laws of thermodynamics & carnot cycle

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

1. Energy fundamentals-First and Second law Made By- Surbhi Jain M.Sc. (environmental science) Dr. Bhim Rao Ambedkar University ,Agra

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

3. SMALL ISSUES SYSTEM- the part of universe SURROUNDING IMAGINARY OR REAL SURFACE SURROUNDING SYSTEM REAL SURFACE

4. SYSTEM TYPES- Matter ex. Energy ex. Open Close Isolated

5. State of systems State variables P V T M

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

7. PROPERTIES 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 and work 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. Derivation When small change in ex. Pressure is= P ex Then small change in volume = dV And work dW = - P ex .dV ( – ve bcz of opposing force)

12. INTERNAL ENERGY 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 (?>>,?<<)

13. FUNCTIONS State F.- It doesn’t matter which way choose to reach from E 1 to E 2 ex- internal energy Path F- Depends up on path ex- work,heat E 1,P,V,T E 2,P’,V’,T’

14. 1 st law 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

15. Change in internal energy at constant pressure= E= Q p -P. V Change at constant volume= E= Q v - P*0 E= Q v So heat given at constant volume is equal to direct change in internal energy

16. Enthalpy It is equal to the internal energy of the system plus the product of pressure and volume.

17. 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= Q p -P. V+P. V ( bcz E= Qp -P V at constant pressure) H= Qp so heat given at constant pressure =direct change in enthalpy

18. 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 Q p Then C p= Q p /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 Q p /n. T- Qv /n. T= Qp-Qv /n. T = Cp- Cv we know that H= Qp (enthalpy), and E= Qv (internal energy).

19. H- E/n. T = Cp- Cv …………2 But we know H= E- P V,put in eq n 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

20. 2 nd law 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.

21. ENTROPY Randomness/ Dis-orderness solid liquid gas Low entropy=low randomness N 2(g)+ 3H 2(g)=2NH 3(g) 1 + 3 = 2 ( randomness decrease S= - ve )

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

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

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

25. 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

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

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

28. REFERENCES 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