Chap02 solutions

Physical Chemistry

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2 The First Law: the concepts Solutions to exercises Discussion questions E2.1(b) Work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular Interpretation 2.1 for a more detailed discussion. E2.2(b) Rewrite the two expressions as follows: (1) adiabatic p 1 /V γ (2) isothermal p 1 The physical reason for the difference is that, in the isothermal expansion, energy Fows into the system as heat and maintains the temperature despite the fact that energy is lost as work, whereas in the adiabatic case, where no heat Fows into the system, the temperature must fall as the system does work. Therefore, the pressure must fall faster in the adiabatic process than in the isothermal case. Mathematically this corresponds to γ> 1. E2.3(b) Standard reaction enthalpies can be calculated from a knowledge of the standard enthalpies of forma- tion of all the substances (reactants and products) participating in the reaction. This is an exact method which involves no approximations. The only disadvantage is that standard enthalpies of formation are not known for all substances. Approximate values can be obtained from mean bond enthalpies. See almost any general chemistry text, for example, Chemical Principles , by Atkins and Jones, Section 6.21, for an illustration of the method of calculation. This method is often quite inaccurate, though, because the average values of the bond enthalpies used may not be close to the actual values in the compounds of interest. Another somewhat more reliable approximate method is based on thermochemical groups which mimic more closely the bonding situations in the compounds of interest. See Example 2.6 for an illustration of this kind of calculation. Though better, this method suffers from the same kind of defects as the average bond enthalpy approach, since the group values used are also averages. Computer aided molecular modeling is now the method of choice for estimating standard reaction enthalpies, especially for large molecules with complex three-dimensional structures, but accurate numerical values are still dif±cult to obtain. Numerical exercises E2.4(b) Work done against a uniform gravitational ±eld is w = mgh (a) w = ( 5 . 0kg ) × ( 100 m ) × ( 9 . 81 m s 2 ) = 4 . 9 × 10 3 J (b) w = ( 5 . ) × ( 100 m ) × ( 3 . 73 m s 2 ) = 1 . 9 × 10 3 J E2.5(b) Work done against a uniform gravitational ±eld is w = mgh = ( 120 × 10 3 kg ) × ( 50 m ) × ( 9 . 81 m s 2 ) = 59 J E2.6(b) Work done by a system expanding against a constant external pressure is w =− p ex ±V ( 121 × 10 3 Pa ) × ( 15 cm ) × ( 50 cm 2 ) ( 100 cm m 1 ) 3 ! = 91 J
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22 INSTRUCTOR S MANUAL E2.7(b) For a perfect gas at constant temperature ±U = 0 so q =− w For a perfect gas at constant temperature, ±H is also zero d H = d (U + pV) we have already noted that U does not change at constant temperature; nor does pV if the gas obeys Boyle’s law. These apply to all three cases below.
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Chap02 solutions - 2 The First Law: the concepts Solutions...

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