Thermodynamics

$ \begin{aligned} &\begin{aligned} & \log _{10} \mathrm{~K}=-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2.303 \mathrm{RT}}+\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2.303 \mathrm{R}} \\ & \text{Compare this with the straight line form } y=mx+c \text{ where } y=\log_{10}K \text{ and } x=\frac{1}{T}. \\ & \text{Then the constant term is the } y\text{-intercept: } \frac{\Delta \mathrm{S}^{\mathrm{o}}}{2.303 \mathrm{R}} \\ & \text{And the coefficient of } \frac{1}{T} \text{ is the slope: } -\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2.303 \mathrm{R}} \end{aligned} \end{aligned} $

For an ideal gas undergoing isothermal reversible expansion:

• Temperature is constant $\Rightarrow \Delta U = 0$ (for ideal gas, $U$ depends only on $T$)

• Hence $\Delta H = 0$ (since $H$ also depends only on $T$ for ideal gas)

Step 1: Given data (convert units)

• $V_1 = 20.0\,\text{dm}^3 = 20.0 \times 10^{-3}\,\text{m}^3 = 0.020\,\text{m}^3$

• $P_1 = 0.5\,\text{MPa} = 0.5 \times 10^6\,\text{Pa} = 5.0\times 10^5\,\text{Pa}$

• $P_2 = 0.2\,\text{MPa}$

• $T = 600\,\text{K}$

Step 2: Find number of moles using $PV = nRT$

$ n=\frac{P_1V_1}{RT} =\frac{(5.0\times 10^5)(0.020)}{(8.314)(600)} =\frac{10000}{4988.4} \approx 2.0\ \text{mol} $

Step 3: Work done in reversible isothermal expansion

For reversible isothermal process:

$ w = -nRT\ln\left(\frac{V_2}{V_1}\right) = -nRT\ln\left(\frac{P_1}{P_2}\right) $

Convert to base-10 log form (NCERT style):

$ w = -2.303\,nRT\,\log\left(\frac{P_1}{P_2}\right) $

Now,

$ \frac{P_1}{P_2}=\frac{0.5}{0.2}=2.5 $

Given:

$ \log(2.5)=\log\left(\frac{5}{2}\right)=\log 5-\log 2=0.6989-0.3010=0.3979 $

So,

$ w=-2.303(2.0)(8.314)(600)(0.3979) $

First calculate $nRT$:

$ nRT = 2.0\times 8.314\times 600 \approx 9976.8\ \text{J} $

Then,

$ w \approx -2.303 \times 9976.8 \times 0.3979 \approx -9130\ \text{J} \approx -9.1\ \text{kJ} $

Step 4: Heat and energy changes

For isothermal ideal gas:

$ \Delta U = 0 $

First law (NCERT sign convention):

$ \Delta U = q + w $

So,

$ 0 = q + (-9.1) \Rightarrow q = +9.1\ \text{kJ} $

Also,

$ \Delta H = 0 $

Correct option

Option A:

$ w=-9.1\,\text{kJ},\quad \Delta U=0,\quad \Delta H=0,\quad q=+9.1\,\text{kJ} $

For methane:

$ \Delta H_\text{atom}(CH_4)=x=\;4D(\text{C–H}) \;\Rightarrow\; D(\text{C–H})=\frac{x}{4} $

For ethane:

$ \Delta H_\text{atom}(C_2H_6)=y=6D(\text{C–H})+D(\text{C–C}) $

$ \Rightarrow\; D(\text{C–C})=y-6\left(\frac{x}{4}\right)=y-\frac{6x}{4} =\frac{4y-6x}{4}\quad(\text{kJ mol}^{-1}) $

Energy per molecule needed to break the C–C bond:

$ E=\frac{D(\text{C–C})\times 1000}{N_A}\quad(\text{J}) $

Longest wavelength corresponds to minimum photon energy:

$ \frac{hc}{\lambda}=E \Rightarrow \lambda=\frac{N_Ahc}{1000\,D(\text{C–C})} $

$ \lambda=\frac{N_Ahc}{1000\left(\frac{4y-6x}{4}\right)} =\frac{N_Ahc}{250(4y-6x)} $

Correct option: D.

