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SINHA�s I. I.T. CHEMISTRY - 1 -

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Solution to THERMODYNAMICS TEST PAPER 1. Solutions (a) True: at standard pressure, the Celsius scale uses 100°C as a reference point. (b) True: the pressure of an ideal gas is directly proportional to its temperature. (c) False: Avogadro's hypothesis refers to the number of molecules or atoms of a given volume. Two equal volumes under the same conditions need not have the same mass (m) even though they have the same number of molecules. (d) True: see (c) (e) False: see (c) (f) True: see (c) 2. Solutions:

(a) Since the process is cyclic, ÄU, ÄH, ÄS, ÄA, and ÄG are zero. Since the process is reversible, ÄStot is equal to zero. (b) Since no heat enters or leaves the system and the volume is fixed (no PV work), ÄU is zero by the First Law. ÄH is not zero since burning hydrogen will change the pressure of the container. ÄA and ÄG are not zero since this process is not at equilibrium. ÄS and ÄStot are not zero since the change in entropy of the reaction is none zero and ÄStot is equal to ÄS since qact is zero making the change in entropy of the surroundings zero. (c) The Joule-Thompson expansion performed with the enthalpy constant so ÄH is zero.Since the gas is non-ideal, ÄU is not zero. Since there is a change in volume and temperature, ÄS is not zero. Since ÄStot is equal to ÄS since qact is zero making the change in entropy of the surroundings zero. ÄA and ÄG are not zero since this process is not at equilibrium. (d) ÄG is zero since this process is in equilibrium at constant pressure. ÄA is approximately zero since this process is at equilibrium and the volume change between the changes in state is small. ÄStot is zero since the process is reversible under the conditions stated. ÄU, ÄH, and ÄS are not zero since heat must be added to melt the ice. 3.

4.Solution: Since the process is isothermal and reversible:

if the gas is ideal (assumed). Note: f i i f p p = V V since the temperature is constant.

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Therefore: 5. Solution. First, note that ÄV, which is a state function, is the same for each path: V2 = (10/1) × (1 L) = 10 L, so ÄV = 9 L. For path (a), w = �(1 atm)× (9 L) = �9 L-atm. For path (b), the work is calculated for each stage separately: w = �(5 atm) × (2�1 L) � (1 atm) × (10�2 L) = �13 L-atm For path (c) the process would be carried out by removing all weights from the piston in Fig. 1 so that the gas expands to 10 L against zero external pressure. In this case w = (0 atm) × 9 L = 0; that is, no work is done because there is no force to oppose the expansion. 6. Solution: In this process the volume of liquid remains practically unchanged, so ÄV = �24.5 L. The work done is w = �PÄV = �(1 atm)(�24.5 L) = 24.6 L-atm (The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = 101.33 J mol�1 and substituting in Eq. 3 (above) we obtain ÄU= q +PÄV = �(75300 J) + [101.33 J/L-atm) × (24.5 L-atm)] = �72.82 kJ In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system�s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of ÄU to �72.82 kJ. 7.Water mixing with water represents no entropy change, but moles of A mixing with moles of B surely does. And we have 0.1 moles of A and 0.15 moles of B for a total of 0.25 moles of solute. That's n. And the final mole fraction of A is XA = 0.1/0.25 = 0.4 meaning that XB = 0.6. mixS = - n R (XAlnXA + XBlnXB) = - ¼ mol 8.314 J/mol K (0.4ln0.4 + 0.6ln0.6) mixS = + 1.40 J/K mixH = 0.00 by definition of ideal mixture. overallG° = + 23.8 - 31.9 = - 8.1 kJ/mol K = e-

overallG° / RT = e8100 J/mol / ( 8.314 J/mol K × 310 K ) = 23.2 > 1

9.Use a P,T form of the adiabatic expression . If we want a p,T adiabatic expression, we can get it by substitution of the ideal gas expression Vm = RT/p into TV1/c fixed. TV1/c = T(RT/p)1/c = R1/c T1+(1/c) p-1/c = fixed



Since R is fixed regardless, we can fold it into whatever constant "fixed" stands for. Likewise we can raise the entire expression to the cth power and the constant similarly. This means that the expression we seek is Tc+1 / p = T(C

V / R) + 1 / p = TC

P / R / p = fixed

(since for an ideal gas, CV+R = CP) But it's temperature in which we're interested, so we should rearrange the expression by raising it to the R/CP power to get T p=R/C

