Question

A 1.89 mole sample of Ar undergoes an isothermal reversible expansion from an initial volume of 2.00 L to a final volume of 85.0 L at 308 K .

Part B

Calculate the work done in this process using the van der Waals equation of state.

Express your answer using three significant figures.

Part C

What percentage of the work done by the van der Waals gas arises from the attractive potential?

Express your answer using two significant figures.

Answer 1

Vf= final volume and Vi = initila volume

for isothermal reversibel expansion, work done =-n*T*ln(Vf/Vi)= 1.89*308* ln (85/2)= -179joules. During expansion work is done by the system on surroundings. Hence -ve/

2. For Argon a= 1.355 BarL2/mole2 =1.355*0.9869 atm.L2/mole2= `1.34 atmL2/mole2 and b= 0.03201 L/mole

work done by vanderwall gas =- { nRT ln (Vf-nb)/(Vi-nb) +n2a*(1/Vf-1/Vi) =-{1.89*0.0821* 308*ln [(85-1.85*0.0320)/ (2-1.85*0.0320) +1.34*1.89*1.89*(1/85-1/2)=-[180.599-2.34) =178.262

2.34 Joules is the work from attractive potentia l

hence % work = 100*(2.34/178.262)=1.31%