<p>CHM 3400 – Problem Set 2 Due date: Wednesday, January 30th Do all of the following problems. Show your work. </p><p>1) The molar volume of a real gas can be written in terms of a power series expansion in pressure. This allows us to improve on the ideal gas law in describing the behavior of a real gas. A particular real gas has a molar volume given by the expression</p><p>2 Vm = V/n = (RT/p) [ 1 + ap + bp ] (1.1) </p><p> where a and b are temperature independent constants found using experimental data. Find the value for (Vm/T)p and (Vm/p)T for a substance whose equation of state is given by eq 1.1.</p><p>2) Consider 1.00 mole of argon gas. Assume that for the conditions in the problem argon obeys the ideal gas law. 3 5 Since argon is a monatomic gas, CV,m = ( /2) R and Cp,m = ( /2) R. Find the value for q, w, U, and H for the following processes.</p><p> a) The temperature of the gas is changed from an initial value Ti = 300.0 K to a final value Tf = 400.0 K at a constant pressure p = 1.00 atm.</p><p> b) The temperature of the gas is changed from an initial value Ti = 300.0 K to a final value Tf = 400.0 K at a constant volume V = 40.0 L.</p><p> c) The gas is expanded isothermally and reversibly from an initial pressure pi = 5.00 atm to a final pressure pf = 1.50 atm, at T = 300.0 K. </p><p>3) 1.000 mole of an ideal gas undergoes an adiabatic irreversible process. The temperature of the gas decreases from an initial value Ti = 320.0 K to a final value Tf = 270.0 K. Over this temperature range the constant pressure molar heat capacity of the gas is given by the expression</p><p>2 Cp,m = a + bT a = 18.43 J/molK b = 0.028 J/molK </p><p>Find q, w, U, and H for the process, or explain why there is not sufficient information to find one or more of these quantities.</p><p>4) The van der Waals equation of state is</p><p> p = nRT - an2 (4.1) (V – nb) V2 where a and b are constants. a) Find a general expression for w, the work, when n moles of a gas obeying the van der Waals equation of state undergoes an isothermal reversible change in volume from an initial volume Vi to a final volume Vf. Give your result in terms of a, b, T, Vi, Vf, and/or other constants. b) Use your answer in a to find the value for work when the volume occupied by 2.000 moles of ethane (C2H6) changes from an initial value 20.00 L to a final volume 1.00 L by a reversible and isothermal process, at T = 300.0 K. The van der Waals coefficients for ethane are a = 5.507 L2atm/mol2, b = 0.0651 L/mol. c) Find the value for w for the same process as part b of this problem, but assuming that ethane obeys the ideal gas law. Compare your result to that found in part b. 5) The constant pressure molar heat capacity of a substance is often written as</p><p>2 Cp,m = a + bT + c/T (5.1) where a, b, and c are constants. a) Using the above expression and the data in Table 2.2, p 59 of Atkins, find q when the temperature of</p><p>1.00 mol of graphite (C(s)) is changed at constant pressure from an initial value T i = 300.0 K to a final value Tf = 400.0 K. b) Find w, U, and H for the above process, or briefly explain why we lack the information to find them.