Thermodynamics

(A) Given, Temperature coefficient $\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}=\frac{\mathrm{R}}{\mathrm{F}}$

$ \mathrm{M}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{M}^{2+}(\mathrm{aq}) $

Here, number of electron $(n)=2$ transferred

Since,

$ \begin{aligned} \Delta S & =\left(\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}\right) n \mathrm{~F} \\\\ \Delta S & =\frac{\mathrm{R}}{\mathrm{F}} \times 2 \times \mathrm{F} \\\\ \Delta \mathrm{S} & =2 \mathrm{R} \end{aligned} $

Option (A) is incorrect.

(B)

$ \begin{aligned} & \mathrm{E}_{\text {cell }}=\frac{-2.303 \mathrm{RT}}{\mathrm{F}} \log \frac{0.01}{0.1}=\frac{2.303 \mathrm{RT}}{\mathrm{F}} \\\\ & \frac{\mathrm{dE}_{\text {cell }}}{\mathrm{dT}}=\frac{2.303 \mathrm{R}}{\mathrm{F}} \\\\ & \therefore \Delta \mathrm{S}=\mathrm{nF} \frac{\mathrm{dE}}{\mathrm{dT}}>0 \end{aligned} $

It is an entropy driven process.

(C) It is correct

During racemisation of optically active compound, disorder increases and hence, entropy increases.

(D) For $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+3 \mathrm{en} \rightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]^{+3}+6 \mathrm{H}_2 \mathrm{O}$,

Entropy increases when bidentate ligands replace monodentate ligands due to increase in the number of molecules on the product side.

Hence, (B, C, D) are correct.

From state I to II Reversible isothermal expansion takes place. So, following changes take place,

● pressure decreases.
● Volume increases.
● Temperature remains constant.
● Enthalpy, H remains constant.
● Entropy, S for expansion increases.

So, all options follows the above mentioned conditions, so all graphs are correct for state I and II.

From state II to III Reversible adiabatic expansion takes place. So, following changes take place.

● pressure decreases.
● Volume increases.
● Temperature decreases.
● Enthalpy, H decreases.
● Entropy, S remains constant.
● H increases instead of decreasing, so only option (c) is incorrect.

All other options, i.e., (a), (b) and (d) follows the above mentioned conditions.

Therefore, correct graphical representations are (a), (b) and (d).

Given, $w = - \int {{p_{ext}}dV} $

For 1 mole van der Walls' gas

$p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)$

For reversible process, p_ext = p_gas

$w = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV} $

But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.

The standard enthalpy of formation is defined as standard enthalpy change for formation of 1 mole of a substance from its elements, present in their most stable state of aggregation.

${3 \over 2}$O₂(g) $ \to $ O₃(g);

${1 \over 8}$S₈(s) + O₂(g) $ \to $ SO₂(g)

In the above two reactions standard enthalpy of reaction is equal to standard enthalpy of formation.

The explanation of given statements are as follows :

(a) U_rms is inversely proportional to the square root of its molecular mass.

${U_{rms}} = \sqrt {{{3RT} \over M}} $

Hence, option (a) is correct.

(b) When temperature is increased four times then U_rms become doubled.

${U_{rms}} = \sqrt {{{3R} \over M} \times 4T} $

${U_{rms}} = 2 \times \sqrt {{{3RT} \over M}} $

Hence, option (b) is correct.

(c) and (d) E_av is directly proportional to temperature but does not depends on its molecular mass at a given temperature as ${E_{av}} = {3 \over 2}KT$. If temperature raised four times than E_av becomes four time multiple.

Thus, option (c) is incorrect and option (d) is correct.