P is fixed, or Tf pf-R/C

P = Ti pi-R/C

P or Tf = Ti (pf / pi)R/C

P Tf = 298 K ( 5 bar / 1 bar)8.315 J mol-1 K-1 / 29.17 J mol-1 K-1 Tf = 298 K 5 0.2850 = 471 K well above boiling! In fact, there are devices made which use sudden adiabatic compression to generate heat high enough to light kindling. 10. nH2O = 1 kg × (1 mol H2O / 0.018 kg) = 55.5 mol H = nCPT + nvapH H = 55.5 mol [ 0.07529 kJ mol-1 K-1 × (75 K) + 40.66 kJ/mol ] H = 55.5 mol (46.31 kJ/mol) = 2570 kJ CH4 + 2 O2 CO2 + 2 H2O rH° = fH°[CO2] + 2 fH°[H2O] - fH°[CH4] rH° = (-393.51) + 2(-285.83) - (-74.81) = - 890.36 kJ/mol Thus, we need nCH4 = 2570/890.36 = 2.886 mol V = nRT/p V = 2.886 mol × 0.08206 atm L mol-1 K-1 × 273 K / (1/1.01325) atm V = 65.5 L CH4 to boil 1 L H2O 11. d(G/T)dT = - H/T2 Integrating, (G/T) = + H/T G/T = G/T + H×[(1/T) - (1/T)] 0 = G = xG + H(1-x) where x = T/T x = H/(H - G) = -45/(-45+16) = 1.55 T = 1.55×T = 1.55(298 K) = 462 K (You get the same answer if you set G = H - TS = 0 and solve for T taking H and S at their standard conditions values.) S° = (H° - G°)/T (T=298 K) G° = - RT ln(K25°C) = - 8.314 J/mol K (298 K) ln(1.774×10-5) G° = + 27.1 kJ/mol d [ln(K)] / d [1/T] = - H° / R H° = - R d [ln(K)] / d [1/T] ~ - R ln(K) / (1/T) H° = - 8.314 J/mol K (10.9764-10.9141) / (3.299-3.411)×10-3 K-1 H° = + 4.61 kJ/mol S° = (4,610 - 27,100) J/mol / 298 K = - 75 J/mol K fG°(diamond) = +2.900 kJ/mol ... from graphite, of course. The densities are [graphite] = 2.260 g/cc and [diamond] = 3.513 g/cc. Ignoring their isothermal compressibilities (because I can find only diamond's, 1.87×10-7 atm-1), at room temperature, what pressure would be required to make diamonds? G = [integral p° to p] V dp ~ V p



= M / V V = M / Vgraphite = 12.01 g/mol / 2.260 g/cc = 5.314 cc/mol = 5.314×10-6 m3/mol Vdiamond = 12.01 g/mol / 3.513 g/cc = 3.419 cc/mol = 3.419×10-6 m3/mol V = Vdiamond - Vgraphite = - 1.895×10-6 m3/mol G = V p = -2.900×103 J/mol p = -2.900×103 J/mol / ( - 1.895×10-6 m3/mol ) = 1.530×109 J m-3 p = 1.530×104 105 Pa = 15,300 bar = 15,101 atm p = p - p° = 15,100 atm 14. VTC

V/R = constant for rev. adiab. exp.

CV=CP-R = 1.007 J/g K(28.96 g/mol) - 8.315 J/mol K = 20.85 J/mol K Tf = Ti(Vi/Vf)

R/CV = 293 K (½)8.315/20.85 = 222 K

15.

17.

From the standard enthalpy of fusion at one atmosphere, calculate the standard molar entropy change upon melting of water.

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18.

19. Requirement 1.ÄS ≥ 0 2.ÄA ≤ 0 3.ÄG ≤ 0 20. Ans We have three processes, all carried out on an ideal gas in the same initial state. The three processes are shown in p, V space. Note that the first two processes, (a) and (b), have the same initial and final states. The final state for the last process, (c), ends at the same pressure but a different volume (because it is adiabatic).

Because the first two processes are isothermal and connect the same initial and final states of the system,

In process (a), the gas expands reversibly and isothermally to a final pressure of 0.1bar.



In process (b), the gas expands against a constant external pressure (the pressure in the system is, of course, changing over the course of the expansion).

The final process, (c), is a reversible adiabatic expansion, and does not end in the same state as the first two processes. For an adiabatic process on an ideal gas, the following relations hold: 21.

This is the equation one needs to find the temperature. It is necessary to find an equation to relate the heat capacity at constant volume to the heat capacity at constant pressure. According to the problem, one is to assume that the air is ideal. Then, one may use the relationship of equation 2.24 to find the heat capacity at constant pressure:

monatomic ideal gases. The chief components of air are oxygen and nitrogen, neither one of which is a

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monatomic gas!] Rearrangement of the equation above gives the following equation for the temperature at the bottom of the mountain:

solution bansal pattern thermodynamics practice test by s.k.sinha see chemistry animations at...

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