Thermodynamics

In reality, work can be converted to heat entirely but not vice versa. This is in accordance with second law of thermodynamics. So, no process is possible in practice where heat is entirely converted to work.

Take for example the heat engine. It has to reject some heat to the sink and convert the remainder into work. Thus, entire heat supplied is not converted into work. On the other hand, in a refrigerator, work has to be supplied to transfer heat from a colder body to hotter surroundings-in this case, the entire work input is converted to heat.

In practice, reversible isothermal processes are impossible because heat transfer always takes place across a finite temperature difference. This is an irreversibility. Thus, the entropy of the system will always increase if there is a heat transfer to the system. In that case, the amount of heat supplied to the system cannot entirely be converted to work (unlike reversible isothermal processes where heat is entirely converted to work).

So, both the statements are clearly true but the second one does not explain the first one.

Process :

$ \text{H}_2\text{O}(l,\,1\,\text{bar},\,373\,\text{K}) \to \text{H}_2\text{O}(g,\,1\,\text{bar},\,373\,\text{K}) $

Step 1. Identify what temperature and pressure mean

At 373 K and 1 bar, this is the normal boiling point of water (approximately, at 1 atm ≈ 1.013 bar).

So, at this condition, liquid water and water vapor are in equilibrium.

Step 2. Thermodynamic definitions

At equilibrium between two phases at a given $T$ and $p$:

$ \Delta G = 0 $

because the two phases have equal molar Gibbs energies.

So for a reversible phase change at equilibrium:

$ \boxed{\Delta G = 0} $

Step 3. Entropy change

When liquid water vaporizes,

• Disorder increases,

• Number of microstates increases,

so

$ \boxed{\Delta S > 0} $

(positive entropy change).

✅ Therefore, the correct set is:

$ \boxed{\Delta G = 0, \ \Delta S = +\text{ve}} $

✅ Correct option: A

$(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ )$

$ = - 54.07 \times 1000 - 298 \times 10$ J/mol

$ = - 54070 - 2980 = 57050$

$\Delta G^\circ = - 2.303\,RT\,{\log _{10}}\,K$

$ - 57050 = - 2.303 \times 298 \times 8.314\,{\log _{10}}K$

$\log K = 10$

Assume the pressure of a monoatomic gas is P and volume is V at the temperature T .

As given, $\frac{\mathrm{P}}{\mathrm{V}}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

From equation (i), it can be concluded that,

$ P=V \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

According to the first law of thermodynamics,

$ d q=\mathrm{C}_v d \mathrm{~T}+p d \mathrm{~V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Here, $d q$ stands for the differential increase in the amount of added heat in the system, $\mathrm{C}_v$ is the molar heat capacity at the constant volume, $d \mathrm{~T}$ is a small temperature difference, $p$ is the pressure and $d \mathrm{~V}$ represents the small change in volume.

For one mole of an ideal gas,

PV = RT

The above relation can be written as,

$ \mathrm{P} d \mathrm{~V}+\mathrm{V} d p=\mathrm{R} d \mathrm{~T}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) $

Also, from equation (i),

$ \mathrm{P} d \mathrm{~V}=\mathrm{V} d p \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

Substituting in equation (iv), we get,

$ 2 \mathrm{P} d \mathrm{~V}=\mathrm{R} d \mathrm{~T} $

On rearranging,

$ \mathrm{P} d \mathrm{~V}=\frac{\mathrm{R}}{2} d \mathrm{~T} $

Substitute this value of $\mathrm{P} d \mathrm{~V}$ in equation (iii),

$ \begin{aligned} d q & =\mathrm{C}_v d \mathrm{~T}+\frac{\mathrm{R}}{2} d \mathrm{~T} \\ & =\left(\mathrm{C}_v+\frac{\mathrm{R}}{2}\right) d \mathrm{~T} \end{aligned} $

On differentiating and rearranging,

$ \int \frac{d q}{d \mathrm{~T}}=\mathrm{C}_v+\frac{\mathrm{R}}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)$

For a monoatomic gas,

$ C_v=\frac{3}{2} R $

Substitute the value of $\mathrm{C}_v$ in equation (vi),

$ \begin{aligned} & & \int \frac{d q}{d \mathrm{~T}} & =\frac{3}{2} \mathrm{R}+\frac{\mathrm{R}}{2} \\ \therefore & & \int \frac{d q}{d \mathrm{~T}} & =\frac{4}{2} \mathrm{R} \end{aligned} $

Carried by give path as shown in the diagram. Hence, by addition of $\Delta S$ for $A$ to $C, C$ to $D$ and $D$ to $B$, the value of $\Delta S_{(A \rightarrow B)}$ can be obtained.

$ \begin{aligned} & \Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{~B})}=\Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{C})}+\Delta \mathrm{S}_{(\mathrm{C} \rightarrow \mathrm{D})}+\Delta \mathrm{S}_{(\mathrm{D} \rightarrow \mathrm{~B})} \\ & \therefore \Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{~B})}=\Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{C})}+\Delta \mathrm{S}_{(\mathrm{C} \rightarrow \mathrm{D})}-\Delta \mathrm{S}_{(\mathrm{B} \rightarrow \mathrm{D})} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

As given,

$ \begin{aligned} \Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{C})} & =50 \text { e.u. } \\ \Delta \mathrm{S}_{(\mathrm{C} \rightarrow \mathrm{D})} & =30 \text { e.u. } \\ \Delta \mathrm{S}_{(\mathrm{B} \rightarrow \mathrm{D})} & =20 \text { e.u. } \end{aligned} $

substituting the values in equation (i),

$ \begin{aligned} \therefore \quad \Delta \mathrm{S}_{(\mathrm{A} \rightarrow \mathrm{~B})} & =50+30-20 \\ \therefore \quad \Delta \mathrm{~S}_{(\mathrm{A} \rightarrow \mathrm{~B})} & =60 \text { e.u. } \end{aligned} $