Chap03 solutions

Physical Chemistry

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3 The First Law: the machinery Solutions to exercises Discussion questions E3.1(b) The following list includes only those state functions that we have encountered in the Frst three chapters. More will be encountered in later chapters. Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefFcient, isothermal compressibility, and Joule–Thomson coefFcient. E3.2(b) One can use the general expression for π T given in Justifcation 3.3 to derive its speciFc form for a van der Waals gas as given in Exercise 3.14(a), that is, π T = a/V 2 m . (The derivation is carried out in Example 5.1.) ±or an isothermal expansion in a van der Waals gas d U m = (a/V m ) 2 . Hence 1U m =− a( 1 /V m , 2 1 m , 1 ) . See this derivation in the solution to Exercise 3.14(a). This formula corresponds to what one would expect for a real gas. As the molecules get closer and closer the molar volume gets smaller and smaller and the energy of attraction gets larger and larger. E3.3(b) The solution to Problem 3.23 shows that the Joule–Thomson coefFcient can be expressed in terms of the parameters representing the attractive and repulsive interactions in a real gas. If the attractive forces predominate then expanding the gas will reduce its energy and hence its temperature. This reduction in temperature could continue until the temperature of the gas falls below its condensation point. This is the principle underlying the liquefaction of gases with the Linde Refrigerator which utilizes the Joule–Thomson effect. See Section 3.4 for a more complete discussion. Numerical exercises E3.4(b) A function has an exact differential if its mixed partial derivatives are equal. That is, f(x,y) has an exact differential if ∂x ± ∂f ∂y ² = ± ² ( a ) = 3 x 2 y 2 and ± ² = 6 x 2 y = 2 x 3 y and ± ² = 6 x 2 y Therefore, exact. ( b ) ∂s = t e s + 1 and ∂t ± ² = e s = 2 t + e s and ± ² = e s Therefore, exact. E3.5(b) d z = ∂z d x + d y = d x ( 1 + y) 2 2 x d y ( 1 + 3 E3.6(b) ( a )d z = d x + d y = ( 3 x 2 2 y 2 ) d x 4 xy d y ( b ) 2 z ∂y∂x = ( 3 x 2 2 y 2 ) 4 y and 2 z ∂x∂y = ( 4 xy) 4 y
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THE FIRST LAW: THE MACHINERY 45 E3.7(b) d z = ∂z ∂x d x + ∂y d y = ( 2 xy + y 2 ) d x + (x 2 + 2 xy) d y 2 z ∂y∂x = ( 2 xy + y 2 ) = 2 x + 2 y and 2 z ∂x∂y = (x 2 + 2 = 2 x + 2 y E3.8(b) ± ∂C p ∂p ² T = " ∂p ± ∂H ∂T ² p # T = 2 H ∂p∂T ³ ± ∂p ² T ´ p Because ± ∂p ² T = 0 for a perfect gas, its temperature derivative also equals zero; thus ± p ∂p ² T = 0. E3.9(b) ± ∂U ² p = (∂H/∂V ) p (∂U/∂V ) p = µ ∂(U + pV) ∂V p p = p + p p = 1 + p p E3.10(b) d p = ± ∂p ² T d V + ± ∂p ² V d T dln p = d p p = 1 p ± ∂p ² T d V + 1 p ± ∂p ² V d T We express ± ∂p ² T in terms of the isothermal compressibility κ T κ T =− 1 V ± ∂p ² T ³ V ± ∂p ² T ´ 1 so ± ∂p ² T 1 κ T V We express ± ∂p ² V in terms of κ T and the expansion coefFcient α = 1 V ± ² p ± ∂p ² V ± ² p ± ∂p ² T 1s o ± ∂p ² V (∂V/∂T ) p (∂V/∂p) T = α κ T so d ln p 1 pVκ T + α T = 1 T ± α d T d V V ² E3.11(b) U = ± 3 2 ² nRT so ± ∂p ² T = 0
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Chap03 solutions - 3 The First Law: the machinery Solutions...

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