Tag: Solutions

Problem 5.51 When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables, , , Rewrite the van der Waals equation in terms of these variables, and notice that the constants and disappear. Solution: Problem 5.51 Solution (Download)

Problem 5.49 Use the result of the previous problem and the approximate values of and given in the text to estimate , , and for , , and (Tabulated values of and are often determined by working backward from the measured critical temperature and pressure.) Solution: Problem 5.49 Solution (Download)

Problem 5.48 As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that , , and Solution:…

Problem 5.45 In Problem 1.40 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process. (a) Use the first law of thermodynamics to show that, as condensation forms during…

Problem 5.32 The density of ice is (a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative.(b) How much pressure would you have to put on an ice cube to make it melt at -1°C?(c) Approximately how deep under a glacier would you have to…

Problem 5.27 Graphite is more compressible than diamond. (a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?(b) The isothermal compressibility of graphite is about while that of diamond is more than ten times less and hence negligible…

Problem 5.23 By subtracting from , , , or , one can obtain four new thermodynamic potentials. Of the four, the most useful is the grand free energy (or grand potential), (a) Derive the thermodynamic identity for , and the related formulas for the partial derivatives of with respect to , , and (b) Prove…

Problem 5.21 Is heat capacity extensive or intensive? What about specific heat ? Explain briefly Solution: Problem 5.21 Solution (Download)

Problem 5.20 The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of since for this given value of…

Problem 5.16 A formula analogous to that for relates the isothermal and isentropic compressibilities of a material: (Here is the reciprocal of the adiabatic bulk modulus considered in Problem 1.39.) Derive this formula. Also check that it is true for an ideal gas. Solution: Problem 5.16 Solution (Download)