Question

1. Explain the entropy change by three different kinds of cases as expansion, phase change and heating process.
2. Define and explain the 3rd law of thermodynamics as you think and by different expressions as you want.
3. Define and explain the criteria for the spontaneous process in the aspect of G, A, H, U and S.

Answer 1

Answers)

i)

a) Entropy change due to expansion:-

Free expansion of a gas is irrevesible function, taking place in completely isolated system and during the expansion temperature of the gas remain constant.

Now, we know that entropy is a state function and change is entropy is denoted by-

nCv ln(T2/T1) + nR ln(V2/V1)

ESY

Now, as temperature remain constant during free expansion, change in entropy will be-

nR ln(V2/V1)

ESY

So, entropy increases during free expansion.

b) Entropy change due to phase change:-

Entropy can be termed as lack of order or predictibility.

The atoms of solids are closely packed compared to liquids and gases, so have very less entropy compared to liqids and gases.

The atoms of liquids are more closely packed than gases but less closely packed the gases so, have less entropy than gases.

So, if the phase change is towards higher internal energy ( eg, melting), entropy of system increases while if the phase change is towards lower internal energy (eg, freezing), entropy of the system decreases.

c) Entropy change due to heating process:-

If you increase temperature, entropy of system increases due to following reasons-

• More energy put into a system excites the molecules and the amount of random activity.
• As a gas expands in a system, entropy increases.If an atom has more space to bounce around, it will bounce more. Gases and plasmas have large amounts of entropy when compared to liquids and solids.

ii)

The third law of thermodynamics can be stated as follws-

In a closed system, the entropy of a system approaches a constant value as its temperature approaches absolute zero (0K).

As per mathematical expression-

As per statistical mechanics, the entropy of a system can be expressed via the following equation:

S – S0 = ?B ln?

Where,

• S is the entropy of the system.
• S0 is the initial entropy.
• ?B denotes the Boltzmann constant.
• ? refers to the total number of microstates that are consistent with the system’s macroscopic configuration.

Now, for a perfect crystal that has exactly one unique ground state, ? = 1. Therefore, the equation can be rewritten as follows:

S – S0 = ?B ln(1) = 0 [because ln(1) = 0]

When the initial entropy of the system is selected as zero, the following value of ‘S’ can be obtained:

S – 0 = 0 ⇒ S = 0

Thus, the entropy of a perfect crystal at absolute zero is zero.

iii)

Spontaneous Process- A spontaneous process can be defned as the process which is capable of proceeding in a given direction without any need of outside source of energy. For a spontaneous process, standard change in free energy is negative and energy is released (i.e., exothermic process)

Spontaneous process in respect of G,H and S:-

As we know from the defination that standard change in free enegy change is negative for spontaneous process. So,

if $\Delta G$ < 0, then process is spontaneous, and

if$\Delta G$ > 0, then process is non spontaneous.

Now, $\Delta G = \Delta H - T\Delta S$

So,

if value of $\Delta H$ <0 and $\Delta S$ >0, then process is spontaneous,

if value of $\Delta H$ >0 and $\Delta S$ <0, then process is non-spontaneous,

if value of $\Delta H$ >0 and $\Delta S$ >0, then the process can either be spontaneous or non- spontaneous depending on the value of temperature (spontaneous at high temperatures and non-spontaneous at low temperatures), and

if value of $\Delta H$ <0 and $\Delta S$ <0, then the process can either be spontaneous or non- spontaneous depending on the value of temperature (spontaneous at law temperatures and non-spontaneous at high temperatures).

Spontaneous process in respect of U (energy) and A:-

If during a process, energy is released, process is spontaneous and if energy is absorbed, process is non spontaneous.