Trying to rearrange the VdW EoS exactly to get V as a function of T and P so you can then differentiate it explicitly results in a mess (you need to solve a 3rd order polynomial in V). An easier method is to find dT/dV)p, and then take the inverse. From the EoS, we have: T = (1/R)*(P+a/V^2)*(V-b) dT/dV)p = (P*V^3 - a*V +2*a*b)/(R*V^3) dV/dT)p = (R*V^3)/(P*V^3 - a*V +2*a*b) This is an exact result. Unfortunately, it\'s probably not the result your professor is looking for. I\'ll derive an approximate result below, which is what one needs to prove the second part of this question. -------------- The J-T coefficient is defined as dT/dP)H (i.e., dT/dP at constant enthalpy). From the basic relationships among partial derivatives: dT/dP)H = -dH/dP)T / dH/dT)p But we know from its definition that for a 1- component system: dH = TdS + VdP so dT/dP)H = [ -T*dS/dP)T - V*dP/dP)T]/[T*dS/dT)p + V*dP/dT)p] = [-T*dS/dP)T - V]/[T*dS/dT)p] From one of the Maxwell relations, we know that dS/dP)T = -dV/dT)p, and we also know that Cp = T*dS/dT)p, so: dT/dP)H = [T*dV/dT)p - V]/Cp We now need to evalutate this expression for a VdW gas. The expression given in the question for the J-T coefficient for the VdW gas is *not* an exact result, which you can easily show by plugging the expression for dV/dT)p found above into the equation for dT/dP)H. In order to get the desired result, we need back up and make some approximations in the derivation of dV/dT)p Going back to the EoS, we can rewrite it as: P*V = R*T - a/V + b/P + a*b/V^2 If we ignore the small second order terms in the VdW coefficients (i.e., a*b/V^2), then V ~= R*T/P - a/(P*V) + b If we now make another appproximation and assume P*V ~= R*T (i.e., assume the gas behaves close to ideally), we can write: V ~= R*T/P - a/(R*T) + b Taking the derivative with respect to T at constant P: dV/dT)p ~= R/P + a/(R*T^2) Plugging this new, approximate expression for dV/dT)p into the J-T coefficient equation, we have: dT/dP)H = [T*dV/dT)p - V]/Cp ~= [R*T/P + a/(R*T) - R*T/P + a/(R*T) - b]/Cp dT/dP)H = ~= [2*a/(R*T) - b]/Cp which is the desired result. Solution Trying to rearrange the VdW EoS exactly to get V as a function of T and P so you can then differentiate it explicitly results in a mess (you need to solve a 3rd order polynomial in V). An easier method is to find dT/dV)p, and then take the inverse. From the EoS, we have: T = (1/R)*(P+a/V^2)*(V-b) dT/dV)p = (P*V^3 - a*V +2*a*b)/(R*V^3) dV/dT)p = (R*V^3)/(P*V^3 - a*V +2*a*b) This is an exact result. Unfortunately, it\'s probably not the result your professor is looking for. I\'ll derive an approximate result below, which is what one needs to prove the second part of this question. -------------- The J-T coefficient is defined as dT/dP)H (i.e., dT/dP at constant enthalpy). From the basic relationships among partial derivatives: dT/dP)H = -dH/dP)T / dH/dT)p But we know from its definition that for a 1- component system: dH = TdS + VdP so dT/dP)H = [ -T*dS/dP)T - V*dP/dP)T.

