Question

Explain how to calculate thermodynamic properties for a gas under adiabatic conditions

Define the Third Law of Thermodynamics.

Use tabulated values of absolute entropies to calculate the entropy change for a reaction.

Recognize that all substances, even elements in their standard state, have absolute entropies that are greater than zero at temperatures above absolute zero.

Write the thermodynamic conditions required for a spontaneous change both in terms of entropy and Gibbs energy.

Write the mathematical definitions of Gibbs energy and how to calculate it these properties for a system. And explain how these two variables differ in what they measure.

Answer 1

Third Law Of Thermodynamics

Explain how to calculate thermodynamic properties for a gas under adiabatic conditions

Ans: The thermodynamic properties for a gas under adiabatic conditions can be calculated by using Pressure-Volume Diagram. The properties may be read from the classical behaviour of an ideal gases.

Consider, pressure is on vertical axis and Volume is on horizontal axis. The conclusions can be drawn as follows:

1. Every adiabat approaches both the P and V axis.
2. Each adiabat intersects with each isotherm just once.
3. An adiabat looks similar to an isotherm, except that during an expansion.
4. If adiabats and isotherms are graphed at regular intervals of entropy and temperature, respectively, then as the eye moves towards the axes, it sees the density of isotherms remains constant, and the density of adiabats grow.

Define the Third Law of Thermodynamics.

Ans: The entropy of any perfectly ordered, crystalline substance at absolute zero is zero.

Use tabulated values of absolute entropies to calculate the entropy change for a reaction

Ans: The entropy change ΔS for a reaction can be calculated by using tabulated values of the standard molar entropy (S°) i. e. The entropy of 1 mol of a substance at standard temp. 298 K.

The absolute entropy values are obtained by measuring the entropy change that occurs between the 0 K [corresponding to S = 0] and 298 K.

The substances with approx. same molar mass and the number of atoms, S° values will follows the order as S°(gas) > S°(liquid) > S°(solid).

The substances with similar molecular structures have similar S° values.

ΔS°reaction = ∑mS°(products) − ∑nS°(reactants)

m and n are the stoichiometric coefficients of each product and each reactant in the Balanced Equation.

Recognize that all substances, even elements in their standard state, have absolute entropies that are greater than zero at temperatures above absolute zero.

Ans: All translational and rotational motions are ceased at absolute zero. Hence Above absolute zero, they have motions, so the entropy increases.