chemistry-class-11-unit-06-chapter-03-chemical-thermodynamics-l-3-8-djx7unb7hxm

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Recap </primary>
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- Essential concepts and definitions
- Heat, work \& energy, internal energy
- First Law of thermodynamics
- Calculation of work, heat in different process for an ideal gas
- Enthalpy and heat capacity
- Determination of $\Delta \mathrm{U}$ and $\Delta \mathrm{H}$
- Change in enthalpy in different processes / reactions

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<footer style = "font-size: 18px"> $\rightarrow$ Chemical Thermodynamics $\rightarrow$ <span style = "color:blue"> Recap </span> $\rightarrow$ First law of thermodynamics $\rightarrow$  </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> First law of thermodynamics </primary>
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- The total energy $E$ of a body $\mathrm{E}-\mathrm{K}+\mathrm{V}+\mathrm{U}$ where $K$ and $V$ are the macrosoopic (not molecular) kinetic and potential energies of the body (due to motion of the body through space and the presence of fields that act on the body) and $U$ is the internal energy of the body (due to molecular motions and inter-molecular interactions)
- $\Delta \mathrm{E}-\Delta (\mathrm{K}+\mathrm{V}+\mathrm{U})$
- In absence of any varying external field acting on the system $(\Delta \mathrm{V}-0)$ and the system at rest $(\mathrm{AK}-0)$
- $\Delta \mathrm{E}-\Delta \mathrm{U}$
- First Law of thermodynamics
- The internal energy of an isolated system is constant
- For isolated system, $\Delta{U}=0$

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<footer style = "font-size: 18px">Chemical Thermodynamics $\rightarrow$ Recap $\rightarrow$ <span style = "color:blue"> First law of thermodynamics </span> $\rightarrow$  $\rightarrow$ Concept of internal energy </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Concept of internal energy </primary>
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- U is the internal energy of the body (due to molecular motions and intermolecular interactions)
- **Internal Energy U** is the total energy within a system
- For closed system
- $U=U_{tr}+U_{rot}+U_{vib}+U_{el}+U_{internal}+U_{rest}$ 
- $U_{tr}+U_{rot}+U_{vib}+U_{el}$=$f(T)$ , $U_{internal}+U_{rest}$ =$f(T,V)$
- $U_{rest}$= rest mass energy of electrons & nuclei =$m_{rest}C^2$=constant
- For closed system U= f(V,T) U=f(P,T)
- For ideal gas, closed system U=f(T)
 
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<footer style = "font-size: 18px">First law of thermodynamics $\rightarrow$  $\rightarrow$ <span style = "color:blue"> Concept of internal energy </span> $\rightarrow$ Internal energy (U) $\rightarrow$ First law of thermodynamics </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> First law of thermodynamics </primary>
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- Closed system $\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}$
- $q=$ increase in energy of the system due to thermal exchange through diathermal wall (heat transferred to the system)
- $w=$ increase in energy of the system due to mechanical exchange through non-rigid wall (work done to the system)
- The signs of $w$ and $q$ are positive if system gains some energy, and are negative if system loses some energy
- So, when the surrounding does some work on system (compression), w $>0$ and when the system does some work on surroundings (expansion), w $<0$ similarly, when the system receives some energy as heat from surroundings, $q>0$ and when the system looses some energy as heat to surroundings, $q<0$

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<footer style = "font-size: 18px">Concept of internal energy $\rightarrow$ Internal energy (U) $\rightarrow$ <span style = "color:blue"> First law of thermodynamics </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **Question 1** Express the changes in internal energy of a system when
- (ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings What type of wall does the system have?
- **Solution**
- w=0
- $\Delta u = w-q = -q$
- non-adiatic / diathermal rigid wall

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<footer style = "font-size: 18px">First law of thermodynamics $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ First law of thermodynamics </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **Question 2** For each process state whether each of q, w, and $\Delta$ u is positive, zero, or negative
- (a) Combustion of benzene in a sealed container with rigid, adiabatic walls

- **Solution**

- w=0

- q=0 $\Delta u=0$

- (b) Combustion of benzene in a sealed container that is immersed in a water bath at $25^oC$ and has rigid, thermally conducting walls

