Florida International University s9

CHM 3400 – Problem Set 2 Due date: Wednesday, January 30th Do all of the following problems. Show your work.

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

2 Vm = V/n = (RT/p) [ 1 + ap + bp ] (1.1)

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.

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.

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.

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.

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.

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

2 Cp,m = a + bT a = 18.43 J/molK b = 0.028 J/molK

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.

4) The van der Waals equation of state is

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

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

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.

6) Hydrogenation is the process in which hydrogen is added to unsaturated hydrocarbons. Consider the following hydrogenation rection

C2H2(g) + 2 H2(g)  C2H6(g) (6.1)

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.

Solutions. 2 1) Since Vm = (RT/p) [ 1 + ap + bp ] = (RT/p) + (aRT) + (bRTp)

2 2 Then (Vm/T)p = /T)p { (RT/p) [ 1 + ap + bp ] } = (R/p) [ 1 + ap + bp ]

2 2 (Vm/p)T = /p)T {(RT/p) + (aRT) + (bRTp)} = - RT/p + bRT = RT [b – (1/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.

a) The gas is ideal, and so

f f U = i nCV,m dT H = i nCp,m dT

Since both CV,m and Cp,m are constant, we can take them outside the above integrals, to get

f f U = i nCV,m dT = nCV,m i dT = nCV,m (Tf – Ti)

= (1.00 mol) (12.471 J/molK) (400.0 K – 300.0 K) = + 1247. J

f f H = i nCp,m dT = nCp,m i dT = nCp,m (Tf – Ti)

= (1.00 mol) (20.785 J/molK) (400.0 K – 300.0 K) = + 2078. J

The process is carried out at constant pressure, and so q = H = + 2078. J

Finally, since U = q + w, w = U – q = (1247. J) – (2078. J) = - 831. J

f NOTE: We could also find w directly using w = - i pex dV

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.

b) The gas is ideal, and so

f f U = i nCV,m dT H = i nCp,m dT

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).

So U = + 1247. J H = 2078. J

The process is carried out at constant volume, and so q = U = + 1247. J.

f Finally, since the process is constant volume, w = - i pex dV = 0.

c) For an isothermal process on an ideal gas U = H = 0. From the first law, U = q + w = 0, and so q = -w.

f But w = - i pex dV

The process is reversible, and so pex = p = nRT/V. Substituting, we get

f w = - i (nRT/V) dV

The process is isothermal, and so T is constant, and can be taken outside the integral.

f So w = - (nRT) i (dV/V) = - (nRT) ln(Vf/Vi)

For an isothermal process on an ideal gas piVi = pfVf (Boyle’s law), so (Vf/Vi) = (pi/pf).

So w = - nRT ln(pi/pf) = - (1.000 mol)(8.314 J/molK)(300.0 K) ln[5.00/1.50] = - 3003. J

q = -w = + 3003. J

3) The process is adiabatic, and so q = 0.

Since the gas is ideal we can say the following

f f U = i nCV,m dT H = i nCp,m dT CV,m = Cp,m - R = (a - R) + bT

f f 2 2 So H = i nCp,m dT = n i (a + bT) dT = n { a(Tf – Ti) + (b/2)(Tf – Ti ) }

= (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] }

= (1.000 mol) { (- 921.5 J) + ( - 413.0 J) } = - 1334. J

f f 2 2 U = i nCV,m dT = n i (a - R + bT) dT = n { (a - R)(Tf – Ti) + (b/2)(Tf – Ti ) }

= (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] }

= (1.000 mol) { (- 505.8 J) + ( - 413.0 J) } = - 918.8. J

Since U = q + w, and q = 0 , w = U = - 918.8 J

4) a) For a van der Waals gas p = nRT - an2 (V - nb) V2

f In general w = - i pex dV. Since the process is reversible, pex = p, so

f 2 2 w = - i [ (nRT)/(V - nb) ] - (an /V ) dV

The process is isothermal, and so T is constant. If we do the integral, we get

2 w = - nRT ln[ (Vf - nb)/(Vi - nb) ] - an [ (1/Vf) - (1/Vi) ]

We may manipulate the terms on the right to get rid of the - signs, to get

2 w = nRT ln [(Vi - nb)/(Vf - nb) ] + an [ (1/Vi) - (1/Vf) ]

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.

b) Using the above general result found in part a

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)]

- (5.507 L2atm/mol2)(2.000 mol)2 [ (1/20.00 L) – (1/1.00 L) ]

= 15607. J – (20.927 Latm)(101.325 J/Latm)

= 15607. J – 2120. J = 13487. J

c) If we used the result obtained from the ideal gas law (which is given in part a of this problem), then

w = nRT ln(Vi/Vf) = (2.000 mol)(8.314 J/molK)(300.0 K) ln(20.00/1.00)

= 14944. J

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.

5) a) The heating is reversible and at constant pressure, and so

f f f 2 q = i n Cp,m dT = n i Cp,m dT = n i [ a + bT + (c/T ) ]dT

2 2 = n { a (Tf – Ti) + (b/2) (Tf – Ti ) – c [ (1/Tf) – (1/Ti) ] }

Using the information from the problem and from Table 2.2, we get

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 ]

- (- 8.54 x 105 J.K/mol) [ (1/400.0 K) – (1/300.0 K) ]

= (1.00 mol) { (1686. J/mol.K) + (167.0 J/mol.K) – 711.7 J/mol.K) }

= + 1141. J

b) Since the process is carried out at constant pressure, H = q = + 1141. J.

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.

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

f w = - i pex dV = - pex (Vf – Vi)  0, and so U  q = 1141. J.

6) We may find Hrxn from the data in Appendix D of Atkins

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

To find Urxn we may use the approximate relationship

Hrxn = Urxn + ngRT ; where ng = change in the number of moles in the gas phase per mole of reaction

-3 So Urxn  Hrxn - ngRT = - 311.41 kJ/mol – ( -2) (8.314 x 10 kJ/molK)(298.15 K)

= - 306.45 kJ/mol

Notice that Urxn and Hrxn are approximately the same, as is typical for chemical reactions.