1st and 2nd law of thermodynamics law , enthalpy, entropy,molar specific heat capacity,carnot cycle

2.  2 PARTS-
 Thermo – Heat
 Dynamics – Study of motion
(Study of motion of heat)

3.  SYSTEM- the part of universe
 SURROUNDING
 IMAGINARY OR REAL SURFACE
SURROUNDING SYSTE
M
REAL
SURFACE

4.  SYSTEM TYPES-
Matter ex. Energy ex.
1. Open
2. Close
3. Isolated

5.  State variables
 P
 V
 T
 M

6.  Any action which can change the state of the
system
 Types
 Reversible
o Slow
o Can be reverted at any point by small operation
force
 Irreversible
o Fast
o Can’t be reverted

7.  Intrinsic / Intensive- does not depend on
the amount of system.
 Extrinsic / extensive- depended on the
amount of system
PVT P,T,V/2 P,T,V/2

8.  Heat –
 Heat added to the system (+)
 Heat release by the system(-)
 Work-
 Work done on the system (+)
 Work done by the system (-)

9. piston

10.  change in energy is depends on effect of
heat (Q) and effect of work(W).
E=Q+W
 No work , no heat exchange then
change in system energy
E=0

11. 1. When small change in ex. Pressure is=Pex
2. Then small change in volume =dV
3. And work
dW= -Pex.dV ( –ve bcz of opposing force)

13.  The max. work which a system can do
w/o any external energy take.
 Can’t be calculated
 But change in energy can be calculated
(?>>,?<<)

14. FUNCTIONS
 State F.-
 It doesn’t matter which
way choose to reach
from E1 to E2
ex- internal energy
 Path F-
 Depends up on path
ex- work,heat
E1,P,V,T
E2,P’,V’,T’

15.  Energy is conserved.
 Energy can’t be created , can’t be
destroy ,it can only be converted one
form to another form.
 For isolated system - E=0
 For open and closed system-
E =Q+W

16.  Change in internal energy at constant
pressure=
E=Qp-P. V
 Change at constant volume=
E=Qv-P*0
E=Qv
So heat given at constant volume is equal to direct
change in internal energy

18. enthalpy
 It is a thermodynamic state function
 Like internal energy it can’t be exactly
calculated
 H=E+PV
H= E+P. V (at constant pressure)
H=Qp-P. V+P. V ( bcz E=Qp-P V at
constant pressure)
H=Qp so heat given at constant
pressure =direct change in enthalpy

19. Molar/specific heat capacity
Amount of heat in joules required to rise 1 mole
of a substance 1 Kelvin.
at constant pressure If we have n mole gas, increased
it temp T and heat expended Qp
Then Cp= Qp/n. T
Similarly to constant volume = Cv=Qv/n. T
(when a system absorbs heat, the temp increases the heat
capacity of a system is defined as the ratio of change in heat to
the change in temp).
we have to prove that Cp-Cv=R……1
Qp/n. T-Qv/n. T= Qp-Qv/n. T = Cp-Cv
we know that H=Qp (enthalpy), and E=Qv (internal energy).

20. H- E/n. T = Cp-Cv…………2
But we know H= E- P V,put in eqn 2
P v/n. T=Cp-Cv……………...3
We know PV=nRT
P( V) = nR( T) ( we are working at cons. pressure )
P V/n. T= R (put the value in eqn 3)
So Cp-Cv=R where R is 2 cal/mol/k

21.  Kelvin statement- In a cyclic process
,heat can not be 100% converted to
work.
 Clausiu’s statement- Heat does not
spontaneously flow from cold to hot.

22.  Randomness/ Dis-orderness
solid liquid gas
 Low entropy=low randomness
 N2(g)+3H2(g)=2NH3(g)
1 + 3 = 2 (randomness decrease S= -ve)

23.  S=Qreversible/T
 Entropy =0 (at truly reversible process) &
 >0 for an irreersible
Gas in
container
Reversible process at Q
heat
Constant temp

24.  Ideal cycle
 Initial state = final state
 No human involved
 4 steps – 2 isothermal (T cons) ,2 adiabatic ( no heat
exchange)
 1st step (thermal expension)
P
v
a
b
Power step

25.  Step 2-adiabatic expansion( Q=0)
 Volume increases then Temp. reduces

26.  Why not compress the gas is isothermally and bring it back
to initial position?
If we do that work net =0
 Why adiabatic expansion?
we want to perform less work during compression to make
cycle viable .
 One of the way is to do that is to reduce pressure i.e.- reduce
temp.
 In first two step system did the work.. After 2 step we will
perform.
a
b

27.  Step 3-isothermal compression
 Compression will increase heat so its use
reservoir.
 Constant heat is to reduce pressure. Because
 low pressure =less work .
T2

28.  Volume is close to initial but temp. is low..
 Step 4- gas is compressed to bring it back to
originally state.

29.  Introductory chemistry for the environmental science (second
edition)
by R.M. HARISON
 Chemistry of Atmosphere
by P.S. SINDHU
 https://chem.libretexts.org
 www.edu.thermodynamics
 www.study.com
 www.brightubengineering.com
 Web.mit.edu