--%>

Describe Thermodynamics Properties

The free energy property leads to convenient expressions for the volume and pressure dependence of internal energy, enthalpy and the heat capacities.

All the properties of a chemical system, a sample of a substance, or a mixture of substances have some fixed, definite values when the state of the system is set by the selection of, for example, a temperature and a pressure.

The properties that we have been with have the symbols V, U, H, S and G. these properties are all interrelated, as you know by thinking of the defining equations such as H = U + PV and G = H - TS.

Suppose the state of the system is changed. The values of the properties of the system change. These property changes must be interrelated.

An example of Maxwell's equations: the dependence of free energy on pressure and that on temperature are given by the partial derivatives,

(∂G/∂P)T = V and (∂G/∂T)P = -S

Since the free energy is a property, the change in free energy will be the same regardless of the order of differentiation with respect to pressure and temperature. We can write

[∂/∂P (∂G/∂T)P]T = [(∂/∂T) (∂G/∂P)T]P

With the equations for the derivatives of G with respect to T and P, this gives us 

(∂S/∂P)T = -(∂V/∂T)P

This derivative relation, who in itself is not at all revealing, is useful in leading us to other relations that give us unexpected insights. It is one of the expressions known as Maxwell's equations.

Pressure and volume dependence of U: for any process, the change in the energy dU of the system is related to the change in the energies of the thermal and mechanical surroundings by

dU = -dUtherm - dU
mech

For a process in which only the mechanical energy is involved, dUmech = P dV. For a reversible process dUtherm = -T dS. By considering this special process we arrive at the relation

dU = T dS - P dV

For a given change in S and V, there will be a particular change in U. thus although we arrived by considering a particular process, it is generally applicable.

Division of equation by dP followed by specification of constant temperature gives

(∂U/∂P)T = T(∂S/∂P)T - P(∂V/∂P)T

The pressure dependence of internal energy on volume can be obtained first writing the relation

(∂U/∂P)T = (VU/∂T)T - (∂V/∂P)T

The corresponding dependence of internal energy on volume can be obtained first writing the relation

(∂U/∂V)T = (∂U/∂P)T (∂P/∂V)T = -(∂V/∂T)P (∂P/∂V)T - P(∂V/∂P)T (∂P/∂V)T

= - T(∂V/∂T)T (∂P/∂V)T - P

The (∂V/∂T) P term can be expressed from dV = (∂V/∂T) P dT + (∂V/∂T)T dP by specifying constant volume, and rearranging to

(∂V/∂T)= - (∂V/∂P)T (∂P/∂T)V    

Now the equation for (∂U/∂V)T becomes

(∂U/∂V)T = T(∂P/∂T)V - P

Energy of an ideal gas

The internal energy U of a sample of an ideal gas depends on only the temperature, not on the pressure or volume of the sample. This ideal was justified by the kinetic molecular theory. We can show that it holds without stepping out of classical thermodynamics.

We can use conformity to the equation PV = nRT as a definition of ideal gas behaviour. If this relation is used to evaluate the terms, we arrive at

(∂U/∂P)T = 0 and (UV/∂V)T = 0

Thus, without any stipulation other than PV = nRT, arrive at the conclusion that the internal energy of an ideal gas depends on only the temperature.

   Related Questions in Chemistry

  • Q : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

  • Q : Hydroxide is highly insoluble in

     : 1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to

  • Q : Hydrocarbons list and identify

    list and identify differences between the major classes of hydrocarbons

  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

  • Q : Means of molal solution Choose the

    Choose the right answer from following. A molal solution is one that contains one mole of a solute in: (a) 1000 gm of the solvent (b) One litre of the solvent (c) One litre of the solution (d) 22.4 litres of the solution

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o

  • Q : Utilization of glacial acetic acid What

    What is the utilization of glacial acetic acid? Briefly describe the uses.

  • Q : Chemistry brief discription of relative

    brief discription of relative lowering of vapour pressure

  • Q : Chemists have not created a periodic

    Explain the reason behind that the chemists have not created a periodic table of compounds?

  • Q : Calculating number of moles from

    Choose the right answer from following. If 0.50 mol of CaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ca3 (PO2)2 which can be formed: (a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10