1. Trying to rearrange the VdW EoS exactly to get V as a function of T and P so you
can then differentiate it explicitly results in a mess (you need to solve a 3rd order polynomial in
V). An easier method is to find dT/dV)p, and then take the inverse. From the EoS, we have: T =
(1/R)*(P+a/V^2)*(V-b) dT/dV)p = (P*V^3 - a*V +2*a*b)/(R*V^3) dV/dT)p =
(R*V^3)/(P*V^3 - a*V +2*a*b) This is an exact result. Unfortunately, it's probably not the
result your professor is looking for. I'll derive an approximate result below, which is what one
needs to prove the second part of this question. -------------- The J-T coefficient is defined as
dT/dP)H (i.e., dT/dP at constant enthalpy). From the basic relationships among partial
derivatives: dT/dP)H = -dH/dP)T / dH/dT)p But we know from its definition that for a 1-
component system: dH = TdS + VdP so dT/dP)H = [ -T*dS/dP)T - V*dP/dP)T]/[T*dS/dT)p +
V*dP/dT)p] = [-T*dS/dP)T - V]/[T*dS/dT)p] From one of the Maxwell relations, we know that
dS/dP)T = -dV/dT)p, and we also know that Cp = T*dS/dT)p, so: dT/dP)H = [T*dV/dT)p -
V]/Cp We now need to evalutate this expression for a VdW gas. The expression given in the
question for the J-T coefficient for the VdW gas is *not* an exact result, which you can easily
show by plugging the expression for dV/dT)p found above into the equation for dT/dP)H. In
order to get the desired result, we need back up and make some approximations in the derivation
of dV/dT)p Going back to the EoS, we can rewrite it as: P*V = R*T - a/V + b/P + a*b/V^2 If
we ignore the small second order terms in the VdW coefficients (i.e., a*b/V^2), then V ~=
R*T/P - a/(P*V) + b If we now make another appproximation and assume P*V ~= R*T (i.e.,
assume the gas behaves close to ideally), we can write: V ~= R*T/P - a/(R*T) + b Taking the
derivative with respect to T at constant P: dV/dT)p ~= R/P + a/(R*T^2) Plugging this new,
approximate expression for dV/dT)p into the J-T coefficient equation, we have: dT/dP)H =
[T*dV/dT)p - V]/Cp ~= [R*T/P + a/(R*T) - R*T/P + a/(R*T) - b]/Cp dT/dP)H = ~= [2*a/(R*T)
- b]/Cp which is the desired result.
Solution
Trying to rearrange the VdW EoS exactly to get V as a function of T and P so you
can then differentiate it explicitly results in a mess (you need to solve a 3rd order polynomial in
V). An easier method is to find dT/dV)p, and then take the inverse. From the EoS, we have: T =
(1/R)*(P+a/V^2)*(V-b) dT/dV)p = (P*V^3 - a*V +2*a*b)/(R*V^3) dV/dT)p =
(R*V^3)/(P*V^3 - a*V +2*a*b) This is an exact result. Unfortunately, it's probably not the
result your professor is looking for. I'll derive an approximate result below, which is what one
needs to prove the second part of this question. -------------- The J-T coefficient is defined as
dT/dP)H (i.e., dT/dP at constant enthalpy). From the basic relationships among partial
derivatives: dT/dP)H = -dH/dP)T / dH/dT)p But we know from its definition that for a 1-
component system: dH = TdS + VdP so dT/dP)H = [ -T*dS/dP)T - V*dP/dP)T]/[T*dS/dT)p +

2. V*dP/dT)p] = [-T*dS/dP)T - V]/[T*dS/dT)p] From one of the Maxwell relations, we know that
dS/dP)T = -dV/dT)p, and we also know that Cp = T*dS/dT)p, so: dT/dP)H = [T*dV/dT)p -
V]/Cp We now need to evalutate this expression for a VdW gas. The expression given in the
question for the J-T coefficient for the VdW gas is *not* an exact result, which you can easily
show by plugging the expression for dV/dT)p found above into the equation for dT/dP)H. In
order to get the desired result, we need back up and make some approximations in the derivation
of dV/dT)p Going back to the EoS, we can rewrite it as: P*V = R*T - a/V + b/P + a*b/V^2 If
we ignore the small second order terms in the VdW coefficients (i.e., a*b/V^2), then V ~=
R*T/P - a/(P*V) + b If we now make another appproximation and assume P*V ~= R*T (i.e.,
assume the gas behaves close to ideally), we can write: V ~= R*T/P - a/(R*T) + b Taking the
derivative with respect to T at constant P: dV/dT)p ~= R/P + a/(R*T^2) Plugging this new,
approximate expression for dV/dT)p into the J-T coefficient equation, we have: dT/dP)H =
[T*dV/dT)p - V]/Cp ~= [R*T/P + a/(R*T) - R*T/P + a/(R*T) - b]/Cp dT/dP)H = ~= [2*a/(R*T)
- b]/Cp which is the desired result.