JOULE-THOMSON CO-EFFICIENT - PowerPoint Presentation

1. JOULE-THOMSON CO-EFFICIENTName of the instructor:U.Nithya M.Sc.,M.Phil.,

2. Joule – Thomson co-efficient: Definition: The relationship between the fall in temperature ΔT and fall of pressure ΔP is called the joule thomson coefficient. It is denoted as μ. we know H =f(T,P)H is the state function and dH is a complete differential dH = (∂H/∂T)pdT + (∂H/∂P)Tdp --------------------- 1 (∂H/∂T)P= CP---------------------------------------— 2 dH= CpdT + (∂H/∂P)TdPJoule – Thomson expansion dH=0

3. CPdT + (∂H/∂P)TdP = 0(or) CPdT = - (∂H/∂P)TdP dT/dP = - (∂H/∂P)T/CPie, (∂T/∂P)H= - (∂H/∂P)T/CP(∂T/∂P)H is called joule-Thomson coefficient and it is denoted By μJT. ΔT = -(∂H/∂P)T/CP. ΔPFor ideal gas : (∂T/∂P)H= 0For real gas : (∂T/∂P)H= positive or negativeThe pressure is decreased the temperature may decrease or increase

4. μJTis positive – decrease in pressure and decrease in temperatureμJTis negative – decrease in pressure and increase in temperature The temperature at which μJTchanges its sign is known as inversion temperature(Ti) above Ti then μJT is negative below Tithen μJTis Positive

5. Relation between μJTand vander wall’s constant and Tiand vander wall’s constant μJT= (∂T/∂P)H= 1/Cp(2a/RT) Ti= 2a/RbDerivation: vandeer wll’s equation (P+an2/v2) (v-nb) =nRTFor one of gas(P+a/v2) (v-b) =RTExpand the equationPV –Pb + aV/V2-ab/V2 = RT

6. Pv – Pb + a/v – ab/v2-RT =0. ---—------------- 1 ab/v2 is neglect a&b are smallEqu 1 becomes PV – Pb + a/v – RT =0(or) Pv =Pb-a/v +RT. ------------------------------ 2We know PV =RT V=RT/P ------------------------------ 33 in 2 in the RHS only Pv = Pb – a/RT/P +RT PV = Pb – ap/RT +RT÷ p Pv/p = Pb/P –aP/RTP + RT/P V = b-a/RT +RT/P ------------------------- 4

7. Differentiate with respect to temperature at constant pressure (∂v/∂T)P= a/RT2+R/P ------------------- 5Rearrange 4 RT/P = V-b + a/RT RT = P(v-b+a/RT) RT = Pv – Pb + Pa/RT RT = P(v-b)+Pa/RT÷PT RT/PT = P(v-b)/PT +Pa/PTRT R/P = v-b /T + a/RT2--------------------66 in 5 (∂v/∂T)P= a/RT2+ v-b/T+a/RT2

8. (∂v/∂T)P=v-b/T+2a/RT2 (∂v/∂T)P= V/T –b/T +2a/RT2Rearrange (∂v/∂T)P-V/T = 2a/RT2-b/TxT. T(∂v/∂T)P- vT/T = 2aT/RT2-bT/T T(∂v/∂T)P- v= 2a/RT – b----------------------- 7 (or) - v= 2a/RT- b-T(∂v/∂T)P----------—------8We know. V= T(∂v/∂T)P+ (∂H/∂P)T -------------------9Rearrange 9. (∂H/∂P)T = V- T(∂v/∂T)P---------------10We know. μJT = (∂T/∂P)H = - (∂H/∂P)Tx1/Cp-------------------1110 in 11. (∂T/∂P)H = - [V – T(∂v/∂T)Px1/Cp (∂T/∂P)H = [V+T(∂v/∂T)Px1/Cp--------------12

9. 12 in 8 (∂T/∂P)H = [ 2a/RT-b- T(∂v/∂T)p– T(∂v/∂T)P] x 1/Cp μJT=(∂T/∂P)H = [2a/RT-b]x1/Cp----------------------13Equation 13 Relationship between μJT and vanderwall’s constant Joule thomson effect is positive. If 2a/RT is greater than b Joule thomson effect is negative if 2a/RT is greater than ba,b and R are constants and sign of μJT. The temperature at which μJTchanges its sign is known as inversion temperature μJT=0 T = Ti [2a/RTi-b] 1/Cp=0

10. 2a/RTi- b =0 2a/RTi= b (or) Ti=2a/Rb. -----------------------14Equation 14 relationship between inversion temperature and vander wall’s constant Significance: (inversion temperature) Inversion temperature value of a gas as determined from its vander waals constant values gives us an idea about the temperature to which a gas must be cooled by a method other than subjecting the gas to joule thomson effect when we liquefy the gas. Significance of Joule –Thomson experiment: joule –thomson experiment helped to find out the fact that each gas possessed an inversion temerature above which the gas can never be cooled by subjecting it to joule-thomson effect. Thus joule Thomsons experiment enabled us to liquefy so called permanent gases like H2,He and air.

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