- q<0, w=0 $\Delta u  <0$
 
- (c) Adiabatic expansion of a nonideal gas into vacuum
 - q=0, w=0, $\Delta u=0$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Calculation of work $\rightarrow$ Calculation of work </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Calculation of work </primary>
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- Expansion against constant pressure (P), w= -P$\Delta V$
- Free expansion expansion against zero opposing force, $P_{ex}=0$, w=0
- Reversible, isothermal process, ideal gas
- $w=-\int_{V_i}^{V_f} P d V=-\int_{V_i}^{V_f} \frac{n R T}{V} d V=-n R T \int_{V_i}^{V_f} \frac{d V}{V}=-n R T \ln \frac{V_f}{V_i}$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Calculation of work $\rightarrow$ <span style = "color:blue"> Calculation of work </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
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- **Question 3**. Ideal gas, closed system, reversible process
- (i) 10 Pa, 1$m^3$, T $\xrightarrow{isothermal}$ 1 Pa, 10$m^3$, T
- **Solution**
- w=-nRT In $\frac{10m^3}{1m^3}$
- w=-PV In 10
- w=-10 P a $m^3$ $\times$ 2.303
- w=-23.303 J

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<footer style = "font-size: 18px">Calculation of work $\rightarrow$ Calculation of work $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Calculation of work </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
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- (ii) 10 Pa, 1$m^3$, T $\xrightarrow{isochoric}$ 1 Pa, 1 $m^3$,$T_2$ $\xrightarrow{isobaric}$ 1 Pa, 10$m^3$, T
- w=$w_1+w_2$
- w=0-P($V_2-V_1$)
- w=-1 Pa(10-1)$m^3$
- w=-9 J

- (iii) 10 Pa, 1$m^3$, T $\xrightarrow{ isobaric}$ 10 Pa, 10 $m^3$,$T_3$ $\xrightarrow{isochoric}$ 1 Pa, 10$m^3$, T
- w=$w_1+w_2$
- w=-10 Pa(10-1)$m^3$+0
- w=-90 J

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<footer style = "font-size: 18px">Calculation of work $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Calculation of work $\rightarrow$ Enthalpy </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Calculation of work </primary>
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- Ideal gas, closed system $\quad$ Calculate $w_f$ and $w_b$
- 1 step $\quad$ 1 Pa, 10 $m^3$, T $\xrightarrow {irreversible} $ 10 Pa, 1 $m^3$, T
- 2 step $\quad$ 1 Pa, 10 $m^3$, T $\xrightarrow {irreversible} $ 5 Pa,2 $m^3$, T $\xrightarrow {irreversible}$ 10 Pa, 1 $m^3$, T<p>
- Infinite no of steps 1 Pa, 10 $m^3$, T $\xrightarrow {irreversible} $ 10 Pa,1 $m^3$, T
- What happens for the reverse process?

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Calculation of work </span> $\rightarrow$ Enthalpy $\rightarrow$ Enthalpy </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **Question** 5. A process
- $H_2O (C) (1 mol) \rightarrow H_2O(g)$ (1 mol) at 1bar, $100^oC$

- **Solution**

- $\Delta H= \Delta_{vap}H$ = 41 KJ $mol^{-1}$
- $\Delta H= \Delta U+ \Delta_{ng}RT$
- $\Delta U= \Delta H- \Delta_{ng}RT$
- $\Delta U$= 41 KJ $mol^{-1} \times 1 mol-1 mol\times 8.314 J mol^{-1} K^{-1}\times373k $
- $\Delta U$= 37.9 KJ
- $\Delta U$= 37.9 KJ $mol^{-1}$

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<footer style = "font-size: 18px">Enthalpy $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Heat (q) </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-3 </Success>
### <primary style="font-size:24px"> Heat (q) </primary>
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- General process
 - $(P_1, V_1, T_1) (I) \rightarrow (P_2, V_2, T_2) (II)$
- $\Delta H, \Delta u$ are state variables-
 - $(P_1, V_1, T_1) (I) \xrightarrow {const. V} (V_1, T_2, P_2) \xrightarrow {const. T} (T_2, V_2, P_2)(II)$
 - $\Delta u = \Delta u_1 + \Delta u_2$
- An ideal gas; u=f(T)
 - $\Delta u_2 =0$
 - $\Delta u_1 = C_v \Delta T$
 - $\Delta u = C_v \Delta T+0 = C_v \Delta T$
 - $\Delta U= C_v \Delta T$

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<footer style = "font-size: 18px">Heat (q) $\rightarrow$ Heat (q) $\rightarrow$ <span style = "color:blue"> Heat (q) </span> $\rightarrow$ Heat (q) $\rightarrow$ Heat (q) </footer>