For any process, NCERT uses the expression for work (done on the system):

$w=-\int_{V_i}^{V_f} P_{\text{ext}}\,dV$

A. Reversible isothermal expansion (ideal gas)

In reversible isothermal change, $P_{\text{ext}}=P=\dfrac{nRT}{V}$.

$ w=-\int_{V_i}^{V_f}\frac{nRT}{V}\,dV =-nRT\ln\left(\frac{V_f}{V_i}\right) $

So, A $\rightarrow$ II.

B. Free expansion

In free expansion, $P_{\text{ext}}=0$.

$ w=-\int_{V_i}^{V_f} 0\,dV=0 $

So, B $\rightarrow$ I.

C. Irreversible expansion (against constant external pressure)

Here $P_{\text{ext}}$ is constant.

$ w=-P_{\text{ext}}(V_f-V_i) $

So, C $\rightarrow$ III.

D. Irreversible compression (against constant external pressure)

Compression means work done on the system is positive. Using the same NCERT form:

$ w=-P_{\text{ext}}(V_f-V_i)=P_{\text{ext}}(V_i-V_f) $

This corresponds to the given form IV (written as $-P_{\text{ext}}(V_i-V_f)$ in the list, which becomes positive when $V_i>V_f$ for compression).

So, D $\rightarrow$ IV.

Final matching

$A\text{-}II,\;B\text{-}I,\;C\text{-}III,\;D\text{-}IV$

Correct option: C

$\underset{5^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\ell)} \longrightarrow \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\ell) \rightleftarrows} \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{g})}$

When water is heated in the microwave, its temperature increases from $5^{\circ}\mathrm{C}$ to $100^{\circ}\mathrm{C}$. After reaching $100^{\circ}\mathrm{C}$, it starts changing from liquid to vapour (boiling), so liquid water and water vapour can exist together.

During boiling, the water changes into steam and the volume increases (expansion).

Because of expansion, the system does work on the surroundings, so work done by the system is negative.

$ \mathrm{w}=-\mathrm{ve} $

Since the microwave supplies energy to the water, heat is absorbed by the system, so $q$ is positive.

Also, water vapour has more internal energy than liquid water at the same temperature, so the internal energy of the system increases. Therefore, $\Delta \mathrm{U}$ is positive.

So, $q=+\mathrm{ve}$ and $\Delta \mathrm{U}=+\mathrm{ve}$.

Option (A)

This is a reversible isothermal expansion of an ideal gas, so we use the NCERT formula for isothermal reversible work:

$ \begin{aligned} & \mathrm{W}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\ & =\frac{-2 \times 8.314 \times 300}{1000} \times \ln \left(\frac{20}{2}\right) \mathrm{kJ} \\ & =-11.5 \mathrm{~kJ} \end{aligned} $

The negative sign shows work is done by the gas during expansion. The magnitude of work is $11.5 \ \mathrm{kJ}$.

Option (B)

This is an irreversible isothermal expansion against a constant external pressure, so work is calculated by:

$ \begin{aligned} & \mathrm{W}=-\mathrm{P}_{\text {ext }}\left[\mathrm{V}_2-\mathrm{V}_1\right] \\ & =-3[3-1] \\ & =-6 \mathrm{~kJ} \end{aligned} $

The negative sign again means work is done by the gas. The magnitude of work is $6 \ \mathrm{kJ}$.

Option (C)

For an ideal gas, internal energy depends only on temperature. For any process (including adiabatic), we use:

$ \begin{aligned} & \Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T} \\ & =1 \times \frac{3}{2} \times \frac{8.314 \times 320}{1000} 100 \mathrm{~kJ} \\ & =3.99 \end{aligned} $

So the change in internal energy comes out to approximately $3.99 \ \mathrm{kJ}$ (about $4 \ \mathrm{kJ}$).

Option (D)

For an ideal gas, enthalpy also depends only on temperature. At constant pressure, we use:

$ \begin{aligned} \Delta \mathrm{H} & =\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{~T} \\ & =1 \times \frac{5}{2} \times \frac{8.314 \times 337}{1000} \mathrm{~kJ} \\ & =7 \mathrm{~kJ} \end{aligned} $

So, $\Delta H = 7 \ \mathrm{kJ}$.