For the reaction at equilibrium

$ \mathrm{A} \rightleftharpoons \mathrm{P} $

At any temperature:

Concentration of $A$ at equilibrium $[A]_{e q}<5$ Concentration of $\mathrm{B}$ at equilibrium $[\mathrm{P}]_{e q}>5$

Equilibrium constant $\left(\mathrm{K}_{e q}\right)=\frac{[\mathrm{P}]_{e q}}{[\mathrm{~A}]_{e q}}$

$ \mathrm{K}_{e q}>1 $

$ \text{Also,}~~ \frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}}>\frac{\mathrm{T}_2}{\mathrm{~T}_1} $

$ \begin{aligned} & \text{If}~\mathrm{T}_2 >\mathrm{T}_1 \\\\ & \text{Hence}~\frac{\ln \mathrm{K}_{\mathrm{T}_1}}{\ln \mathrm{K}_{\mathrm{T}_2}} >\frac{\mathrm{T}_2}{\mathrm{~T}_1}>1 \\\\ & \frac{\mathrm{K}_{\mathrm{T}_1}}{\mathrm{~K}_{\mathrm{T}_2}} >1 \\\\ & \mathrm{~K}_{\mathrm{T}_1} >\mathrm{K}_{\mathrm{T}_2} \end{aligned} $

It can be clearly seen from diagram that concentration of reactant is lesser and that of product is higher at $T_1$ compared to $T_2$. Hence,

equilibrium constant is higher at $\mathrm{T}_1$ than $\mathrm{T}_2$. The reaction is exothermic.

i.e., $\Delta \mathrm{H}^{\circ}<0$ for forward reaction

The expression for the spontaneity and Gibbs free energy of reaction:

$ \begin{aligned} \Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\\\ \Delta \mathrm{S}^{\circ} & =\frac{\Delta \mathrm{H}^{\circ}-\Delta \mathrm{G}^{\circ}}{\mathrm{T}} \end{aligned} $

Since, both $\Delta \mathrm{G}^{\circ}$ and $\Delta \mathrm{H}^{\circ}$ are less than zero $\left\{\Delta \mathrm{H}^{\circ}<0\right.$ and $\left.\Delta \mathrm{G}^{\circ}<0\right\}$

If $\Delta \mathrm{H}^{\circ}>\Delta \mathrm{G}^{\circ}$, then $\Delta \mathrm{S}^{\circ}>0$

Otherwise,

If $\Delta \mathrm{H}^{\circ}<\Delta \mathrm{G}^{\circ}$, then $\Delta \mathrm{S}^{\circ}<0$

we know,

$ \begin{aligned} \Delta \mathrm{G}^{\circ} & =-\mathrm{RT} \ln \mathrm{K} . \\\\ \Delta \mathrm{G}_1^{\circ} & =-\mathrm{RT}_1 \ln \mathrm{K}_{\mathrm{T}_1} \\\\ \text { and } \Delta \mathrm{G}_2^{\circ} & =-\mathrm{RT} \ln \mathrm{K}_{\mathrm{T}_2} \end{aligned} $

Its given, $\ln \mathrm{K}_{\mathrm{T}_1}>\ln \mathrm{K}_{\mathrm{T}_2}$

$ \begin{gathered} \Delta \mathrm{G}_1^{\circ}<\Delta \mathrm{G}_2^{\circ} \\\\ \Delta \mathrm{H}_{\mathrm{T}_1}^{\circ}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_1}^{\mathrm{o}}<\Delta \mathrm{H}_{\mathrm{T}_2}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{S}_{\mathrm{T}_2}^{\mathrm{o}} \end{gathered} $

This is possible only if $\Delta S^{\circ}<0$

Hence,

$ \begin{array}{r} \Delta \mathrm{G}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0 \\\\ \Delta \mathrm{H}^{\circ}<0, \Delta \mathrm{S}^{\circ}<0 \end{array} $

In the given curve AC represents isochoric process as volume at both the points is same i.e., V₁

Similarly, AB represents isothermal process (as both the points are at T₁ temperature) and BC represents isobaric process as both the points are at p₂ pressure.

Now, (i) for option (a)

q_AC = $\Delta$U_BC = nC_v(T₂ $-$ T₁)

where, n = number of moles

C_v = specific heat capacity at constant volume

However, W_AB $\ne$ p₂(V₂ $-$ V₁) instead

${W_{AB}} = nR{T_1}\ln \left( {{{{V_2}} \over {{V_1}}}} \right)$

So, this option is incorrect.

