Thermodynamics

In an adiabatic process, where no heat is exchanged, the relationship between temperature (T) and volume (V) for an ideal gas is given by:
T V^{(γ - 1)} = constant, Here, γ (gamma) Adiabatic constant of gas.

T2 =546 K

In free expansion against a vacuum, there is no external pressure resisting the expansion, so no work is done (W = 0). Since the process is isothermal for an ideal gas, there is also no net change in internal energy, so the heat exchanged (Q) is zero.

If $∆\mathrm{H} >0, ∆\mathrm{S}>0$ then process is spontaneous at high temperature
If $∆\mathrm{H} >0, ∆\mathrm{S}<0$ process is always non spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}>0$, then process is always spontaneous
If $∆\mathrm{H} <0, ∆\mathrm{S}<0$, then process is spontaneous at low temperature.

In an adiabatic process, where no heat is exchanged, the relationship between temperature (T) and volume (V) for an ideal gas is given by:
T V^{(γ - 1)} = constant, Here, γ (gamma) Adiabatic constant of gas.

T2 =546 K

For isothermal reversible expansion:

$W=-nRT\ln \left(\frac{P_1}{P_2}\right)$

$$\begin{gathered} W=-1\times 2\times 298\times \ln \left(\frac{20}{10}\right)=-596\times \ln \left(2\right) \\ =-596\times \left(0.693\right)=–413.14\text{cal} \end{gathered}$$

For closed insulated container q =0

$$\begin{gathered} ∆\mathrm{E} =\mathrm{q}+\mathrm{w} \\ ∆\mathrm{E} = \mathrm{W} \end{gathered}$$

Extensive properties are mass-dependent and intensive properties are independent of the amount of mass.

Mass , Volume, and internal energy are all extensive properties

• A: Liquid → vapour = increase in randomness ⇒ entropy increases

• B: Cooling to 0 K decreases entropy

• C: Solid decomposes → gases form ⇒ entropy increases

• D: 1 mole gas → 2 moles gas ⇒ entropy increases

“ΔG is positive for a spontaneous reaction” is incorrect.
For spontaneity ΔG < 0; a positive ΔG indicates non-spontaneity.

“ΔG is positive for a spontaneous reaction” is incorrect.
For spontaneity ΔG < 0; a positive ΔG indicates non-spontaneity.

For closed insulated container q =0

$$\begin{gathered} ∆\mathrm{E} =\mathrm{q}+\mathrm{w} \\ ∆\mathrm{E} = \mathrm{W} \end{gathered}$$

The given reaction is an endothermic process that is non-spontaneous at the freezing point of water but becomes spontaneous at the boiling point of water. This indicates that the process follows Gibbs free energy equation

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

It is only possible when $\Delta H$ and $\Delta S$ both are positive.

Using ΔG° = -RT ln Kc, with Kc = 10¹², R = 8.314 J/mol·K, T = 298 K, we get ΔG° = -68.47 kJ/mol. Correct answer: (C).

$$\begin{gathered} \mathrm{For} \mathrm{monoatomic} \mathrm{gas} \mathrm{\gamma } =5/3 \\ \mathrm{As} \mathrm{we} \mathrm{know} \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{H}}{∆\mathrm{U}} (\mathrm{at} \mathrm{constant} \mathrm{P}, \mathrm{H} =\mathrm{Q} ) \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{Q}}{∆\mathrm{U}}, \mathrm{hence} \frac{∆\mathrm{Q}}{∆\mathrm{U}}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=5/3 \end{gathered}$$

$\begin{array}{l}\mathrm{\Delta H}=\mathrm{\Delta U}+{\mathrm{\Delta n}}_{\mathrm{g}}\mathrm{RT}\\ 15000=\mathrm{\Delta U}+(-2)\times 8.314\times 300\end{array}$

$\begin{array}{l}\mathrm{\Delta U}=15000+4988.4\\ \mathrm{\Delta U}=19988.4\mathrm{J}\end{array}$

Using ΔH = ΔU + ΔnRT, with ΔU = -3263.9 kJ/mol, Δn = -1.5, R = 8.314 × 10⁻³ kJ/mol·K, and T = 298 K,

$$\mathrm{\Delta }H=-3263.9-3.717=-3267.6\text{ kJ/mol}$$

Correct answer: (D).

The given reactions are:

$$\begin{gathered} S\left(g\right)+\frac{3}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+2xkcal \\ S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to S{O}_3\left(g\right)+ykcal \end{gathered}$$

The heat of formation of $S{O}_2\left(g\right)$ can be derived by subtracting the heat of the second reaction from the first:

$\Delta H_{\text{formation }}\left(S{O}_2\right)=y-2x$

Thus, the enthalpy of formation of $S{O}_2$ is $y-2x$.

