?The cycle involved in the operation of an internal combustion engine is called the Otto

Calculate \(\Delta S\) (for the system) when the state of 2.00 mol of gas molecules, for which \(C_{p, m}=\frac{7}{2} R\), is changed from \(25^{\circ} \mathrm{C}\) and 1.50 atm to \(135^{\circ} \mathrm{C}\) and 7.00 atm. Text Transcription: deltaS C_p,m = 7/2R 25 degree C 135 degree C

Calculate the change in entropy of the system when 10.0 g of ice at \(-10.0^{\circ} \mathrm{C}\) is converted into water vapour at \(115.0^{\circ} \mathrm{C}\) and at a constant pressure of 1 bar. The molar constant-pressure heat capacities are: \(C_{p, m}\left(H_{2} \mathrm{O}(\mathrm{s})\right)=37.6 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\); \(C_{p, \mathrm{m}}\left(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right)=75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\); and \(C_{p, m}\left(H_{2} \mathrm{O}(\mathrm{g})\right)=33.6 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\). The standard enthalpy of vaporization of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is \(40.7 \mathrm{kJ} \mathrm{mol}^{-1}\), and the standard enthalpy of fusion of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is \(6.01 \mathrm{kJ} \mathrm{mol}^{-1}\), both at the relevant transition temperatures. Text Transcription: -10.0 degree C 115.0 degree C C_p,m(H_2O(s)) = 37.6 J K^?1 mol^?1 C_p,m(H_2O(l)) = 75.3 J K^?1 mol^?1 C_p,m(H_O(g)) = 33.6 J K^?1 mol^?1 H_2O(l) 40.7 kJ mol^?1 H_2O(l) 6.01 kJ mol^?1

Calculate the change in entropy of the system when 15.0 g of ice at \(-12.0 \ ^{\circ} \mathrm{C}\) is converted to water vapour at \(105.0 \ ^{\circ} \mathrm{C}\) at a constant pressure of 1 bar. For data, see the preceding exercise. Text Transcription: -12.0 degree C 105.0 degree C

Consider a process in which \(1.00 \mathrm{mol} \ \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) at \(-5.0 \ ^{\circ} \mathrm{C}\) solidifies to ice at the same temperature. Calculate the change in the entropy of the sample, of the surroundings and the total change in the entropy. Is the process spontaneous? Repeat the calculation for a process in which \(1.00 \mathrm{mol} \ \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) vaporizes at \(95.0 \ ^{\circ} \mathrm{C}\) and 1.00 atm. The data required are given in Exercise E3B.7(a). Text Transcription: -5.0 degree C 95.0 degree C 1.00molH_wO(l) 1.00molH_wO(l)

Show that a process in which liquid water at \(5.0 \ ^{\circ} \mathrm{C}\) solidifies to ice at the same temperature is not spontaneous (Hint: calculate the total change in the entropy). The data required are given in Exercise E3B.7(a). Text Transcription: 5.0 degree C

The molar heat capacity of trichloromethane (chloroform, \(\mathrm{CHCl}_{3}\)) in the range 240 K to 330 K is given by \(C_{p, \mathrm{m}} /\left(\mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)=91.47+7.5 \times 10^{-2}(T / \mathrm{K})\). Calculate the change in molar entropy when \(\mathrm{CHCl}_{3}\) is heated from 273 K to 300 K. Text Transcription: C_p,m/(J K^?1 mol^?1) = 91.47 + 7.5 × 10^?2(T/K) CHCl_3 CHCl_3

The molar heat capacity of \(\mathrm{N}_{2}(\mathrm{g})\) in the range 200 K to 400 K is given by \(C_{p, \mathrm{m}} /\left(\mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)=28.58+3.77 \times 10^{-3}(\mathrm{T} / \mathrm{K})\). Given that the standard molar entropy of \(\mathrm{N}_{2}(\mathrm{g})\) at 298 K is \(191.6 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1}\), calculate the value at 373 K. Repeat the calculation but this time assuming that \(C_{p, \mathrm{m}}\) is independent of temperature and takes the value 2\(29.13 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1}\). Comment on the difference between the results of the two calculations. Text Transcription: N_2(g) C_p,m/(J K^?1 mol^?1) = 28.58 + 3.77 × 10^–3(T/K) N_2(g) 191.6 J K^?1 mol^?1 C_p,m 29.13 J K^?1 mol^?1

Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature \(T_{\mathrm{h}}\) and the other at \(T_{\mathrm{c}}\), are brought into thermal contact and allowed to reach equilibrium. Evaluate the change in entropy for two blocks of copper, each of mass 500 g, with \(C_{p, \mathrm{m}}=24.4 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1}\), taking \(T_{\mathrm{h}}=500 \mathrm{K}\) and \(T_{\mathrm{c}}=250 \mathrm{K}\). Text Transcription:\ T_h T_c C_p,m = 24.4 J K^?1 mol^?1 T_h = 500 K T_c = 250 K