</p><p>6) Hydrogenation is the process in which hydrogen is added to unsaturated hydrocarbons. Consider the following hydrogenation rection</p><p>C2H2(g) + 2 H2(g)  C2H6(g) (6.1)</p><p>Find Urxn and Hrxn for the above hydrogenation reaction at T = 25.0 C. Note that useful thermodynamic data are given in Appendix D of Atkins.</p><p>Solutions. 2 1) Since Vm = (RT/p) [ 1 + ap + bp ] = (RT/p) + (aRT) + (bRTp)</p><p>2 2 Then (Vm/T)p = /T)p { (RT/p) [ 1 + ap + bp ] } = (R/p) [ 1 + ap + bp ]</p><p>2 2 (Vm/p)T = /p)T {(RT/p) + (aRT) + (bRTp)} = - RT/p + bRT = RT [b – (1/p )]</p><p>3 5 2) Note that CV,m = ( /2) R = 12.471 J/molK, and Cp,m = ( /2) R = 20.785 J/molK.</p><p> a) The gas is ideal, and so </p><p> f f U = i nCV,m dT H = i nCp,m dT</p><p>Since both CV,m and Cp,m are constant, we can take them outside the above integrals, to get </p><p> f f U = i nCV,m dT = nCV,m i dT = nCV,m (Tf – Ti) </p><p>= (1.00 mol) (12.471 J/molK) (400.0 K – 300.0 K) = + 1247. J</p><p> f f H = i nCp,m dT = nCp,m i dT = nCp,m (Tf – Ti) </p><p>= (1.00 mol) (20.785 J/molK) (400.0 K – 300.0 K) = + 2078. J</p><p>The process is carried out at constant pressure, and so q = H = + 2078. J</p><p>Finally, since U = q + w, w = U – q = (1247. J) – (2078. J) = - 831. J</p><p> f NOTE: We could also find w directly using w = - i pex dV </p><p> f Since external pressure is constant, then w = - pex i dV = - pex (Vf – Vi). Since we have sufficient information to find both Vi and Vf we could also use this method to find w.</p><p> b) The gas is ideal, and so </p><p> f f U = i nCV,m dT H = i nCp,m dT</p><p>The process has the same initial and final temperature as in part a, and the gas and number of moles of gas are also the same, and so U and H are also the same (this is not true in general, but is true here because internal energy and enthalpy for an ideal gas depend only on temperature).</p><p>So U = + 1247. J H = 2078. J</p><p>The process is carried out at constant volume, and so q = U = + 1247. J.</p><p> f Finally, since the process is constant volume, w = - i pex dV = 0.</p><p> c) For an isothermal process on an ideal gas U = H = 0. From the first law, U = q + w = 0, and so q = -w.</p><p> f But w = - i pex dV</p><p>The process is reversible, and so pex = p = nRT/V. Substituting, we get</p><p> f w = - i (nRT/V) dV</p><p>The process is isothermal, and so T is constant, and can be taken outside the integral.</p><p> f So w = - (nRT) i (dV/V) = - (nRT) ln(Vf/Vi)</p><p>For an isothermal process on an ideal gas piVi = pfVf (Boyle’s law), so (Vf/Vi) = (pi/pf). </p><p>So w = - nRT ln(pi/pf) = - (1.000 mol)(8.314 J/molK)(300.0 K) ln[5.00/1.50] = - 3003. J</p><p> q = -w = + 3003. J</p><p>3) The process is adiabatic, and so q = 0. </p><p>Since the gas is ideal we can say the following</p><p> f f U = i nCV,m dT H = i nCp,m dT CV,m = Cp,m - R = (a - R) + bT</p><p> f f 2 2 So H = i nCp,m dT = n i (a + bT) dT = n { a(Tf – Ti) + (b/2)(Tf – Ti ) }</p><p>= (1.000 mol) { (18.43 J/molK)(270.0 K – 320.0 K) + [(0.028 J/molK2)/2][(270.0 K)2 – (320.0 K)2] }</p><p>= (1.000 mol) { (- 921.5 J) + ( - 413.0 J) } = - 1334. J</p><p> f f 2 2 U = i nCV,m dT = n i (a - R + bT) dT = n { (a - R)(Tf – Ti) + (b/2)(Tf – Ti ) }</p><p>= (1.000 mol) { (10.116 J/molK)(270.0 K – 320.0 K) + [(0.028 J/molK2)/2][(270.0 K)2 – (320.0 K)2] }</p><p>= (1.000 mol) { (- 505.8 J) + ( - 413.0 J) } = - 918.8. J</p><p>Since U = q + w, and