Thermodynamic processes

1. Isothermal process:

$T=$ constant

$\mathrm{dT}=0$

$\Delta \mathrm{T}=0$

2. Isochoric process:

$V=$ constant

$\mathrm{d} V=0$

$\Delta \mathrm{V}=0$

3. Isobaric process:

$P =$ constant

$\mathrm{dP}=0$

$\Delta \mathrm{P}=0$

4. Adiabatic process: $q=0$

or heat exchange with the surrounding $=0$ (zero)

IUPAC Sign convention about Heat and Work :

Work done on the system = Positive

Work done by the system $=$ Negative

$1^{\text {st }}$ Law of Thermodynamics

$\Delta U=\left(U_{2}-U_{1}\right)=q+w$

Law of equipartition of energy :

$U=\frac{f}{2} n R T \quad$ (only for ideal gas)

$\Delta \mathrm{E}=\frac{\mathrm{f}}{2} \mathrm{nR}(\Delta \mathrm{T})$

where $f=$ degrees of freedom for that gas. (Translational + Rotational)

$\mathrm{f}=3$ for monoatomic

$f =5$ for diatomic or linear polyatomic

$f=6$ for non - linear polyatomic

Calculation of heat $(q)$ :

Total heat capacity :

$$ \mathrm{C}_{\mathrm{T}}=\frac{\Delta \mathrm{q}}{\Delta \mathrm{T}}=\frac{\mathrm{dq}}{\mathrm{dT}}=\mathrm{J} /{ }^{\circ} \mathrm{C} $$

Molar heat capacity :

$$ C=\frac{\Delta q}{n\Delta T}=\frac{d q}{n d T}=J{mole^{-1}} K^{-1} $$

$$ C_{P}=\frac{\gamma R}{\gamma-1} $$

$$ C_{V}=\frac{R}{\gamma-1} $$

Specific heat capacity (s) :

$$ \mathrm{S}=\frac{\Delta \mathrm{q}}{\mathrm{m} \Delta \mathrm{T}}=\frac{\mathrm{dq}}{\mathrm{mdT}}=\mathrm{Jgm}^{-1} \mathrm{~K}^{-1} $$

WORK DONE (w) :

Isothermal reversible expansion/compression of an ideal gas :

$$ W = -nRT \ln (V_f / V_i) $$

Reversible and irreversible isochoric processes.

Since $d V=0$

So $d W=-P_{\text {ext }} \cdot d V=0$.

Reversible isobaric process :

$$ W=P\left(V_{f}-V_{i}\right) $$

Adiabatic reversible expansion :

$$ \Rightarrow T_2 ~V_2^{\gamma-1} = T_1 ~V_1^{\gamma-1} $$

Reversible Work :

$$ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} = \frac{nR(T_2 - T_1)}{\gamma - 1} $$

Irreversible Work :

$$ W = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} = \frac{nR(T_2 - T_1)}{\gamma - 1} $$

$$ = nC_v(T_2-T_1)= -P_{ext}(V_2-V_1) $$

and use $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

Free expansion-Always going to be irrerversible and since $P_{\text {ext }}=0$

so $\mathrm{dW}=-\mathrm{P}_{\mathrm{ext}}$. $\mathrm{dV}=0$

If no heat is supplied $q=0$

then $\Delta \mathrm{E}=0 \quad$ so $\quad \Delta \mathrm{T}=0$.

Application of $I^{st}$ Law :

$\Delta U = \Delta Q + \Delta W $ $\Rightarrow \Delta W=-P \Delta V $

$\therefore \Delta U = \Delta Q - P\Delta V $

Constant volume process

Heat given at constant volume $=$ change in internal energy

$\therefore \mathrm{du}=(\mathrm{dq})_{\mathrm{v}}$

$\mathrm{du}=\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$

$C_{v}=\frac{1}{n} \cdot \frac{d u}{d T}=\frac{f}{2} R$

Constant pressure process:

$$\mathrm{H} = Enthalpy \text {(state function and extensive property)}$$

$$\mathrm{H}=\mathrm{U}+\mathrm{PV}$$

$$ \Rightarrow C_{p}-C_{v}=R \text { (only for ideal gas) } $$

Second Law Of Thermodynamics :

$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surrounding} > 0 $ for a spontaneous process.

