There are six problems on the exam. Do all of the problems. Show your work

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1 CHM 3400 Fundamentals o Physical Chemistry First Hour Exam There are six problems on the exam. Do all o the problems. Show your work R = L. atm/mole. K N A = x R = L. bar/mole. K 1 L. atm = J R = J/mole. K 1 atm = bar = x 10 5 N/m 2 1. Real gases behave dierently than ideal gases under conditions o low temperature and/or high pressure. This is one reason other equations o state that better model the behavior o real gases have been developed. Consider a lecture bottle (a small gas cylinder) with volume V = cm 3. The lecture bottle contains 68.2 g o nitrogen gas (N 2, M = g/mol), at a temperature T = 20.0 C. a) Find the pressure o gas in the lecture bottle assuming nitrogen obeys the ideal gas law. Give your inal value or pressure in units o atm (atmospheres). b) Find the pressure o gas in the lecture bottle assuming nitrogen obeys the van der Waals equation o state. Give your inal value or pressure in units o atm (atmospheres). For nitrogen a = L 2 atm/mol 2, b = L/mol. c) As discussed in class, the van der Waals a and b coeicients can be used to estimate the values or the critical constants o the gas. Using the inormation in part b o this problem, estimate the value or T c, the critical temperature, or nitrogen. 2. The main component in the atmosphere o Jupiter is hydrogen gas (H 2, M = 2.02 g/mol, = 0.27 nm 2 ). Consider a sample o hydrogen gas at p = torr, T = 90.0 K. For these conditions o pressure and temperature you may assume H 2 is an ideal gas. Find the ollowing a) c rms, the rms average speed o an H 2 molecule in the gas. Give your answer in units o m/s. b), the mean ree path o an H 2 molecule in the gas. Give your answer in units o nm. 3. One virial equation or a real gas is p = (nrt/v) [ 1 + (nb/v) ] (3.1) where B, the irst virial coeicient, is a constant ound by itting eq 3.1 to experimental data. a) Find an expression or ( p/ V) T, the partial derivative o pressure with respect to volume, at constant temperature, or a gas whose equation o state is given by eq 3.1. b) n moles o a gas whose equation o state is given by eq 3.1 undergoes an isothermal reversible expansion rom an initial volume V i to a inal volume V. Find an expression or w, the work associated with the isothermal expansion.

2 moles o an ideal gas undergoes a reversible expansion at a constant pressure p = atm. Experimentally, it is ound that q = J and T = (T - T i) = 22.0 K or the process. What are w, U, H, and V (where V = V V i) or the process? Give w, U, and H in units o J (joule) and V in units o L (liters). 5. Anthracene (C 14H 10(s)), M = g/mol) is an example o a polycyclic aromatic hydrocarbon (PAH). Because PAHs are urban air pollutants, there is interest in their thermodynamic properties. The experimental value or the enthalpy o combustion or anthracene, measured at T = 25.0 C, is H comb(c 14H 10(s)) = kj/mol. The constant pressure molar heat capacity o anthracene, measured at T = 25.0 C, is J/mol K. a) Give the balanced combustion reaction or anthracene. b) Give the balanced ormation reaction or anthracene. c) Based on the inormation in this problem and in Appendix D o Atkins, ind U comb(c 14H 10(s)), the energy o combustion o anthracene, and H (C 14H 10(s)), the enthalpy o ormation o anthracene, at T = 25.0 C. d) Based on the inormation in this problem and in Appendix D o Atkins, estimate the value or H comb(c 14H 10(s)), the enthalpy o combustion o anthracene, at T = C. 6. The pressure o mol o an ideal gas is changed rom an initial value p i = atm to a inal value p = atm, at a constant temperature T = K. Note that the temperature o the surroundings is also constant and equal to K. The constant pressure molar heat capacity o the gas is C p,m = J/mol K. q or the process is J. What are w, U, H, S syst, S surr, and S univ or the above process?