q = 0 , w = U = - 918.8 J</p><p>4) a) For a van der Waals gas p = nRT - an2 (V - nb) V2 </p><p> f In general w = - i pex dV. Since the process is reversible, pex = p, so</p><p> f 2 2 w = - i [ (nRT)/(V - nb) ] - (an /V ) dV</p><p>The process is isothermal, and so T is constant. If we do the integral, we get</p><p>2 w = - nRT ln[ (Vf - nb)/(Vi - nb) ] - an [ (1/Vf) - (1/Vi) ]</p><p>We may manipulate the terms on the right to get rid of the - signs, to get</p><p>2 w = nRT ln [(Vi - nb)/(Vf - nb) ] + an [ (1/Vi) - (1/Vf) ] </p><p>As a check, if we set a = b = 0, we get w = nRT ln(Vi/Vf) = - nRT ln(Vf/Vi) , the ideal gas result. That doesn't prove our result is correct, but does show it is consistent with what we expect.</p><p> b) Using the above general result found in part a</p><p> w = (2.000 mol)(8.314 J/molK)(300.0 K) ln[(20.00 L – (2.000 mol))(0.0651 L/mol)]/ [(1.00 L – (2.000 mol))(0.0651 L/mol)]</p><p>- (5.507 L2atm/mol2)(2.000 mol)2 [ (1/20.00 L) – (1/1.00 L) ]</p><p>= 15607. J – (20.927 Latm)(101.325 J/Latm) </p><p>= 15607. J – 2120. J = 13487. J</p><p> c) If we used the result obtained from the ideal gas law (which is given in part a of this problem), then</p><p> w = nRT ln(Vi/Vf) = (2.000 mol)(8.314 J/molK)(300.0 K) ln(20.00/1.00)</p><p>= 14944. J</p><p>This is larger than the result found in part b (using the van der Waals equation) by 1457. J, or ~ 11 % larger. As is generally the case, the result found using the van der Waals equation is likely closer to the true value for w than the result found using the ideal gas law.</p><p>5) a) The heating is reversible and at constant pressure, and so </p><p> f f f 2 q = i n Cp,m dT = n i Cp,m dT = n i [ a + bT + (c/T ) ]dT</p><p>2 2 = n { a (Tf – Ti) + (b/2) (Tf – Ti ) – c [ (1/Tf) – (1/Ti) ] }</p><p>Using the information from the problem and from Table 2.2, we get</p><p> q = (1.00 mol) { (16.86 J/mol.K) (400.0 K – 300.0 K) + (4.77 x 10-3 J/mol.K2)/2 [ (400.0 K)2 – (300.0 K)2 ]</p><p>- (- 8.54 x 105 J.K/mol) [ (1/400.0 K) – (1/300.0 K) ]</p><p>= (1.00 mol) { (1686. J/mol.K) + (167.0 J/mol.K) – 711.7 J/mol.K) } </p><p>= + 1141. J</p><p> b) Since the process is carried out at constant pressure, H = q = + 1141. J.</p><p>From the first law U = q + w. We have an expression for q, so if we could find either U or w we could use the first law to find the missing quantity. However, we do not have sufficient information to find either U or w, and so neither of them can be found.</p><p>NOTE: While the above is true, we know that the volume occupied by a solid does not change very much with temperature. Since pressure is constant for the above process</p><p> f w = - i pex dV = - pex (Vf – Vi)  0, and so U  q = 1141. J.</p><p>6) We may find Hrxn from the data in Appendix D of Atkins</p><p>Hrxn = Hf(C2H6(g)) – [Hf(C2H2(g)) + 2 Hf(H2(g))] = ( - 84.68 kJ/mol) – [ (226.73 kJ/mol) + 2 (0.0 kJ/mol) ] = - 311.41 kJ/mol</p><p>To find Urxn we may use the approximate relationship</p><p>Hrxn = Urxn + ngRT ; where ng = change in the number of moles in the gas phase per mole of reaction</p><p>-3 So Urxn  Hrxn - ngRT = - 311.41 kJ/mol – ( -2) (8.314 x 10 kJ/molK)(298.15 K)</p><p>= - 306.45 kJ/mol</p><p>Notice that Urxn and Hrxn are approximately the same, as is typical for chemical reactions. </p>