If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume V1 to a final volume V2 , the expression for the work done is

Question

If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume V1 to a final volume V2 , the expression for the work done is (A) RT ln (V2/V1) +a (V2-V1) (B) -nRT ln (V2-b/V1-b)+a ((1/V1)-(1/V2)) (C) RT ln (P2/P1) (D) RT ln (V2-b/V1-b)-a ((1/V1)-(1/V2))

Tardigrade Admin · Accepted Answer

For the van der Waalsâ gas w=-∫ limitsV1V2 p textexternal dV =-∫ limitsV1V2 ((RT/Vm-b)-(a/Vm2))dV =-∫ limitsV1V2((R T/Vm-b)) d V+∫ limitsV1V2((a/Vm2)) d V The first integral is solved by making the substit y=Vm-b =-∫ limitsV1V2((R T/Vm-b)) d V=-∫ limitsV1V2((R T/y*)) d y =-R T[ ln (V2+b)- ln (V1+b)] Therefore, the work is given by w=-nRT ln ((V2-b/V1-b))+a((1/V1)-(1/V2))

Q. If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume $V_{1}$ to a final volume $V_{2}$ , the expression for the work done is

$RT l n \frac{V _{2}}{V _{1}} + a (V_{2} - V_{1})$

$- n RT l n \frac{V _{2} - b}{V _{1} - b} + a (\frac{1}{V _{1}} - \frac{1}{V _{2}})$

$RT l n \frac{P _{2}}{P _{1}}$

$RT l n \frac{V _{2} - b}{V _{1} - b} - a (\frac{1}{V _{1}} - \frac{1}{V _{2}})$

Q. If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume V1​ to a final volume V2​ , the expression for the work done is

RTlnV1​V2​​+a(V2​−V1​)

−nRTlnV1​−bV2​−b​+a(V1​1​−V2​1​)

RTlnP1​P2​​

RTlnV1​−bV2​−b​−a(V1​1​−V2​1​)

Q. If one mole of a van der Waals’ gas undergoes reversible isothermal transformation from an initial volume $V_{1}$ to a final volume $V_{2}$ , the expression for the work done is

$RT l n \frac{V _{2}}{V _{1}} + a (V_{2} - V_{1})$

$- n RT l n \frac{V _{2} - b}{V _{1} - b} + a (\frac{1}{V _{1}} - \frac{1}{V _{2}})$

$RT l n \frac{P _{2}}{P _{1}}$

$RT l n \frac{V _{2} - b}{V _{1} - b} - a (\frac{1}{V _{1}} - \frac{1}{V _{2}})$