According to Newton’s law of cooling, the rate of change of temperature is proportional to the temperature difference between the system and its surroundings: \(\frac{\mathrm{d} T}{\mathrm{~d} t}=-\alpha\left(T-T_{\text {sur }}\right)\) where \(T_{\text {sur }}\) is the temperature of the surroundings and ? is a constant. (a) Integrate this equation with the initial condition that \(T=T_{i}\) at t = 0. (b) Given that the entropy varies with temperature according to \(S(T)-S\left(T_{i}\right)=C \ln \left(T / T_{\mathrm{i}}\right)\), where \(T_{\mathrm{i}}\) is the initial temperature and C the heat capacity, deduce an expression entropy of the system at time t. Text Transcription: dT/dt = alpha(T-T_sur) T_sur T = T_i S(T) ? S(T_i) = C ln(T/T_i)

A block of copper of mass 500 g and initially at 293 K is in thermal contact with an electric heater of resistance \(1.00 \mathrm{k} \Omega\) and negligible mass. A current of 1.00 A is passed for 15.0 s. Calculate the change in entropy of the copper, taking \(C_{p, \mathrm{m}}=24.4 \ \mathrm{JK}^{-1} \ \mathrm{mol}^{-1}\). The experiment is then repeated with the copper immersed in a stream of water that maintains the temperature of the copper block at 293 K. Calculate the change in entropy of the copper and the water in this case. Text Transcription: 1.00komega C_p,m = 24.4 JK^-1mol^-1

A block of copper \(\left(C_{p, \mathrm{m}}=24.44 \ \mathrm{J} \mathrm{K}^{-1} \ \mathrm{mol}^{-1}\right)\) of mass 2.00 kg and at \(0^{\circ} \mathrm{C}\) is introduced into an insulated container in which there is \(1.00\mathrm{mol} \ \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C}\) and 1.00 atm. Assuming that all the vapour is condensed to liquid water, determine: (a) the final temperature of the system; (b) the heat transferred to the copper block; and (c) the entropy change of the water, the copper block, and the total system. The data needed are given in Exercise E3B.7a. Text Transcription: (C_p,m = 24.44 J K^?1 mol^?1) 0 degree C 100 degree C 1.00molH_2O(g)

The protein lysozyme unfolds at a transition temperature of \(75.5^{\circ} \mathrm{C}\) and the standard enthalpy of transition is 509 \(\mathrm{kJ} \ \mathrm{mol}^{-1}\). Calculate the entropy of unfolding of lysozyme at \(25.0^{\circ} \mathrm{C}\), given that the difference in the molar constant-pressure heat capacities upon unfolding is \(6.28 \ \mathrm{kJ} \ \mathrm{K}^{-1} \mathrm{mol}^{-1}\) and can be assumed to be independent of temperature. (Hint: Imagine that the transition at \(25.0^{\circ} \mathrm{C}\) occurs in three steps: (i) heating of the folded protein from \(25.0^{\circ} \mathrm{C}\) to the transition temperature, (ii) unfolding at the transition temperature, and (iii) cooling of the unfolded protein to \(25.0^{\circ} \mathrm{C}\). Because the entropy is a state function, the entropy change at \(25.0^{\circ} \mathrm{C}\) is equal to the sum of the entropy changes of the steps.) Text Transcription: 75.5 degree C 25.0 degree C 25.0 degree C 25.0 degree C 25.0 degree C 25.0 degree C 6.28 kJ K^?1 mol^?1