3 lutions. 1) Note that the moles o N 2 is n = 68.2 g 1 mol = mol g a) For the ideal gas law result p = nrt = (2.436 mol)( L atm/mol K)( K) = atm V ( L) b) For the result predicted rom the van der Waals equation p = nrt - an 2 (V nb) V 2 = (2.436 mol)( L atm/mol K)( K) - (1.352 L 2 atm/mol 2 )(2.436 mol) 2 [ L (2.436 mol)( L/mol) ] ( L) 2 = atm atm = atm c) T c = 8a = 8(1.352 L 2 atm/mol 2 ) = 126. K 27Rb 27( L atm/mol K)( L/mol) 2) a) c rms = [3RT/M] 1/2 = [3(8.314 J/mol K)(90.0 K)/(2.02 x 10-3 kg/mol)] 1/2 = m/s b) p = 2 torr 1 atm x 10 5 Pa = Pa 760 torr 1 atm = RT = (8.314 J/mol K)(90.0 K) (2) 1/2 N A p (2)1 /2 (6.022 x mol -1 )(0.27 x m 2 )(266.6 Pa) = 1.22 x 10-5 m = nm Note that some editions o Atkins do not contain the actor o 2 1/2 in the equation or. A key to making these calculations simple is to put all quantities in the equations in MKS units, which means one can be (reasonably) sure that the results will also be in MKS units. 3) p = (nrt/v) [ 1 + (nb/v) ] = nrt + n 2 RTB V V 2 Writing p in this way will make doing the problem more simple a) ( p/ V) T = / V) T nrt + n 2 RTB = - nrt - 2n 2 RTB = - (nrt/v 2 )[ 1 + (2nB/V) ] V V 2 V 2 V 3 b) w = - i p ex dv The process is reversible, and so p ex = p w = - i p dv = - i [ (nrt/v) + (n 2 RTB/V 2 ) ] dv The process is isothermal, and so T can be taken outside the integral, to give = - { nrt ln V n 2 RTB (1/V) } i = - { nrt ln(v /V i) - n 2 RTB [ (1/V ) (1/V i) ] } = - nrt { ln(v /V i) nb [ (1/V ) (1/V i) ] }

4 4) Process is constant pressure, and so q = H = J The gas is ideal, and so U = i n C V,m dt = n i C V,m dt However, since the gas is ideal, C p,m C V,m = R ; so C V,m = C p,m R U = n i n (C p,m R ) dt = n i C p,m dt - n i R dt = n i C p,m dt - nr( T) But H = = n i C p,m dt U = H nr( T) = J (1.000 mol)(8.314 J/mol K)(22.0 K) = J U = q + w ; so w = U q = (1277. J) (1460. J) = J Finally w = - i p ex dv The process is reversible, and so p ex = p. Since p is a constant it can be taken outside the integral, and so w = - i p dv = - p i dv = - p( V) and so V = - w = - ( J) (1 L atm) = 3.01 L p (0.600 atm) (101.3 J) 5) a) combustion C 14H 10(s) + 33 / 2 O 2(g) 14 CO 2(g) + 5 H 2O( ) b) ormation 14 C(s) + 5 H 2(g) C 14H 10(s) c) Since H U + n grt U = H - n grt = ( kj/mol) (- 2.5)(8.314 x 10-3 J/mol K)(298. K) = kj/mol Using the general relationship or the enthalpy change or a chemical reaction, we may say H comb = [ 14 H (CO 2(g)) + 5 H (H 2O( )) ] [ H (C 14H 10(s)) ] H (C 14H 10(s)) = [ 14 H (CO 2(g)) + 5 H (H 2O( )) ] [ H comb ] = [ 14 ( ) + 5 ( ) ] ( ) = kj/mol

5 d) H comb(t 2) = H comb(t 1) + T1 T2 C p dt I we assume C p is independent o temperature (a reasonably good approximation or a small change in temperature) then H comb(t 2) = H comb(t 1) + ( C p) ( T) But C p = [ 14 C p,m(co 2(g)) + 5 C p,m(h 2O( )) ] [ C p,m(c 14H 10(s)) + 33 / 2 C p,m(o 2(g)) ] = [ 14 (37.11) + 5 (75.291) ] [ (210.5) + 33 / 2 (29.355) ] = J/mol K T = 100. C 25. C = 75. C = 75. K (true because the size o a degree Centigrade and a degree Kelvin is the same) H comb(100. C) = H comb(25. C) + (201.1 x 10-3 kj/mol K) (75. K) = kj/mol kj/mol = kj/mol 6) The process is isothermal and the gas is ideal, and so U = H = 0. From the irst law, U = q + w. Since U = 0, w = -q = J To ind S syst, we need a reversible process with the same initial and inal state as the unspeciied process in the problem. A reversible isothermal expansion rom p i = atm to p = 1.00 atm is such a process. S syst = i (dq) rev/t. The process being used to ind S syst is isothermal and reversible, and so we can take T outside the integral, to get S syst = (1/T) i (dq) rev = q rev/t For an isothermal reversible expansion o an ideal gas, q rev = nrt ln(p i/p ) And so S syst = (1/T) [ nrt ln(p i/p ) ] = nr ln(p i/p ) = (1.00 mol) (8.314 J/mol K) ln(10.0/1.00) = J/K S surr = - q syst/t = J/320.0 K = J/K (Note that moving J o heat rom the system to the surroundings is reversible rom the point o view o the surroundings, even i the process taking place is irreversible rom the point o view o the system.) S univ = S syst + S surr = J/K = ( J/K) = J/K