Given:
Insulated thermally means q = 0

Acc. to 1st law,

$$\begin{gathered} ∆\mathrm{U} = \mathrm{q} + \mathrm{w} \\ ∆\mathrm{U} = 0 + \mathrm{w}= \mathrm{w} \\ \mathrm{w} > 0, ∆\mathrm{U} > 0 \end{gathered}$$

(a) Using Hess's Law, we sum the given reactions:
1. C (diamond) $\to C$ (graphite) + X
2. C (diamond) $+O_2\to C{O}_2+Y$
3. C (graphite) $+O_2\to C{O}_2+Z$

Reversing (3):

$C{O}_2\to C\left(\text{ graphite }\right)+O_2-Z$

Adding to (2):
C(diamond) → C(graphite) + (Y–Z)
Comparing with (1): X = Y – Z

For Isochoric process $\mathrm{P} \mathrm{\alpha } \mathrm{T} \mathrm{at} \mathrm{constant} \mathrm{V}$

$$\begin{gathered} \mathrm{Work} \mathrm{done} \mathrm{in} \mathrm{isochoric} \mathrm{process} \mathrm{W} =-\mathrm{P}∆\mathrm{V}, ∆\mathrm{V} =0 , \mathrm{hence} \mathrm{W}= 0 \\ \mathrm{Now} \mathrm{internal} \mathrm{energy} ∆\mathrm{U} ={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}, \mathrm{Hence} \mathrm{as} \mathrm{temperature} \mathrm{increse} , \mathrm{internal} \mathrm{energy} \mathrm{increses} \end{gathered}$$

$$(\Delta G)_{P,T} > 0 \;\Rightarrow\; \text{Non-spontaneous process}$$

For Isochoric process $\mathrm{P} \mathrm{\alpha } \mathrm{T} \mathrm{at} \mathrm{constant} \mathrm{V}$

$$\begin{gathered} \mathrm{Work} \mathrm{done} \mathrm{in} \mathrm{isochoric} \mathrm{process} \mathrm{W} =-\mathrm{P}∆\mathrm{V}, ∆\mathrm{V} =0 , \mathrm{hence} \mathrm{W}= 0 \\ \mathrm{Now} \mathrm{internal} \mathrm{energy} ∆\mathrm{U} ={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}, \mathrm{Hence} \mathrm{as} \mathrm{temperature} \mathrm{increse} , \mathrm{internal} \mathrm{energy} \mathrm{increses} \end{gathered}$$

For closed insulated container q =0

$$\begin{gathered} ∆\mathrm{E} =\mathrm{q}+\mathrm{w} \\ ∆\mathrm{E} = \mathrm{W} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{desired} \mathrm{reaction} \mathrm{is} \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ}-----\left(1\right) \\ \mathrm{Now} \mathrm{C}\left(\mathrm{diamond}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{y} \mathrm{KJ}---------\left(2\right) \\ \mathrm{C}\left(\mathrm{Graphite}\right) +{\mathrm{O}}_2\to {\mathrm{CO}}_2+ \mathrm{z} \mathrm{KJ}---------\left(3\right) \\ \mathrm{By} \mathrm{substraction} \mathrm{equation} \left(3\right) \mathrm{from} \left(2\right) \mathrm{we} \mathrm{get} \mathrm{equation} \left(1\right) \\ \mathrm{C}\left(\mathrm{diamond}\right) \to \mathrm{C}\left(\mathrm{graphite}\right) + \mathrm{x} \mathrm{KJ} \\ \mathrm{Now} \mathrm{x}=\mathrm{y}-\mathrm{z} \mathrm{KJ} \end{gathered}$$.

Extensive properties are mass-dependent and intensive properties are independent of the amount of mass.

Mass , Volume, and internal energy are all extensive properties

The standard enthalpy of formation $\left({∆}_{\mathrm{f}}{\mathrm{H}}^{\mathrm{\theta }}\right)$ for an element in its most stable form at any specified temperature is defined as zero. Oxygen's most stable form is the diatomic gas, O₂(g), at all temperature where it remains a gas.

$$\begin{gathered} \mathrm{For} \mathrm{monoatomic} \mathrm{gas} \mathrm{\gamma } =5/3 \\ \mathrm{As} \mathrm{we} \mathrm{know} \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{H}}{∆\mathrm{U}} (\mathrm{at} \mathrm{constant} \mathrm{P}, \mathrm{H} =\mathrm{Q} ) \\ \mathrm{\gamma }=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=\frac{∆\mathrm{Q}}{∆\mathrm{U}}, \mathrm{hence} \frac{∆\mathrm{Q}}{∆\mathrm{U}}=\frac{{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_{\mathrm{V}}}=5/3 \end{gathered}$$

Negative value of electron gain enthalpy indicates the amount of energy released when an electron is added to an isolated gaseous system.

More negative value of electron gain enthalpy means large amount of energy is released and it also signifies greater ionic character.

B has most negative electron gain enthalpy hence it will form a compound with greater ionic character with E followed by A , D and C.

• Energy supplied (heat), $Q = 48 \, \text{J}$Q=48J

• Gas: monoatomic, $n = 1 \, \text{mol}$n=1mol

• Temperature change: \Delta T = 2\,^\circ \text{C} = 2 \, \text{K}

We are asked to find work done, W.

For any process:

$$Q = \Delta E + W$$Q=ΔE+W

Where:

• $\Delta E = n C_V \Delta T$ΔE=nCV​ΔT (for a monoatomic ideal gas, $C_V = \frac{3}{2} R$CV​=23​R)

• $W =$W= work done by gas

• $R = 8.314 \, \text{J/mol·K}$R=8.314J/mol\cdotpK

$$C_V = \frac{3}{2} R = \frac{3}{2} \cdot 8.314 \approx 12.471 \, \text{J/mol·K}$$CV​=23​R=23​⋅8.314≈12.471J/mol\cdotpK $$\Delta E = n C_V \Delta T = 1 \cdot 12.471 \cdot 2 \approx 24.942 \, \text{J} \approx 25 \, \text{J}$$ΔE=nCV​ΔT=1⋅12.471⋅2≈24.942J≈25J

$$W = Q - \Delta E = 48 - 24.942 \approx 23 \, \text{J}$$W=Q−ΔE=48−24.942≈23J