a) Use the equation, dH = TdS + Vdp, divide by dp at constant temperature, and t

Question

a) Use the equation, dH = TdS + Vdp, divide by dp at constant temperature, and the use one of Maxwell’s equation to derive the thermodynamic equation of state: b) Use the equation derived in (a) to derive an expression for (H/p)T for a perfect gas. c) By substituting the expansion coefficient in the equation in part (a), the expression will be suitable for a liquid or solid. Integrate this expression and calculate the change in enthalpy when the pressure on 1 mole of benzene is increased from 1 atm to 100 atm at 298 k. You can assume that the volume and the expansion coefficient does not change significantly in this pressure range.

Explanation / Answer

dH= T*dS+VdP

at constant temperature

(dH/dP)T= T*(dS/dP)T+ V, but from Maxwell relation , (dS/dP)T= -(dV/dT)P

(dH/dP)T= -T*(dV/dT)P+V (1)

for perfect gas PV= RT, V= RT/P, (dV/dT)P= R/P = V/T

hence (dH/dP)T= -T*V/T+V= V-V=0

but coefficinet of volume expansion, alpha= (1/V)*(dV/dT)P, V*alpha = (dV/dT)P

Eq.1 now becomes, (dH/dP)T=-T*V*alpha+V

(dH/dP)T= V*(1-alpha*T), alpha= 0.001219/K, V= specific volume of benzene = 1/0.88 ml/gm= 1.14ml/gm

1000ml= 1L .hence V= 1.14/1000/gm = 1.14*10-3 L/gm

molar mass of Benzene (C6H12)= 6*12+12= 84 gm/mole

1 mole of benzene contains 84 gm

hence specific volume = 1.14*10-3/(1/84) L/mole =0.09576 L/moles

(dH/dP)T= 0.09576*(1-0.001219*298)

=0.061

dH= 0.061*dP

when integrated, enthalpy change= 0.061*(P2-P1)=0.061*(100-1)= 0.061*99=6.039 L.atm

but 1 L.atm= 101.3 joules

hence enthalpy change= 6.039*101.3 =612 Joules

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