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Expression for the joule–Thomson cooling produced in a Van der Waals gas. Suppose that one mole of a gas is allowed to expand through a porous plug from a pressure P1 and volume V1 to a pressure P2 and volume V2 (Fig.3.3). Let the temperature change from T1 to T2 due to joule –Thomson effect.

Net external woek done by the Gas = P2 V2-P1 V1 		……………..(1) Now, an internal work done by the gas in overcoming the forces of molecular attraction. For a van der Waals Gas the attractive forces between the molecules are equivalent to an internal pressure a/V2 Internal work done by the gas when the gas expands from a volume V1 to V2 is ∫_V1^V2▒〖Pdv=∫_V1^V2▒a/V2〗   dv=  a/V1-a/V2				………………..(2) ∴ Total work done by the gas + external work + internal work ∴ W=(P2V2-P1V1) + a/V1-a/V2								……………..(3) Now, van der Waals equation of state for gas is (P+a/V2) (v-b) = RT Or PV = RT + Pb - a/V			(Neglecting ab/V2) ∴ p v = RT1 + bP1 a/V1 and P2 V2 = RT2 + bP2 - a/V2 Substituting these values for P1 V1 and P2 V2 in Eq.(3) W=R(T2-T1)-b (P1-P2) + 2a/v1-2A/v2 						………….(4) Since a and b are very small PY=RT or V=RTIP Hence v1 = RT1/P1 And V2=RT2/p2 ∴ W = R (T2 – T1) – b (P1 –P2) + 2aP1/RT1-2aP2/RT2 Now T1 ¬ and T2 are equal. Hence we may write T for T1 or T2 Let T1-T2 = δT. Then W = -RδT – B(P1 – P2) + 2a/RT (P1-P2) = (P1-P2){2a/RT-b}-RδT. Or 	δT(Cv+R) =(P1 – P2) {2a/RT-b}. Or 	δT= ((P1 – P2))/CP {2a/RT-b}. Eq. (6) give the fall in temperature, or the cooling produced in a van der Waals gas when subjected to a throttling process. If 2a/RT > b, then δT is positive. Hence there will be a cooling effect. If 2a/RT < b, then δT is negative. Hence there will be a heating effect. If 2a/RT = b, then δT=0 Hence there will be neither a cooling nor a heating effect.

The temperature at which the joule-Thomson effect changes sing is called the temperature, 2a/RTi =b.

∴ Ti =2a/Rb Thus, above the temperature of inversion joule-Thomson effect will be a heating effect and below it a cooling effect. Heating effect for hydrogen and helium. The temperature of inversion for hydrogen and helium are - 80°∁ and -240°∁ respectively. These temperatures are much below the room temperature. Hence at ordinary temperatures, hydrogen and helium show heating effect. if these gases are cooled below their temperature of inversion and then passed through the porous plug, they will also shoe a cooling effect. Relation between Boyle temperature (TB), temperature of inversion (Ti) and critical temperature (Tc) We have,

Thus the Temperature of inversion of gas is much higher then its critical temperature. Example 1. Calculate the change in temperature Ehen gas suffers joule-Thomson expansion at -173°∁ the pressure difference between the two side of the plug being 20 atmospheres. Does the gas shoe a heating effect or a cooling effect in this expansion At-173 °∁ ? give :R = 8.3JK-1 mol-1 and for helium van den Walls constants are a= 0.0341 atom lit2/mole2 and b=002357 liter/mole. Here P1-P2 =20 atmospheres = 20*1.101325*105 pa: T = -173°∁ =100K A0.0341 atom lit2/mole2 =0.0341*[(1.01325*105¬)*10-6] Nm-4 mole -2 =0.003455 Nm4 mole-2 b= 0.02341 liter/mole =0.00002357 M3 mol-1 Cp=5/2R = 20.75JK-1 mole-1. δT = ? δT = (P1-P2)/Cp[2a/RT-b δT = (20*1.01325×10)/20.75[(2×0.003455)/(8.3*100)-0.00002357 =-1.489k. Since δT is negative, there is a rise in temperature equal to 1.489 K. Now, Ti = 2a/RB = (2×0.003455)/(8.3×100)=35.43K =-237.6°∁ Since the initial temperature is -173°∁ which is higher than Ti a heating effect will take place Example 2. The van Waals constants for hydrogen are a = 2.45*10-2 Nm4/mole2 and b= 2.67*10-5m3/mole. Calculate temperature of inversion, critical temperature and Boyle temperature Temperature of inversion=Ti =2a/RB=(2×(2.45×〖10〗^(-2)))/(8.314×(2.67×〖10〗^(-5)) )=220K Critical temperature = Tc =8a/27RB= (8×(2.45×〖10〗^(-2)))/(27×(2.67×〖10〗^(-5)8.314) )=32.7K Boyle temperature=TB =a/RB= ((2.45×〖10〗^(-2)))/(8.314(2.67×〖10〗^(-5)) )=110.4K Example 3. Calculate the drop in temperature produced by adiabatic throttling in the case of oxygen when the pressure is reduced by 50 atmospheres. Initial temperature is 27°∁ van der Waals’ constants are a =0.132 Nm4 /mole2 and b=3.12×10-5 m3/mole. Sol. Here, P1 - P2 =50 ×105n/m3; T=300K cP= (7/2)R=(7/2)×8.314=29.1J mol-1K-1; δT=? δT= ((p_(1-P_(2 ) )))/C_p {2a/RT-b}

=(50×〖10〗^5)/29.1[(2×0.132)/(8.314×300)]-3.12×〖10〗^(-5)] =12.82 k.