The Joule-Thomson coefficient Ujt of a van der Waals gas is given by Ujt=1/Cp{(2

Question

The Joule-Thomson coefficient Ujt of a van der Waals gas is given by Ujt=1/Cp{(2a/RT)-b} . Calculate deltaH for an isothermal gas compression 2.1 g of ethane (considered as a van der Waals gas) from 1 atm and 330K up to 6 atm [Note: For ethane, a= 5.489 L^2 atm mol^-2 and b=0.0638 L mol^-1. The Joule-Thomson coefficient Ujt of a van der Waals gas is given by Ujt=1/Cp{(2a/RT)-b} . Calculate deltaH for an isothermal gas compression 2.1 g of ethane (considered as a van der Waals gas) from 1 atm and 330K up to 6 atm [Note: For ethane, a= 5.489 L^2 atm mol^-2 and b=0.0638 L mol^-1.

Explanation / Answer

Joules thomson coefficient is given by

(dT/dP)H= 1/Cp(2a/RT-b)

H= f(T,P)

dH= (dH/dT)PdT+ (dH/dP)T dP

Since the temperature is consntat

dH= (dH/dP)TdP

we know that (dH/dP)T*(dP/dT)H* (dT/dH)P=-1

(dH/dP)T= -UCp, whee U= (dT/dP)H and Cp = (dH/dT)P

but (dH/dP)T= -UCp = -(2a/RT-b) = -(2*5.489/(0.0821*330- 0.0638)=-0.34 L/mol

dH= -0.34dP

when integrated

dH= -0.34*(6-1)= -1.7 L.atm/mol =-.1.7*101.3 Joules/mole=-172.21 Joules/mole

moles in 2.1 gm of ethane= mass/molar mass = 2.1/30 =0.07

Enthalpy change of Vanderwaal gas = 0.07*(-172.21) =-12.05 Joules

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