(ii) For option (b)

q_BC = $\Delta$H_AC = nC_p(T₂ $-$ T₁)

where, C_p = specific heat capacity at constant pressure

Likewise,

W_BC = $-$p₂(V₁ $-$ V₂)

Hence, this option is correct.

(iii) For option (c)

as nC_p(T₂ $-$ T₁) < nC_v(T₂ $-$ T₁)

so, $\Delta$H_CA < $\Delta$U_CA

and q_AC = $\Delta$U_BC

Hence, this option is also correct.

(iv) For option (d)

Although q_BC = $\Delta$H_AC

but $\Delta$H_CA $\le$ $\Delta$U_CA

Hence, this option is incorrect.

$\Delta {S_{surr}} = - {{\Delta H} \over {{T_{surr}}}}$

For endothermic, if T_surr increases, unfavourable change in entropy of the surroundings decreases.

For exothermic, if T_surr increases, favourable change in entropy of the surroundings decreases.

Option (A): Correct. For compression of gas, $V_2 < V_1$, thus, from $w=-p \Delta V$, work done on the gas will be positive. Below graph shows the pressure-volume curve for work done during reversible and irreversible compression process. Area under curve during irreversible process is greater than the area under curve during reversible process. Therefore, work done on the gas in irreversible process is greater than in reversible process, that is, $w_{\text {irr }}>w_{\text {rev }}$.


Option (B): Correct. For work done in free expansion, $p_{\text {ext }}=0$, therefore, $w=0$. For an adiabatic process, $q=0$ and for isothermal process, $\Delta T=0$. Therefore, from first law of thermodynamics, $\Delta U=q+w=0$.

Option (C): Correct. Below graph represents the expansion under adiabatic (AC) and isothermal condition (AB). Therefore, $\left|w_{\mathrm{AB}}\right|>\left|w_{\mathrm{AC}}\right|$.

When dealing with solutions and thermodynamics, particularly with ideal solutions, several key aspects must be considered to understand the behavior of the system and its interactions with the surroundings. Let's analyze each option in the context of forming an ideal solution from benzene and naphthalene at room temperature:

Option A: $\Delta G$ is positive

For an ideal solution, the Gibbs free energy change ($\Delta G$) upon mixing can be described as follows:

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

An ideal solution has the characteristic that there is no enthalpy change upon mixing ($\Delta H = 0$) because the intermolecular forces between unlike molecules are assumed to be equal to those between like molecules. Additionally, when two or more components form an ideal solution, the entropy of the system ($\Delta S$) increases due to the mixing of different molecules, which is a manifestation of increased randomness or disorder.

Since $\Delta H = 0$ and $\Delta S > 0$ for an ideal solution, applying these to the Gibbs free energy equation yields:

$\Delta G = 0 - T\Delta S = -T\Delta S$

Given that $T > 0$ and $\Delta S > 0$, $\Delta G$ will ultimately be negative, indicating a spontaneous process. Therefore, Option A is false.

Option B: $\Delta S_{system}$ is positive

As mentioned earlier, the entropy of the system increases in the formation of an ideal solution. This is because the different molecules of benzene and naphthalene mix, increasing the randomness or disorder of the system. Therefore, Option B is true.

Option C: $\Delta S_{surroundings}$ = 0

In the formation of an ideal solution, there is no heat exchange with the surroundings since $\Delta H = 0$. Without a heat exchange, there is no change in the entropy of the surroundings, as entropy change in the surroundings is generally associated with heat exchange according to the relation $\Delta S = \frac{q}{T}$ for reversible processes. Therefore, Option C is true.

Option D: $\Delta H$ = 0

In the context of an ideal solution, it is assumed that the enthalpy change ($\Delta H$) is zero because the energy required to break interactions among similar molecules is exactly balanced by the energy released from the formation of new interactions between different types of molecules. This implies no net heat effect. Therefore, Option D is true.