The cycle involved in the operation of an internal combustion engine is called the Otto cycle (Fig. 3.1). The cycle consists of the following steps: (1) Reversible adiabatic compression from A to B, (2) reversible constant-volume pressure increase from B to C due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from C to D, and (4) reversible constant-volume pressure decrease back to state A. Assume that the pressure, temperature, and volume at point A are \(p_{A}\), \(T_{A}\), and \(V_{A}\), and likewise for B–D; further assume that the working substance is 1 mol of perfect gas diatomic molecules with \(C_{V, \mathrm{m}}=\frac{5}{2} R\). Recall that for a reversible adiabatic expansion (such as step 1) \(V_{\mathrm{A}} T_{\mathrm{A}}^{c}=V_{\mathrm{B}} T_{\mathrm{B}}^{c}\), where \(c=C_{V, \mathrm{m}} / R\), and that for a perfect gas the internal energy is only a function of the temperature. (a) Evaluate the work and the heat involved in each of the four steps, expressing your results in terms of \(C_{V, \mathrm{m}}\) and the temperatures \(T_{\mathrm{A}}-T_{\mathrm{D}}\). (b) The efficiency ? is defined as the modulus of the work over the whole cycle divided by the modulus of the heat supplied in step 2. Derive an expression for ? in terms of the temperatures \(T_{\mathrm{A}}-T_{\mathrm{D}}\). (c) Use the relation between V and T for the reversible adiabatic processes to show that your expression for the efficiency can be written \(\eta=1-\left(V_{\mathrm{B}} / V_{\mathrm{A}}\right)^{1 / c}\) (Hint: recall that \(V_{\mathrm{C}}=V_{\mathrm{B}}\) and \(V_{\mathrm{D}}=V_{\mathrm{A}}\).) (d) Derive expressions, in terms of \(C_{V, \mathrm{m}}\) and the temperatures, for the change in entropy (of the system and of the surroundings) for each step of the cycle. (e) Assuming that \(V_{\mathrm{A}}=4.00 \mathrm{dm}^{3}\), \(p_{\mathrm{A}}=1.00 \mathrm{atm}\), \(T_{\mathrm{A}}=300 \mathrm{K}\), and that \(V_{\mathrm{A}}=10 V_{\mathrm{B}}\) and \(p_{\mathrm{C}} / p_{\mathrm{B}}=5\), evaluate the efficiency of the cycle and the entropy changes for each step. (Hint: for the last part you will need to find \(T_{\mathrm{B}}\) and \(T_{\mathrm{D}}\), which can be done by using the relation between V and T for the reversible adiabatic process; you will also need to find \(T_{\mathrm{C}}\) which can be done by considering the temperature rise in the constant volume process.) Text Transcription: C_V,m = 5/2 R V_AT_A&c = V_BT_B^c c = C_V,m/R C_V, m T_A - T_D T_A - T_D eta = 1(V_B/V_A)^1/c VC = VB VD = VA C_V,m V_A = 4.00 dm^3 p_A = 1.00 atm T_A = 300 K V_A = 10V_B p_C/p_B = 5 T_B T_D T_C

When a heat engine is used as a refrigerator to lower the temperature of an object, the colder the object the more work that is needed to cool it further to the same extent. (a) Suppose that the refrigerator is an ideal heat engine and that it extracts a quantity of heat |dq| from the cold source (the object being cooled) at temperature \(T_{c^{\prime}}\). The work done on the engine is |dw| and as a result heat (|dq| + |dw|) is discarded into the hot sink at temperature \(T_{\mathrm{h}}\). Explain how the Second law requires that, for the process to be allowed, the following relation must apply: \(\frac{|\mathrm{d} q|}{T_{\mathrm{c}}}=\frac{|\mathrm{d} q|+|\mathrm{d} w|}{T_{\mathrm{h}}}\) (b) Suppose that the heat capacity of the object being cooled is C (which can be assumed to be independent of temperature) so that the heat transfer for a change in temperature \(\mathrm{d} T_{\mathrm{c}}\) is \(\mathrm{d} q=C \mathrm{d} T_{\mathrm{c}}\). Substitute this relation into the expression derived in (a) and then integrate between \(T_{\mathrm{c}}=T_{\mathrm{i}}\) and \(T_{\mathrm{c}}=T_{\mathrm{f}}\) to give the following expression for the work needed to cool the object from \(T_{\mathrm{i}}\) to \(T_{\mathrm{f}}\) as \(w=C T_{\mathrm{h}}\left|\ln \frac{T_{\mathrm{f}}}{T_{\mathrm{i}}}\right|-\left|C\left(T_{\mathrm{f}}-T_{\mathrm{i}}\right)\right|\) (c) Use this result to calculate the work needed to lower the temperature of 250 g of water from 293 K to 273 K, assuming that the hot reservoir is at 293 K \(\left(C_{p, \mathrm{m}}\left(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right)=75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)\). (d) When the temperature of liquid water reaches 273 K it will freeze to ice, an exothermic process. Calculate the work needed to transfer the associated heat to the hot sink, assuming that the water remains at 273 K (the standard enthalpy of fusion of \(\mathrm{H}_{2} \mathrm{O}\) is \(6.01 \mathrm{kJ} \mathrm{mol}^{-1}\) at the normal freezing point). (e) Hence calculate the total work needed to freeze the 250 g of liquid water to ice at 273 K. How long will this take if the refrigerator operates at 100 W? Text Transcription: T_c T_h dT_c dq = Cdt_c T_c = T_i T_c = T_f T_i T_f (C_p,m(H_2O(l)) = 75.3 J K^?1 mol^?1) H_2O 6.01 kJ mol^-1 |dq|/t_c = |dq| + dw|/t_h w = CT_h |ln t_f/t_i| - |C(T_f - T_i)|

