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

Chemical Thermodynamics L-3

Recap

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

Chemical Thermodynamics L-3

First law of thermodynamics

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, $\Delta{U}=0$

Chemical Thermodynamics L-3

Concept of internal energy

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)

Chemical Thermodynamics L-3

Internal energy (U)

Extensiveproperty
For cyclic process, $\int d U=0$
$\Delta \mathrm{U}$ is independent of path
State function
- How can we change the value of $\mathrm{U}$ of a closed system?
- by exchanging energy with surroundings
work (we will restrict ourselves to P-V work or mechanical work)
heat

Chemical Thermodynamics L-3

First law of thermodynamics

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, $q<0$

Chemical Thermodynamics L-3

Problem

Question 1 Express the changes in internal energy of a system when
(i) No heat is absorbed by the system from surroundings, but work (w) is done on the system What type of wall does the system have?
Solution
q = 0
$\Delta u= q+w$
adiabatic
non-rigid wall

Chemical Thermodynamics L-3

Problem

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

Chemical Thermodynamics L-3

Problem

Question 1 Express the changes in internal energy of a system when
(iii) $\mathrm{w}$ amount of work is done by the system and q amount heat is supplied to the system. What type of system would it be?
Solution
$\Delta u= q-w$
closed

Chemical Thermodynamics L-3

Problem

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, $\Delta u=0$

Chemical Thermodynamics L-3

Calculation of work

P-V work / mechanical work/ work of expansion (or compression), $w_{p v}$ or $w_{\text {exp }}$
non-expansion/ additional work , any other work $w_{non p v}$ or $w_{\text {add}}$
P-V Work/Work
(a) General expression of work (reversible process)
Total work done w= - $\int_{V_i}^{V_f}P_{ex}dV-\int_{V_i}^{V_f}PdV$
(b) General expession of work (irreversible process)
Total work done, $w=-P_{ex}\left(V_f-V_i\right)$

Chemical Thermodynamics L-3

Calculation of work

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
$w=-\int_{V_i}^{V_f} P d V=-\int_{V_i}^{V_f} \frac{n R T}{V} d V=-n R T \int_{V_i}^{V_f} \frac{d V}{V}=-n R T \ln \frac{V_f}{V_i}$

Chemical Thermodynamics L-3

Problem

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

Chemical Thermodynamics L-3

Problem

(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

Chemical Thermodynamics L-3

Calculation of work

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
Infinite no of steps 1 Pa, 10 $m^3$ , T $\xrightarrow {irreversible}$ 10 Pa,1 $m^3$ , T
What happens for the reverse process?

Chemical Thermodynamics L-3

Enthalpy

$H=U+P V$
State variable
Extensive quantity
Absolute value can be not be determined experimentally
$\Delta H$ only depands on initial and final state

Chemical Thermodynamics L-3

Enthalpy

Ist (law, closed system)
- $\Delta u=q+w$
A process at constant value
- $(P_1, T_1, V) \rightarrow(P_2, T_2, V)$
- w=0
- $\Delta u = q_V$
A process at constant Pressure
- $(T_1, V_1, P) \rightarrow(T_2, V_2, P)$
- $q_p=\Delta H$

Chemical Thermodynamics L-3

Enthalpy

Constant pressure process
$\Delta H=\Delta u+P \Delta V$
For liquid/Solid $\triangle V$ is very small $\Delta v \approx 0$
$\Delta H=\Delta U$
Reactions / processes involving gases
$\Delta H=\Delta U+\Delta n g R T \text { (gases are } \text { ideal)}$

Chemical Thermodynamics L-3

Problem

Molar enthalpy change of vapourisation of water at 1 bar and $100^o{C}$ is $41 {kJ} {mol^{-1}}$ ", calculate the internal energy change when (assume water vapour is a perfect gas)
1 mol of water is vaporised at 1 bar and $100^o{C}$
1 mol of water is converted into ice

Chemical Thermodynamics L-3

Problem

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 $mol^{-1}$

Chemical Thermodynamics L-3

Problem

$H_2O (l) (1 mol) \rightarrow H_2O (s)$
generally constant pressure
dH= $\Delta$ U+P $\Delta$ v
dH $\approx$ $\Delta$ U = 41 KJ
$\Delta$ U = 41 KJ $mol^{-1}$

Chemical Thermodynamics L-3

Heat (q)

Heat exchange happens because of temp difference
$q \propto \Delta T$
$q=C \Delta T$
$dq \propto dT$
$dq= CdT$
$q= \int_{T_1}^{T_2}CdT$
if C is constant in temp range between $T_1 at T_2$
$q=C \Delta T$

Chemical Thermodynamics L-3

Heat (q)

$q=C \Delta T$
C= heat capacity (capital letter/ uppercase) extensive quantity.
$q=nC_m \Delta T$
$C_m= \frac{C}{n}$ = molar heat capacity (intensive quantity)
$q=m.c.dT$
c= $\frac{c}{m}$ = specific heat Capacity (small letter / lowercase)

Chemical Thermodynamics L-3

Heat (q)

Heat exchage can be carried out in two ways - content pressure, constant volume
if the process is at constant V
$(P_1, T_1, V) \rightarrow (P_2, T_2, V)$
$\Delta u= q_v$
$\Delta u= C_v \Delta T$
if the process is at Constant P
$\Delta H = qp =C_p \Delta T$

Chemical Thermodynamics L-3

Heat (q)

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)
- $\Delta u_2 =0$
- $\Delta u_1 = C_v \Delta T$
- $\Delta u = C_v \Delta T+0 = C_v \Delta T$
- $\Delta U= C_v \Delta T$

Chemical Thermodynamics L-3

Heat (q)

$\Delta H= C_p \Delta T$ for ideal gas
$\Delta U= C_v \Delta T$
$C_p$ and $C_v$ are related

Chemical Thermodynamics L-3

Heat (q)

for solids and liquids
$\Delta V \approx 0$
$C_p \approx C_v$
in true sense $C_p>C_v$ (with exception where volume reduces on heating)
$C_p=C_v$
$4^o C$ , water,1atm

Chemical Thermodynamics L-3

Ideal gas

$\Delta H =\Delta U+\Delta(P V)$
$\Delta =\Delta U+\Delta(n R T)$
$C_P\Delta T = C_V \Delta T$ +n R $\Delta T$
$C_P-C_V = n R$
$C_{p,m}-C_{v,m} = R$

Chemical Thermodynamics L-3

Problem

Question 6. q, w, $\Delta U, \Delta H$
- Reversible melting of benzene at l atm and normal melting point
Solution (i)
- $q>0 \quad \Delta H >0$
- $\Delta v>0 \quad w=-p\Delta v$
- $w<0$
- $\Delta u=q+w \quad \Delta u >0$

Chemical Thermodynamics L-3

Problem

Solution (ii)
Reversible melting of ice at 1 atm, $0^{\circ} \mathrm{C}$
- $q>0, \quad q_\rho=\Delta H>0$
- $\Delta v<0 \quad w=-p \Delta v$
- $w >0 \text { - very small }$
- $\Delta u=q+w>0$

Chemical Thermodynamics L-3

Problem

Solution (iii)
Reversible isothermal and expansion of an ideal gas
- $\Delta V>0, \omega<0$
- $\Delta T=0, \Delta H=0$
- $\Delta u=0$
- $\Delta u=0= q + w$
- $q>0$