Thermodynamics

Physical Chemistry; Exact differentials, state function, Joule Thomson's experiment, derivation of Joule Thomson's coefficient, its significance and applications, inversion temperature.

1. THERMODYNAMICS Pallavi Kumbhar MSc Part-I

2. CONTENTS 01 Maxwell’s Thermodynamic Relations 02 Common Forms of Maxwell’s Relations 03 State functions & Exact differentials 04 Joule Thomson E xperiment 05 Joule-Thomson Coefficient 06 Significance of Joule-Thomson Coefficient 07 Joule- T homson Coefficient For An Ideal Gas 08 Applications of Joule-Thomson Effect  

3. Maxwells Thermodynamic Relations What are Maxwell’s relations? These are the set of thermodynamics equations derived from a symmetry of secondary derivatives and from thermodynamic potentials. For example dU = TdS – PdV Then, These relations are named after James Clerk Maxwell, who was a 19th-century physicist.  

4. F unction Differential Natural Variables Maxwell Relation U dU = TdS – PdV S, V H dH = TdS + VdP S, P A dA = - PdV – SdT V, T G dG = VdP – SdT P, T F unction Differential Natural Variables Maxwell Relation U dU = TdS – PdV S, V H dH = TdS + VdP S, P A dA = - PdV – SdT V, T G dG = VdP – SdT P, T Common Forms Of Maxwell’s Relations Internal energy (U), Enthalpy (H), Helmholtz free energy (A), and Gibbs free energy (G), entropy (S), pressure (P), volume (V), and temperature (T). 

5. State Functions & Exact Differentials State function is a property which depends only on the initial and final states of the system and is independent of its path. Examples: U, H, S, A, G E xact differential is an infinitesimal quantity, which when integrated, gives a result that is independent of the path between the initial and final states. For example c onsider the function A(x, y) as an exact differential, where x and y are independent variables Then, Or where C is arbitrary constant    

6. Joule Thomson Experiment A B T1 T2 P1>P2 The work done by piston A = P 1 V 1 The work done by piston B = - P 2 V 2 Net work done by the system, W = P 1 V 1 - P 2 V 2 Change in internal energy, ∆U = U 2 - U 1 From 1 st law ∆U = q - w Since q = 0, -∆U = -W Substituting the values of ∆U and W we get, -(P 1 V 1 - P 2 V 2 ) = -(U 2 - U 1 ) P 2 V 2 - P 1 V 1 = U 1 - U 2 Rearranging the terms, we get U 2 + P 2 V 2 = U 1 + P 1 V 1 Since H = U + PV Therefore, H 2 = H 1 and ∆H = 0 Thus, during Joule Thomson effect, enthalpy of the system remains constant. It is also called Iso-enthalpic process

7. Joule-Thomson Coefficient In Joule-Thomson experiment, H is constant Dividing the above equation by dP at constant enthalpy, On rearranging the above equation, we get Since, Cp = therefore, μ JT = where μ JT = The rate of change of temperature with respect to pressure at constant enthalpy is defined as Joule Thomson Coefficient.  

8. Significance Of Joule-Thomson Coefficient If the gas temperature is then  μ JT  is  since ∂P is So the gas above the inversion temperature negative always negative warms below the inversion temperature positive always negative cools The temperature at which µ=0 and the gas shows neither heating effect nor cooling effect on adiabatic expansion is defined as inversion temperature .

9. Temperature Inversion Curve In this curve… Within the region bounded by the red line, a Joule–Thomson expansion produces cooling ( μ JT  > 0). Outside that region, the expansion produces heating. The gas–liquid coexistence curve is shown by the blue line, terminating at the critical point (the solid blue circle). The dashed lines shows the regions where N 2  is neither a supercritical fluid, a liquid, nor a gas.

10. Joule- T homson Coefficient For An Ideal Gas We know that, μ JT = = Therefore, , μ JT = …( Since, H = U + PV) = + ] = + ] For an ideal gas, = 0 Also, PV = constant so, = 0 Therefore, μ JT = 0  

11. Applications of Joule-Thomson Effect The cooling produced in the Joule-Thomson expansion has made it a very valuable tool in refrigeration. This effect is applied in the Linde technique in the petrochemical industry, where the cooling effect is used to liquefy gases. It is also used in many cryogenic applications. For example for the production of liquid nitrogen, oxygen, and argon. The effect can also be used to liquefy even helium.

12. REFRENCES Physical Chemistry Peter Atkins, Julio De Paula Images from Google