In summary, Options B, C, and D are true, while Option A is false for the ideal solution of benzene and naphthalene at room temperature.

For isothermal process T₁ = T₂. Work done in isothermal process is less than adiabatic process. In case of isothermal process, the temperature remains constant so there is no change in the internal energy, whereas in case of adiabatic process expansion occurs through internal energy.

As entropy is a state function and is additive

$\Delta {S_{X \to Z}} = \Delta {S_{X \to Y}} + \Delta {S_{Y \to Z}}$

On moving from Y to Z, the work done is zero as the volume is kept constant (isochoric process), so

${W_{X \to Y \to Z}} = {W_{X \to Y}}$

An intensive property is a property that is independent of the amount of mass or size of the system. In contrast, an extensive property is dependent on the size or amount of mass in the system. Let's analyze each option based on this definition:

• Molar Conductivity: Molar conductivity, denoted as $\Lambda_m$, is a measure of the ionic conductivity of an electrolyte solution divided by the concentration of the electrolyte. Since it is normalized by the amount of substance, it does not depend on the total amount of the substance present in the solution. Thus, molar conductivity is an intensive property.
• Electromotive Force (EMF): EMF is the voltage generated by a battery or by the magnetic force according to Faraday’s Law of electromagnetic induction. It refers to a potential difference and does not depend on the quantity of material or size of the battery. Therefore, EMF is considered an intensive property.
• Resistance: Resistance, denoted by $R$, measures how much a material opposes the flow of electric current. It depends on the material’s length, cross-sectional area, and resistivity. For instance, a longer wire has greater resistance. Therefore, resistance is an extensive property.
• Heat Capacity: Heat capacity is the amount of heat required to change the temperature of a substance by a certain temperature interval. It is dependent on the amount of substance present, and therefore, it is an extensive property. The specific heat capacity, however, is intensive since it is the heat capacity per unit mass.

From the options provided:

• Option A (Molar Conductivity) is intensive.
• Option B (Electromotive Force) is intensive.
• Option C (Resistance) is extensive.
• Option D (Heat Capacity) is extensive.

Therefore, the intensive properties among the given options are Option A (Molar Conductivity) and Option B (Electromotive Force).

State functions are properties of a system that depend only on the current state of the system, not on the path used to get to that state. They are intrinsic properties of the system and include properties like pressure, temperature, volume, internal energy, enthalpy, entropy, and Gibbs free energy.

Option A: Internal Energy

Internal energy, denoted as $U$, is a state function. It is the total energy contained within the system, including kinetic and potential energy at the molecular level. Internal energy changes in a system only depend on the initial and final states of the system, regardless of the process or path taken to achieve these states. Therefore, it fits the criteria of a state function.

Option B: Irreversible Expansion Work

Irreversible expansion work is not a state function. Work, in general, is a path function because it depends on the path taken during a process. Whether the path involves irreversible or reversible processes, work includes energy transfer that depends significantly on how that transfer is carried out. Therefore, irreversible expansion work depends on the specific details of the process and is not solely determined by the initial and final states of the system.

Option C: Reversible Expansion Work

Similarly to irreversible expansion work, reversible expansion work is also a path function and not a state function. Even though reversible processes are ideal and involve quasi-static changes that maintain the system in near equilibrium throughout, the work done (or reversible expansion work) during such processes still depends on the specific path taken. This includes how slowly the process is carried out and the intermediate steps, distinguishing it from a state function.

Option D: Molar Enthalpy

Molar enthalpy, denoted as $H$, is a state function. It is defined as the sum of the internal energy $U$ of a system plus the product of the pressure $P$ and volume $V$ of the system, multiplied by the number of moles $n$, expressed as:

$H = U + PV$

Since $U$, $P$, and $V$ are all state functions, their combination into enthalpy continues to be dependent solely on the state of the system, not on the path taken to reach that state.

Conclusion:

Among the given options, Internal energy (Option A) and Molar enthalpy (Option D) are state functions, while both Irreversible expansion work (Option B) and Reversible expansion work (Option C) are not state functions. They are path functions.