Entropy (S):

$$ \Delta S_{system} = \int_{A}^{B} \frac{d q_{\text {rev }}}{T} $$

Entropy calculation for an ideal gas undergoing a process :

$$ \text{State A} \quad \xrightarrow[\Delta \mathrm{S}_{\mathrm{irr}}]{\mathrm{irr}} \quad \text{State B} $$

$ P_1, V_1, T_1 \quad \quad\quad \quad P_2, V_2, T_2 $

$\Delta S_{\text {system }}=nc_{v} \ln \frac{T_2}{T_1}+ nR \ ln \frac{V_2}{V_1}\quad $(only for an ideal gas)

Third Law Of Thermodynamics :

The entropy of perfect crystals of all pure elements & compounds is zero at the absolute zero of temperature.

Gibb’s free energy (G) : (State function and an extensive property)

$$ G_{system } = H_{system } - TS_{system } $$

Criteria of spontaneity :

PYQ-2023- Chemical equilibrium-Q1

(i) If $\Delta G_{\text {system }}$ is $(-\mathrm{ve})<0 \Rightarrow$ process is spontaneous

(ii) If $\Delta \mathrm{G}_{\text {system }}$ is $>0 \quad \quad \Rightarrow$ process is non spontaneous

(iii) If $\Delta G_{\text {system }}=0 \quad \quad \Rightarrow$ system is at equilibrium.

Physical interpretation of $\Delta \mathbf{G}:$

$\rightarrow$ The maximum amount of non-expansional (compression) work which can be performed.

$\Delta \mathrm{G}=\mathrm{dw}_{\text {non-exp }}=\mathrm{dH}-\mathrm{TdS}$.

Standard Free Energy Change $\left(\Delta \mathbf{G}^{\circ}\right)$

1. $\Delta \mathrm{G}^{\circ}=-2.303 \text{ RT} \log _{10} \mathrm{~K}$

2. At equilibrium $\Delta \mathrm{G}=0$.

3. The decrease in free energy $(-\Delta G)$ is given as :

$$ \quad \quad -\Delta G = W_{net}=2.303 \ nRT \ log_{10} \frac{V_2}{V_1} $$

4. $\Delta \mathrm{G}_{\mathrm{f}}^{0}$ for elemental state $=0$

5. $\Delta G_{f}^{0}=G_{\text {products }}^{0}-G_{\text {Reactants }}^{0}$

Thermochemistry :

Change in standard enthalpy

$ \Delta H^{0} = H_{m, 2}^{0}- H_{m, 1}^{0}$

heat added at constant pressure.

$=C_{P} \Delta T.$

If $ \ H_{products }>H_ {reactants } $

$\rightarrow \quad$ Reaction should be endothermic as we have to give extra heat to reactants to get these converted into products

and if $\ H_{products }>H_ {reactants } $

$\rightarrow \quad$ Reaction will be exothermic as extra heat content of reactants will be released during the reaction.

Enthalpy change of a reaction :

$ \Delta H_{reaction } = H_{products }>H_ {reactants }$

$ \Delta H_{reaction } = H_{products }^{0} > H_ {reactants }^{0}$

$ \quad \quad \quad = positive - endothermic $

$ \quad \quad \quad = negative - exothermic $

Temperature Dependence Of $\Delta \mathrm{H}$ : (Kirchoff’s equation):

For a constant pressure reaction

$\Delta H_{2}{ }^{0}=\Delta H_{1}{ }^{0}+\Delta C_{p}(T_2-T_1)$

where $\Delta C_{P}=C_{P}(products)-C_{P}(reactants).$

For a constant volume reaction

$$ \Delta E_{2}^{0}=\Delta E_{1}^{0}+\int \Delta C_{V}\cdot dT $$

Enthalpy of Reaction from Enthalpies of Formation :

The enthalpy of reaction can be calculated by

$v_{\mathrm{B}}$ is the stoichiometric coefficient.

Estimation of Enthalpy of a reaction from bond Enthalpies:

PYQ-2023- Chemical equilibrium-Q7

$ \Delta H $= (Enthalpy required to break reactants into gaseous atoms )$ - $

(Enthalpy released to form products from the gaseous atoms )

Resonance Energy :

$\Delta H_{resonance }^{0} $

$=\Delta H_{f, experimental}^{0}-\Delta H_{f,calclulated}^{0} $

$=\Delta H_{c,calclulated}^{0}-\Delta H_{c,experimental} ^{0} $