$ \mathrm{C}(\mathrm{~s})+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_4(\mathrm{~g}) $

$-x=\left(\Delta H_{\text {sub }}\right.$ of carbon $)+2 \times($ B.E. of H-H $)-4 \times($ B.E. of $C-H) $

$-x=y+2 z-4($ B.E. of C-H $)$

B.E of $\mathrm{C}-\mathrm{H}=\frac{\mathrm{y}+2 \mathrm{z}+\mathrm{x}}{4}$

Work Done is Area Under the Curve

In a graph where pressure ($P$) is plotted against volume ($V$), the area under the curve shows the amount of work done during the process.

Option B: Zero Work

In option (B), the area under the curve is zero. This means no work is done in this case.

Other Options: Work is Negative (Expansion)

For the other options, the area is below the axis, so the work done is negative. This happens because the gas is expanding.

Magnitude of Work

The answer given by NTA looks only at the amount (magnitude) of work, not if it is negative or positive.

$ \begin{aligned} & W_1 = (44.8) \times \ln 2 = 31.04 \\ & W_3 < 31.04 \\ & W_4 < 31.04 \end{aligned} $

(A) The standard free energy change ($\Delta_\mathrm{r} G^\circ$) is found by subtracting the $\mathrm{G}^\circ$ value of reactant (A) from that of product (B).

(A) Calculation:
$\Delta_\mathrm{r} G^\circ = \mathrm{G}^\circ_B - \mathrm{G}^\circ_A$

In the graph, this value is positive (i.e., $+\text{ve}$).

(B) A positive $\Delta_\mathrm{r} G^\circ$ means the reaction is not spontaneous in the forward direction under standard conditions. But it does not mean that $\mathrm{N_2O_4}$ will not change into $\mathrm{NO_2}$ at all. Some amount will still convert, and equilibrium will be set up with both $\mathrm{N_2O_4}$ and $\mathrm{NO_2}$ present.

(C) The reverse reaction does not go to completion. Because at equilibrium (point E in the graph), both $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ are present. This means the reverse reaction is also only partially complete.

(D) When 1 mole of $\mathrm{N_2O_4}$ is converted to the equilibrium mixture, the free energy change is $\Delta_\mathrm{r} G^\circ = -0.84~\mathrm{kJ~mol}^{-1}$ (as shown on the graph).

(E) When 2 moles of $\mathrm{NO_2}$ change to the equilibrium mixture, the standard free energy change is:

$-5.40~\mathrm{kJ~mol}^{-1} + (-0.84~\mathrm{kJ~mol}^{-1}) = -6.24~\mathrm{kJ~mol}^{-1}$

Given data:

$n = 0.5\ \text{mol},\ P_0 = 2\ \text{bar},\ T = 600\ \text{K}$

The gas is compressed in two steps: first against constant pressure $P_1$ and then against constant pressure of 8 bar. Here $P_1$ can take any value between 2 and 8 bar.

We can find the volumes at each stage using the ideal gas equation $PV = nRT$ :

$V_0 = \frac{nRT}{2}, \quad V_1 = \frac{nRT}{P_1}, \quad V_2 = \frac{nRT}{8}$

Step 1: Work done on the gas during compression against external pressure $P_1$ is given by:

$W_1 = P_1(V_0 - V_1)$

Substituting the values of $V_0$ and $V_1$:

$W_1 = P_1\left(\frac{nRT}{2} - \frac{nRT}{P_1}\right) = nRT\left(\frac{P_1}{2} - 1\right) \qquad \ldots(1)$

Step 2: Work done on the gas when compressed against 8 bar pressure:

$W_2 = P_2(V_1 - V_2) = 8\left(\frac{nRT}{P_1} - \frac{nRT}{8}\right) = nRT\left(\frac{8}{P_1} - 1\right) \qquad \ldots(2)$

Total work done on the gas:

$|W| = W_1 + W_2 = nRT\left(\frac{P_1}{2} - 1\right) + nRT\left(\frac{8}{P_1} - 1\right)$

$|W| = nRT\left(\frac{P_1}{2} + \frac{8}{P_1} - 2\right)$

Finding the condition for minimum work:

We need to minimize $\left(\frac{P_1}{2} + \frac{8}{P_1}\right)$. Using the inequality AM ≥ GM :

$\frac{\left(\frac{P_1}{2} + \frac{8}{P_1}\right)}{2} \ge \sqrt{\frac{P_1}{2} \times \frac{8}{P_1}}$

Simplifying,

$\frac{P_1^2 + 16}{4P_1} \ge 2$

$P_1^2 - 8P_1 + 16 \ge 0$

$ (P_1 - 4)^2 \ge 0 $

This means the minimum value of $|W|$ occurs when $P_1 = 4$ bar.

