--%>

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)P = - (∂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 : What is adsorption and its examples. In

    In a liquid a solid substance a molecule present within the bulk of the substance is being attracted infirmly from all sides by the neighbouring molecules. Hence there is no bet force acting on the molecule or there are no unbalanced forces of the molecule. On the oth

  • Q : Strength of any solution Give me answer

    Give me answer of this question. A solution contains 1.2046 x 1024 hydrochloric acid molecules in one dm3 of the solution. The strength of the solution is: (a) 6 N (b) 2 N (c) 4 N (d) 8 N

  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm

  • Q : Calculating amount of Sodium hydroxide

    Choose the right answer from following. The amount of NaOH in gms in 250cm3 of a0.100M NaOH solution would be : (a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm

  • Q : Basicity order order of decreasing

    order of decreasing basicity of urea and its substituents

  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Molarity of Nacl solution When 5.85 g

    When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1

  • Q : Modern periodic table and Mendeleevs

    Differentiate between the modern periodic table and Mendeleevs table?

  • Q : Question based on vapour pressure and

    Give me answer of this question. The vapour pressure of water at 20degreeC is 17.54 mm. When 20g of a non-ionic, substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular weight of the substances: (a) 210.2 (b) 206.88

  • Q : Problem on solutions The 2N aqueous

    The 2N aqueous solution of H2S04 contains: (a) 49 gm of H2S04 per litre of solution (b) 4.9 gm of H2S04 per litre of solution (c) 98 gm of H2S04