THE FIRST LAW OF THERMODYNAMICS

THE FIRST LA OF THERMODYNAMIS 9 9 (a) IDENTIFY and SET UP: The ressure is constant and the volume increases (b) = d Figure 9 Since is constant, = d = ( ) The -diagram is sketched in Figure 9 The roblem gives T rather than and, so use the ideal gas law to rewrite the exression for EXEUTE: = nrt so = nrt, ; = nrt subtracting the two equations gives ( ) = nr( T T) Thus = nr( T T) is an alternative exression for the work in a constant ressure rocess for an ideal gas Then = nr( T T) = (00 mol)(8 J/mol K)(07 7 ) =+ 0 J EALUATE: The gas exands when heated and does ositive work 9 IDENTIFY: At constant ressure, = Δ = nrδ T SET UP: R = 8 J/mol K ΔT has the same numerical value in kelvins and in 7 0 J EXEUTE: Δ T = = = K Δ TK =Δ T and T = 70 + = 6 nr (6 mol) (8 J/mol K) EALUATE: hen > 0 the gas exands hen is constant and increases, T increases 9 IDENTIFY: Examle 9 shows that for an isothermal rocess = nrtln( / ) = nrt says decreases when increases and T is constant SET UP: T = 8 K = EXEUTE: (a) The -diagram is sketched in Figure 9 (b) = (00 mol)(8 J/mol K)(8 K)ln = 60 J EALUATE: Since decreases, is negative Figure 9 9 IDENTIFY: Use the exression for that is aroriate to this tye of rocess SET UP: The volume is constant EXEUTE: (a) The diagram is given in Figure 9 (b) Since Δ = 0, = 0 9-

9- hater 9 EALUATE: For any constant volume rocess the work done is zero Figure 9 9 IDENTIFY: Examle 9 shows that for an isothermal rocess = nrtln( / ) Solve for SET UP: For a comression ( decreases) is negative, so = 8 J T = 9 K = ln nrt EXEUTE: (a) e e / nrt 069 = = (76 atm) = 088 atm = = 069 (00 mol)(8 J/mol K)(9 K) / nrt e = 8 J nrt (b) In the rocess the ressure increases and the volume decreases The -diagram is sketched in Figure 9 EALUATE: is the work done by the gas, so when the surroundings do work on the gas, is negative Figure 9 96 (a) IDENTIFY and SET UP: The -diagram is sketched in Figure 96 Figure 96 (b) alculate for each rocess, using the exression for that alies to the secific tye of rocess EXEUTE:, Δ = 0, so = 0 is constant; so = Δ = (00 0 Pa)(00 m 000 m ) = 00 0 J ( is negative since the volume decreases in the rocess) tot = + = 00 0 J EALUATE: The volume decreases so the total work done is negative 97 IDENTIFY: alculate for each ste using the aroriate exression for each tye of rocess SET UP: hen is constant, = Δ hen Δ = 0, = 0 = ( ), = 0, = ( ) and = 0 The total work done by the system is EXEUTE: (a) + + + = ( )( ), which is the area in the lane enclosed by the loo (b) For the rocess in reverse, the ressures are the same, but the volume changes are all the negatives of those found in art (a), so the total work is negative of the work found in art (a) EALUATE: hen Δ > 0, > 0and when Δ < 0, < 0 98 IDENTIFY: Aly Δ U = Q SET UP: For an ideal gas, U deends only on T

