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# hw3_sol - 1 One mole of a gas occupies one half of a...

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1 One mole of a gas occupies one half of a container that is separated from the rest of the container by a partition. The initial volume of the gas is V and its temperature is T . After the partition is removed, the gas expands irreversibly and occupies the volume 2 V . The equation of state of the gas is 2 ( / )( ) P a V V b RT + = and its heat capacity V C is temperature-independent. A. What is the temperature of the gas after expansion? We have dU = V C n dT + [T V P T -P] dV Using P = RT/(V-b) – a/V 2 we find 2 2 V V RT RT a dV dU C dT dV C dT a V b V b V V = + + = + Integrating this, 1 1 ( (2) (1)) 0 (2) (1) V U C T T a V V = = or, using V(2)=2V(1)=2V, T(2)=T(1)- 2 V a C V B. What is the change of the entropy of the gas in this process? Use V V V C C P RdV dS dV dT dT T T V b T = + = +

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S=R ln((2V-b)/(V-b)) + /(2 ) ln V V T a C V C T C. In the class, we have derived the relationship ( )/ constant V V C R C PV + = between the pressure and the volume of an ideal gas undergoing a reversible adiabatic process, in which it receives no heat from the surroundings. Derive an analogous relationship for the gas obeying the van der Waals equation of state above (and making the same assumption of temperature independent V C ). Hint: Consider the change of entropy S in such a process. Since it’s a reversible process, we have dS = δ q/T=0 or 0 V C RdV dT V b T + = Integrating this equation gives ln( ) ln constant V R V b c T + = or / ( ) constant V R C V b T = (*) Combining this with the equation of state 2 ( / )( ) P a V V b RT + = we find 1 2 ( / )( ) constant V R C P a V V b + + = D. For the process described in part C, how will the temperature of the gas change if its volume increases from V to 2 V . Compare your result with that from part A. Using Eq. (*) from part C: / 2 1 2 V R C V b T T V b =
This is very different from the result of part A. In particular, this result depends on b and does not depend on a while the result of part A depended on a but not b . This difference is not unexpected: One process is reversible while the other is not.

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