Substitute $P_1 = 4$:

$|W|_{\min} = nRT\left(\frac{4}{2} + \frac{8}{4} - 2\right)$

$|W|_{\min} = 0.5 \times R \times 600 \times (2 + 2 - 2)$

$|W|_{\min} = 600R$

(P) Physisorption: In physisorption, gas molecules get attached to a solid surface with weak van der Waals forces. The process can be shown as:

$ \mathrm{x(g)} \rightarrow \mathrm{x(s)} $

Since a gas changes to a solid, the disorder decreases, so $\Delta S < 0$. Also, energy is released when molecules are adsorbed, so $\Delta H < 0$.

(Q) Diamond → Graphite: When diamond changes to graphite, the structure becomes more disordered because graphite has weak van der Waals forces between its layers. Hence:

$ \Delta S > 0, \quad \Delta H < 0 $

Graphite is more stable under normal conditions because of its lower energy and higher disorder.

(R) Denaturation of Protein: Denaturation means the protein’s regular structure is disrupted when heat or chemicals are added. This breaks bonds within the protein chain and increases disorder. So:

$ \begin{aligned} & \Delta H > 0 \text{ (heat is absorbed)} \\ & \Delta S > 0 \text{ (disorder increases as structure breaks)} \end{aligned} $

(S) Propene → Cyclopropane: Propene is an open-chain compound, while cyclopropane has a closed ring with angular strain. Converting propene to cyclopropane increases stored energy and reduces molecular freedom.

$ \begin{aligned} & \Delta H > 0 \text{ (due to angular strain)} \\ & \Delta S < 0 \text{ (open chain has more rotational freedom)} \end{aligned} $

Therefore, the correct matching is:

$ \mathrm{P}-2; \quad \mathrm{Q}-5; \quad \mathrm{R}-1; \quad \mathrm{S}-4 $

$\mathrm{KCl}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\Delta \mathrm{H} \text { sol. }} \mathrm{K}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{\text {(aq) }}^{-1}$

$\begin{aligned} \Delta \mathrm{H}_{\mathrm{Sol}^{\mathrm{n}} .} & =\mathrm{L} \cdot \mathrm{E} \cdot+(\mathrm{H} \cdot \mathrm{E})_{\mathrm{K}_{(\mathrm{g})}^{+}}+(\mathrm{HE})_{\mathrm{Cl}_{(\mathrm{g})}^{-1}} \\ & =\mathrm{Z}-\mathrm{x}-\mathrm{y} \\ & =\mathrm{z}-(\mathrm{x}+\mathrm{y}) \end{aligned}$

Option A:

It claims that the standard state for a pure gas is defined at 1 bar and 273 K.

In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is 298 K (25°C), not 273 K (0°C).

Thus, Option A is incorrect.

Option B:

It states that the standard enthalpy of formation $\Delta_f H^\circ_{500}$ is zero for $O_2(g)$.

By convention, the standard enthalpy of formation for an element in its most stable form is defined as zero. Even though typical tables list $\Delta_f H^\circ_{298}$ values, if one were to evaluate the formation enthalpy at any temperature where the element remains in its standard state, the reaction

$O_2(g) \rightarrow O_2(g)$

has no change in enthalpy.

Therefore, by definition, $\Delta_f H^\circ_{500}(O_2(g)) = 0$.

This makes Option B correct.

Option C:

It claims that $\Delta_f H^\circ_{298}$ is zero for $O(g)$ (atomic oxygen).

However, the standard state of oxygen is diatomic $O_2(g)$, not atomic oxygen. Since $O(g)$ is not the most stable form of oxygen under standard conditions, its formation enthalpy is not zero.

Hence, Option C is incorrect.

Option D:

It asserts that the term "standard state" implies that the temperature is 0°C (273 K).

In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature—commonly 298 K for tabulated thermodynamic data—not necessarily 0°C.

Thus, Option D is also incorrect.

To summarize:

The only correct statement is Option B.

Therefore, the correct answer is: Option B.