EXEUTE: (a) decreases and is negative (b) Since T is constant, Δ U = 0 and Q= Since is negative, Q is negative (c) Q=, the magnitudes are the same The First Law of Thermodynamics 9- EALUATE: Q < 0 means heat flows out of the gas The lunger does ositive work on the gas The energy added by the ositive work done on the gas leaves as heat flow out of the gas and the internal energy of the gas is constant 99 IDENTIFY: Δ U = Q For a constant ressure rocess, = Δ SET UP: EXEUTE: (a) Q =+ 0 J, since heat enters the gas = Δ = = (80 0 Pa)(00 m 00 m ) 78 0 J (b) Δ U = Q = 0 J 78 0 J = 77 0 J EALUATE: (c) = Δ for a constant ressure rocess and Δ U = Q both aly to any material The ideal gas law wasn t used and it doesn t matter if the gas is ideal or not 90 IDENTIFY: The tye of rocess is not secified e can use Δ U = Q because this alies to all rocesses alculate Δ U and then from it calculate Δ T SET UP: Q is ositive since heat goes into the gas; Q = + 00 J ositive since gas exands; =+ 00 J EXEUTE: Δ U = 00 J 00 J = 900 J e can also use ( ) Δ U = n R Δ T since this is true for any rocess for an ideal gas ΔU ( 900 J) Δ T = = = nr (00 mol)(8 J/mol K) T = T+Δ T = 7 = EALUATE: More energy leaves the gas in the exansion work than enters as heat The internal energy therefore decreases, and for an ideal gas this means the temerature decreases e didn t have to convert Δ T to kelvins since Δ T is the same on the Kelvin and elsius scales 9 IDENTIFY: Aly Δ U = Q to the air inside the ball SET UP: Since the volume decreases, is negative Since the comression is sudden, Q = 0 EXEUTE: Δ U = Q with Q = 0 gives Δ U = < 0so Δ U > 0 Δ U =+ 0 J (b) Since Δ U > 0, the temerature increases EALUATE: hen the air is comressed, work is done on the air by the force on the air The work done on the air increases its energy No energy leaves the gas as a flow of heat, so the internal energy increases 9 IDENTIFY and SET UP: alculate using the equation for a constant ressure rocess Then use Δ U = Q to calculate Q (a) EXEUTE: ( ) for this constant ressure rocess = d = = ( 0 Pa)(0 m 70 m ) = 0 J (The volume decreases in the rocess, so is negative) (b) Δ U = Q Q=Δ U + = + = Q negative means heat flows out of the gas (c) EALUATE: 0 0 J ( 0 J) 0 J ( ) (constant ressure) and U Q = d = Δ = aly to any system, not just to an ideal gas e did not use the ideal gas equation, either directly or indirectly, in any of the calculations, so the results are the same whether the gas is ideal or not 9 IDENTIFY: alculate the total food energy value for one doughnut K = mv SET UP: cal = 86 J EXEUTE: (a) The energy is (0 g)(0 kcal g) + (70 g)(0 kcal g) + (70 g)(90 kcal g) = 9 kcal The time required is (9 kcal) (0 kcal h) = 07 h = 6 min (b) v= K m = (9 0 cal)(86 J cal) (60 kg) = 9 m s = 0 km h EALUATE: hen we set K = Q, we must exress Q in J, so we can solve for v in m/s 9 IDENTIFY: Aly Δ U = Q SET UP: > 0 when the system does work

9- hater 9 EXEUTE: (a) The container is said to be well-insulated, so there is no heat transfer (b) Stirring requires work The stirring needs to be irregular so that the stirring mechanism moves against the water, not with the water (c) The work mentioned in art (b) is work done on the system, so < 0, and since no heat has been transferred, Δ U = > 0 EALUATE: The stirring adds energy to the liquid and this energy stays in the liquid as an increase in internal energy 9 IDENTIFY: Aly Δ U = Q to the gas SET UP: For the rocess, Δ = 0 Q = + 00 J EXEUTE: (a) Since Δ = 0, = 0 since heat goes into the gas nr (b) = nrt says = = constant Since doubles, T doubles Tb = Ta T (c) Since = 0, Δ U = Q=+ 00 J Ub = Ua + 00 J EALUATE: For an ideal gas, when T increases, U increases 96 IDENTIFY: Aly Δ U = Q is the area under the ath in the -lane SET UP: > 0 when increases EXEUTE: (a) The greatest work is done along the ath that bounds the largest area above the -axis in the - lane, which is ath The least work is done along ath (b) > 0 in all three cases; Q=Δ U +, so Q> 0 for all three, with the greatest Q for the greatest work, that along ath hen Q > 0, heat is absorbed EALUATE: ΔU is ath indeendent and deends only on the initial and final states and Q are ath indeendent and can have different values for different aths between the same initial and final states 97 IDENTIFY: Δ U = Q is the area under the ath in the -diagram hen the volume increases, > 0 SET UP: For a comlete cycle, Δ U = 0 EXEUTE: (a) and (b) The clockwise loo (I) encloses a larger area in the - lane than the counterclockwise loo (II) lockwise loos reresent ositive work and counterclockwise loos negative work, so I > 0 and II < 0 Over one comlete cycle, the net work I + II > 0, and the net work done by the system is ositive (c) For the comlete cycle, Δ U = 0 and so = Q From art (a), > 0, so Q > 0, and heat flows into the system (d) onsider each loo as beginning and ending at the intersection oint of the loos Around each loo, Δ U = 0, so Q= ; then, QI = I > 0 and QII = II < 0 Heat flows into the system for loo I and out of the system for loo II EALUATE: and Q are ath deendent and are in general not zero for a cycle 98 IDENTIFY and SET UP: Deduce information about Q and from the roblem statement and then aly the first law, Δ U = Q, to infer whether Q is ositive or negative EXEUTE: (a) For the water Δ T > 0, so by Q= mc Δ T heat has been added to the water Thus heat energy comes from the burning fuel-oxygen mixture, and Q for the system (fuel and oxygen) is negative (b) onstant volume imlies = 0 (c) The st law (Eq9) says Δ U = Q Q < 0, = 0 so by the st law Δ U < 0 The internal energy of the fuel-oxygen mixture decreased EALUATE: In this rocess internal energy from the fuel-oxygen mixture was transferred to the water, raising its temerature 99 IDENTIFY: Δ U = Q For a constant ressure rocess, = Δ SET UP: 6 Q =+ 0 0 J; Q > 0 since this amount of heat goes into the water EXEUTE: (a) = Δ = (0 0 Pa)(08 m 00 0 m ) = 67 0 J 6 6 (b) Δ U = Q = 0 0 J 67 0 J = 0 0 J = 00 atm = 0 0 Pa 6 EALUATE: 0 0 J of energy enters the water 67 0 J of energy leaves the materials through exansion work and the remainder stays in the material as an increase in internal energy 90 IDENTIFY: Δ U = Q SET UP: Q < 0 when heat leaves the gas EXEUTE: For an isothermal rocess, Δ U = 0, so = Q= J EALUATE: In a comression the volume decreases and < 0

