dT at constant T) times (dV over dT at constant P). . For ideal gas, (dU over dV at const T )is zero, thus Cp-Cv is just P times ((dV over dT) at constant P). It is R at constant pressure so Cp-Cv is R.