$\begin{aligned} & \Delta H=1 \times y(0-10)-x \times 1000+1 \times z\left(-10^{\circ}-0^{\circ}\right) \\ & \Delta H=-10(100 x+y+z) \text { Joule. } \end{aligned}$

From the given information

$\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.

$\mathrm{HCl}(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\quad\text{..... (1)}$

$\begin{aligned} & \Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}} \\ & \mathrm{HCl}(\mathrm{~g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\quad\text{..... (2)} \end{aligned}$

$\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$

Hence heat of solution depends upon amount of solvent

By equation ............. (2) $-$ equation ........... (1)

$\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$

So Heat of dilution $=-72.79-(-69.01)$

$=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$

Hence option (3) is incorrect.

For heat of formation reactant should be in elemental form hence option (4) is incorrect

$\begin{aligned} & (10 \mathrm{~L}, 300 \mathrm{~K}) \xrightarrow{\mathrm{n}=1}(20 \mathrm{~L}, 300 \mathrm{~K}) \\ & -\mathrm{q}=\mathrm{w}=-\mathrm{nRT} \ln \frac{\mathrm{~V}_2}{\mathrm{~V}_1} \\ & =-8.3 \times 300 \times \ln \left(\frac{20}{10}\right) \\ & =-1.718 \mathrm{~kJ} \\ & \Rightarrow \mathrm{q}=1.718 \mathrm{~kJ} \\ & \mathrm{w}=-1.718 \mathrm{~kJ} \\ & \Delta \mathrm{U}=0(\because \Delta \mathrm{~T}=0) \end{aligned}$

For reaction to be spontaneous according to $2^{\text {nd }}$ law:

$\begin{aligned} & \Delta \mathrm{G}<0 \\ & \Rightarrow \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S}<0 \\ & \Rightarrow \mathrm{~T}>\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\ & \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}} \end{aligned}$

The correct choice is Option C: Both Statement I and Statement II are true.

Explanation:

Statement I is true because

During the ice–water phase change at 0 °C, any heat added goes into the latent heat of fusion.

The temperature remains fixed at 0 °C until all the ice has melted.

Statement II is also true because

The heat absorbed at the melting point is used to overcome (break) the hydrogen‐bonding forces between water molecules in ice—in other words, it increases the system’s potential energy.

Since no heat goes into raising the kinetic energy of the molecules, the temperature (which measures average kinetic energy) does not increase during melting.

Mathematically, for a mass m of ice, the heat needed is

$ Q = m\,L_f $

where $L_f$ is the latent heat of fusion of ice. This heat does not change the temperature but is entirely consumed in changing the phase (i.e., increasing potential energy).

$\begin{aligned} &\text { For isothermal process }\\ &\begin{aligned} & \left|\mathrm{W}_{\text {reversible }}\right|_{\text {expansion }}=\left|\mathrm{W}_{\text {reversible }}\right|_{\text {compression }} \\ & =-\mathrm{nRT} \ln \frac{\mathrm{~V}_{\mathrm{f}}}{\mathrm{~V}_{\mathrm{i}}} \\ & \left|\mathrm{~W}_{\text {irreversible }}\right|_{\text {expansion }}<\left|\mathrm{W}_{\text {irreversible }}\right|_{\text {compression }} \\ & \mathrm{d}>\mathrm{c}=\mathrm{a}>\mathrm{b} \\ & \left|\mathrm{~W}_{\text {irreversible }}\right|_{\text {expansion }}=-\mathrm{P}_{\text {ext }}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right) \end{aligned} \end{aligned}$

Graphical representation

We can compare work by area of PV graph.

In the Haber process, the chemical reaction is as follows:

$ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3 $

The change in enthalpy ($\Delta \text{H}^\circ$) is negative, indicating that the reaction is exothermic.

The change in entropy ($\Delta \text{S}^\circ$) is also negative because the number of moles of gas decreases during the reaction.

Key points to consider:

As temperature increases, the term $-\frac{\Delta H_R^0}{T}$ decreases.

The free energy change is given by:

$ \Delta \text{G}^\circ = -\text{RT} \ln \text{K}_{\text{eq}} $

Rearranging gives:

$ \text{R} \ln \text{K}_{\text{eq}} = -\frac{\Delta \text{G}^0}{T} $

For an exothermic reaction, as temperature increases, the equilibrium constant ($ \text{K}_{\text{eq}} $) decreases.