The standard molar entropy of \(\mathrm{NH}_{3}(\mathrm{g})\) is \(192.45 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) at 298 K, and its heat capacity is given by eqn 2B.8 with the coefficients given in Table 2B.1. Calculate the standard molar entropy at (a) \(100^{\circ} \mathrm{C}\) and (b) \(500^{\circ} \mathrm{C}\). Text Transcription: NH_3(g) 192.45 JK^-1mol^-1 100 degree C 500 degree C

Account for deviations from Trouton’s rule for liquids such as water, mercury, and ethanol. Is their entropy of vaporization larger or smaller than \(85 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\)? Why? Text Transcription: 25JK^-1mol^-1

Use Trouton’s rule to predict the enthalpy of vaporization of benzene from its normal boiling point, \(80.1^{\circ} \mathrm{C}\). Text Transcription: 80 degree C

Use Trouton’s rule to predict the enthalpy of vaporization of cyclohexane from its normal boiling point, 80.7^{\circ} \mathrm{C}. Text Transcription: 80.7 degree C

The enthalpy of vaporization of trichloromethane (chloroform, \(\mathrm{CHCl}_{3}\)) is \(29.4 \mathrm{kJ} \mathrm{mol}^{-1}\) at its normal boiling point of 334.88 K. Calculate (i) the entropy of vaporization of trichloromethane at this temperature and (ii) the entropy change of the surroundings. Text Transcription: CHCl_3 29.4kJ mol^-1

The enthalpy of vaporization of methanol is \(35.27 \mathrm{kJ} \mathrm{mol}^{-1}\) at its normal boiling point of \(64.1^{\circ} \mathrm{C}\). Calculate (i) the entropy of vaporization of methanol at this temperature and (ii) the entropy change of the surroundings. Text Transcription: 64.1 degree C 35.27 kJ mol^-1

Estimate the increase in the molar entropy of \(\mathrm{O}_{2}(\mathrm{g})\) when the temperature is increased at constant pressure from 298 K to 348 K, given that the molar constant-pressure heat capacity of \(\mathrm{O}_{2}\) is \(29.355 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) at 298 K. Text Transcription: O_2(g) O_2 29.355 J K^-1

Estimate the change in the molar entropy of \(\mathrm{N}_{2}(\mathrm{g})\) when the temperature is lowered from 298 K to 273 K, given that \(C_{p \mathrm{m}}\left(\mathrm{N}_{2}\right)=29.125 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) at 298 K. Text Transcription: N_2(g) C_p,m(N_2) = 29.125JK^-1mol^-1

The molar entropy of a sample of neon at 298 K is \(146.22 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\). The sample is heated at constant volume to 500 K; assuming that the molar constant-volume heat capacity of neon is \(\frac{3}{2} R\), calculate the molar entropy of the sample at 500 K. Text Transcription: 146.22 J K^-1 mol^-1 2/3R

Calculate the molar entropy of a constant-volume sample of argon at 250 K given that it is \(154.84 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\); the molar constant-volume heat capacity of argon is \(\frac{3}{2} R\). Text Transcription: 154.84 JK^-1mol^-1 3/2R

Two copper blocks, each of mass 1.00 kg, one at \(50^{\circ} \mathrm{C}\) and the other at \(0^{\circ} \mathrm{C}\), are placed in contact in an isolated container (so no heat can escape) and allowed to come to equilibrium. Calculate the final temperature of the two blocks, the entropy change of each, and \(\Delta S_{\text {tot }}\). The specific heat capacity of copper is \(0.385 \mathrm{JK}^{-1} \mathrm{g}^{-1}\) and may be assumed constant over the temperature range involved. Comment on the sign of \(\Delta S_{\text {tot }}\). Text Transcription: 50 degree C 0 degree C deltaS_tot deltaS_tot 0.385JK^-1g^-1

Calculate \(\Delta S_{\text {tot }}\) when two iron blocks, each of mass 10.0 kg, one at \(100^{\circ} \mathrm{C}\) and the other at \(25^{\circ} \mathrm{C}\), are placed in contact in an isolated container and allowed to come to equilibrium. The specific heat capacity of iron is \(0.449 \mathrm{JK}^{-1} \mathrm{g}^{-1}\) and may be assumed constant over the temperature range involved. Comment on the sign of \(\Delta S_{\text {tot }}\). Text Transcription: deltaS_tot deltaS_tot 25 degree C 100 degree C 0.449JK^-1g^-1

Calculate \(\Delta S\) (for the system) when the state of 3.00 mol of gas molecules, for which \(C_{p, m}=\frac{5}{2} R\), is changed from \(25^{\circ} \mathrm{C}\) and 1.00 atm to \(125^{\circ} \mathrm{C}\) and 5.00 atm. Text Transcription: deltaS C_p,m = 5/2R 25 degree C 125 degree C