Both $\Delta \text{H}^\circ$ and $\Delta \text{S}^\circ$ are almost constant with respect to temperature.

It is free expansion of gas $\Rightarrow P_{\text {ext }}=0$

Where $\mathrm{w}=0, \mathrm{q}=0$ and $\Delta \mathrm{U}=0$

Given,

${C_{(diamond)}}\buildrel {} \over \longrightarrow {C_{(graphite)}} + X\,kJ\,mo{l^{ - 1}}$ ............ (1)

${C_{(diamond)}} + {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Y\,kJ\,mo{l^{ - 1}}$ ............. (2)

${C_{(graphite)}} + {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Z\,kJ\,mo{l^{ - 1}}$ .............. (3)

Condition for temperature : constant

Hess's law is applied here. (Hess's law of constant heat summation)

The given reactions (2) and (3),

(2) - (3) $\Rightarrow$

${C_{(diamond)}} + {O_2}(g) - \left( {{C_{(graphite)}} + {O_2}(g)} \right)\buildrel {} \over \longrightarrow C{O_2}(g) + Y - \left( {C{O_2}(g) + Z} \right)$

${C_{(diamond)}} + {O_2}(g) - {C_{(grpahite)}} - {O_2}(g)\buildrel {} \over \longrightarrow C{O_2}(g) + Y - C{O_2}(g) - Z$

${C_{(diamond)}} - {C_{(graphite)}} \to Y - Z$

${C_{(diamond)}} \to {C_{(graphite)}} + (Y - Z)$ ............. (4)

Comparing (1) and (4),

$X = Y - Z$

So, the correct answer is option (2) $X = Y - Z$

Amount of energy transpired as hat, $Q=500 \mathrm{~J}$

Number of moles of Argon, $n=0.5 \mathrm{~mol}$

Temperature (Initial), $T_i=298 \mathrm{~K}$

Pressure, $P=1.00 \mathrm{~atm}$

Final temperature, $T_F=$ ?

Change in internal energy, $\Delta V=$ ?

$R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

For a constant pressure process, the heat added is related to the temperature changes as

$Q=n C_p \Delta T$ ....... (1)

$\begin{aligned} &\text { Also, the change in internal energy formula is }\\ &\Delta U=n C_V \Delta T \quad-(2)\quad\text{.... (2)} \end{aligned}$

$\begin{aligned} &\text { For monoatomic gas (Argon), }\\ &\begin{aligned} C_v & =\frac{3}{2} R \\ \text { So, } C_p & =C_v+R \quad\left(C_p-C_v=R\right) \\ & =\frac{3}{2} R+R=\frac{5}{2} R \end{aligned} \end{aligned}$

$\begin{aligned} &\text { In equation (1) } \Rightarrow\\ &\begin{aligned} & Q=n C_p \Delta T \\ & Q=n C_p\left(T_F-T_i\right) \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} R\left(T_F-T_i\right) \quad R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} \times 8.3 \mathrm{~J} K^{-1} \mathrm{~mol} ^{-1}\left(T_F-T_i\right) \end{aligned} \end{aligned}$

$T_F-T_i=\frac{500 \mathrm{~J}}{0.5 \mathrm{mol} \times \frac{5}{2} \times 8.3 \text { J K}^1 \mathrm{~mol}^{-1}}$

$\begin{aligned} & =\frac{500}{0.5 \times \frac{5}{2} \times 8.3} \mathrm{k} \\ & =48.2 \mathrm{k} \\ T_F & =48.2 \mathrm{k}+298 \mathrm{k} \\ & =346.2 \mathrm{k} \\ & \approx 348 \mathrm{k} \end{aligned}$

$\begin{aligned} &\text { In equation (2) } \Rightarrow\\ &\begin{aligned} \Delta U & =n C_V \Delta T \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} R \times 48.2 \mathrm{~K} \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol^{-1}} \times 48.2 \mathrm{~K} \\ & =300.045 \mathrm{~J} \approx 300 \mathrm{~J} \end{aligned} \end{aligned}$

The final temperature is, 348 k

change in internal energy is 300 J

So, correct answer is Option 2) 348 K and 300 J

As internal energy ' $U$ ' is a state function, its cyclic integral must be zero in a cyclic process

$\therefore \Delta U \text { case }(I)=\Delta U \text { case }(I I)=\Delta U \text { case }(\text { III })$

If pressure is made two time then mixture of ice and water will completely convert into water (liquid) form.

To determine the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$, we need to consider the given reactions and their enthalpy changes.

The reactions are:

$\mathrm{S}(\mathrm{~g}) + \frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + 2x \mathrm{kcal}$

$\mathrm{SO}_2(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{SO}_3(\mathrm{~g}) + y \mathrm{kcal}$

For reaction 2, we have:

$ \Delta \mathrm{H}_{\mathrm{r}} = \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_3} - \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} = -y $

Given that:

$ -\mathrm{y} = -2 \mathrm{x} - \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} $

Rearranging this equation gives:

$ \left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_2} = y - 2x $

Thus, the heat of formation for $\mathrm{SO}_2(\mathrm{~g})$ is $y - 2x$ kcal.

The rise in temperature of neutralization reaction will be maximum for maximum number of moles of strong acid and strong base neutralized and lower volume of final solution.

Option (D) and (C) have weak acids and in option (A) only 20 mmol of HCl will be neutralized with 70 mL final volume.

Reaction is spontaneous at relatively high temperature and non-spontaneous at low temperature $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ It is only possible when $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are positive.

Option (1)

Let's analyze each combination using the Gibbs free energy equation:

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

A reaction is spontaneous when $\Delta G < 0.$

Now, consider each case:

(A) $\Delta H > 0$ and $\Delta S < 0$

Here, $\Delta G = +\text{(positive)} - T(-\text{(positive)}) = \text{positive} + T\text{(positive)}$

This is positive for any temperature.

Thus, the reaction is non-spontaneous for any T, which is correct.

(B) $\Delta H > 0$ and $\Delta S > 0$

Now, $\Delta G = +\text{(positive)} - T\text{(positive)}.$

For $\Delta G$ to be negative, the temperature has to be high enough such that $T\Delta S > \Delta H.$

The table incorrectly states that the reaction is spontaneous at low T.

(C) $\Delta H < 0$ and $\Delta S < 0$

In this case, $\Delta G = -\text{(positive)} - T(-\text{(positive)}) = -\text{(positive)} + T\text{(positive)}.$

At very low temperatures, the term $T\Delta S$ is small, making $\Delta G$ negative (spontaneous).

The table mistakenly indicates that the reaction is non-spontaneous at low T.

(D) $\Delta H < 0$ and $\Delta S > 0$

Then, $\Delta G = -\text{(positive)} - T\text{(positive)}.$

This expression is negative at all temperatures, so the reaction is correctly marked as spontaneous.

Thus, the incorrect combinations are:

(B): It should be spontaneous at high T, not low T.

(C): It should be spontaneous at low T, not non-spontaneous.

Therefore, the incorrect combinations are (B) and (C) only.

The correct answer is Option D.

$\begin{aligned} & \Delta \mathrm{S}_{\text {overall }}=\Delta \mathrm{S}_1+\Delta \mathrm{S}_2+\Delta \mathrm{S}_3+\Delta \mathrm{S}_4+\Delta \mathrm{S}_5 \\ & \Delta \mathrm{~S}_2=\frac{\Delta \mathrm{H}_{\mathrm{m} \text { fusion }}}{273} \mathrm{~T}_{\mathrm{f}}=273^{\prime} \mathrm{K}^{\prime} \\ & \Delta \mathrm{S}_3=\int_\limits{273}^{373} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}} \\ & \Delta \mathrm{~S}_4=\frac{\Delta \mathrm{H}_{\mathrm{m} \text { vaporisation }}}{373} \mathrm{~T}_{\mathrm{b}}=373^{\prime} \mathrm{K}^{\prime} \\ & \Delta \mathrm{S}_5=\int_\limits{373}^{383} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}} \\ & \text { Answer }=(2) \end{aligned}$

We want to match each partial derivative in List–I to the appropriate thermodynamic quantity in List–II:

1. $\displaystyle\left(\frac{\partial G}{\partial T}\right)_{P}$

A well-known thermodynamic identity is:

$ \left(\frac{\partial G}{\partial T}\right)_{P} \;=\; -S. $

Hence,

$ (A) \quad \longrightarrow \quad -S \quad (\text{II}). $

2. $\displaystyle\left(\frac{\partial H}{\partial T}\right)_{P}$

By definition of the heat capacity at constant pressure,

$ \left(\frac{\partial H}{\partial T}\right)_{P} \;=\; C_P. $

Hence,

$ (B) \quad \longrightarrow \quad C_P \quad (\text{I}). $

3. $\displaystyle\left(\frac{\partial G}{\partial P}\right)_{T}$

A standard thermodynamic relation (for a single‐component system) is:

$ \left(\frac{\partial G}{\partial P}\right)_{T} \;=\; V. $

Hence,

$ (C) \quad \longrightarrow \quad V \quad (\text{IV}). $

4. $\displaystyle\left(\frac{\partial U}{\partial T}\right)_{V}$

By definition of the heat capacity at constant volume,

$ \left(\frac{\partial U}{\partial T}\right)_{V} \;=\; C_V. $

Hence,

$ (D) \quad \longrightarrow \quad C_V \quad (\text{III}). $

Final Matching

$ (A) \to \text{(II)}, \quad (B) \to \text{(I)}, \quad (C) \to \text{(IV)}, \quad (D) \to \text{(III)}. $

Which corresponds to:

Option C: (A)-(II), (B)-(I), (C)-(IV), (D)-(III).

Let’s analyze the situation step by step:

System: The liquid inside a thermally insulated (adiabatic) and closed vessel.

Process: The liquid is mechanically stirred from outside.

Heat Exchange ($q$)

Because the vessel is thermally insulated, there is no heat exchange between the system and surroundings. Hence,

$ q = 0. $

Work ($w$)

The external mechanical stirring does work on the system by agitating the liquid. Work done on the system is conventionally taken as positive:

$ w > 0. $

Change in Internal Energy ($\Delta U$)

From the First Law of Thermodynamics,

$ \Delta U \;=\; q \;+\; w. $

Since $q = 0$ and $w > 0$, we get

$ \Delta U = w > 0. $

Hence:

$ \boxed{ \Delta U > 0,\quad q = 0,\quad w > 0. } $

Answer: Option B is correct.

Heat at constant pressure,

$ \begin{aligned} & q_p=n C_p \Delta T \\ & \Rightarrow \quad 1 \times 10.314 \times 1=10.314 \mathrm{~J} \end{aligned} $

Enthalpy change $\Delta H$

$ \Delta H=q_p=10.314 \mathrm{~J} / \mathrm{mol} $

Given, $\Delta S=-42.4 \mathrm{JK}^{-1}$

$ \Delta H=-41.2 \mathrm{~kJ}=-41200 \mathrm{~J} $

Using the equation, $\Delta G=\Delta H-T \Delta S$

$ T_{\mathrm{eq}}=\frac{-41200 \mathrm{~J}}{-42.4 \mathrm{JK}^{-1}} \approx 971.8 \mathrm{~K} $

A reversible process is one that can be reversed by an infinitesimal change in a condition, allowing the system and surroundings to be restored to their original states. Such process is occur at equilibrium.

I. Vaporisation of a liquid at its boiling point This is a reversible process because the liquid and vapour are in equilibrium at the boiling point.

II. Expansion of gas into vacuum This is an irreversible, spontaneous process as the system is not in equilibrium and cannot be reversed without external work.

III. Transformation of a solid into liquid at its melting point This is a reversible processes as the solid and liquid phases are in equilibrium at the melting point.

IV. Neutralisation of an acid by a base This is an irreversible, spontaneous reaction that proceeds to completion.

Therefore, only processes I and III are reversible.

$ \text { Gibbs' free energy is given by } $

$ \begin{array}{rlrl} \Delta G^{\circ} & =-R T \ln (K) \\ \Rightarrow \quad-115000 \mathrm{~J} & =-8.314 \times 298 \times \ln K_p \\ \Rightarrow \quad \log _{10}\left(K_p\right) & =\frac{46.42}{2.303} \\ & \log _{10}\left(K_p\right) =+20.15 \end{array} $

$ \begin{aligned} &\text { Work done in isothermal reversible expansion, }\\ &\begin{aligned} W & =-n R T \ln \left(\frac{p_2}{p_1}\right) \\ W & =-(1)(8.3)(300) \ln \left(\frac{2}{20}\right)=5730 \mathrm{~J} / \mathrm{mole} \text { or } 5.73 \mathrm{~kJ} / \mathrm{mol} \\ x & =5.73 \end{aligned} \end{aligned} $