The First Law of Thermodynamics 9-9 IDENTIFY: For a constant ressure rocess, = Δ, Q= nδt and Δ U = n Δ T Δ U = Q and = + R For an ideal gas, Δ = nrδ T SET UP: From Table 9, = 86 J/mol K EXEUTE: (a) The diagram is given in Figure 9 (b) = = nr( T T) = (00 mol)(8 J mol K)(000 K) = 08 J (c) The work is done on the iston (d) Since Eq (9) holds for any rocess, Δ U = n Δ T = (00 mol)(86 J mol K)(000 K) = 7 J (e) Either Q= n Δ T or Q=Δ U + gives Q= 90 J to three significant figures (f) The lower ressure would mean a corresondingly larger volume, and the net result would be that the work done would be the same as that found in art (b) EALUATE: = nrδ T, so, Q and ΔU all deend only on Δ T hen T increases at constant ressure, increases and > 0 ΔU and Q are also ositive when T increases Figure 9 9 IDENTIFY: For constant volume Q= n Δ T For constant ressure, Q= n Δ T For any rocess of an ideal gas, Δ U = n Δ T SET UP: R = 8 J/mol K For helium, = 7 J/mol K and = 078 J/mol K EXEUTE: (a) Q= n Δ T = (0000 mol)(7j mol K)(00 ) = 99 J The -diagram is sketched in Figure 9a (b) Q= n Δ T = (0000 mol)(078 J/mol K)(00 ) = 8 J The -diagram is sketched in Figure 9b (c) More heat is required for the constant ressure rocess ΔU is the same in both cases For constant volume = 0 and for constant ressure > 0 The additional heat energy required for constant ressure goes into exansion work (d) Δ U = n Δ T = 99 J for both rocesses ΔU is ath indeendent and for an ideal gas deends only on Δ T EALUATE: = + R, so > Figure 9 9 IDENTIFY: For constant volume, Q= n Δ T For constant ressure, Q= n Δ T SET UP: From Table 9, = 076 J/mol K and = 907 J/mol K Q EXEUTE: (a) Using Equation (9), Δ T = = 6 J = 679 K n (08 mol)(076 J mol K) The -diagram is sketched in Figure 9a and T = 98 K

9-6 hater 9 Q 6 J (b) Using Equation (9), Δ T = = = 99 K n (08 mol)(907 J mol K) and T = 900 K The -diagram is sketched in Figure 9b EALUATE: At constant ressure some of the heat energy added to the gas leaves the gas as exansion work and the internal energy change is less than if the same amount of heat energy is added at constant volume ΔT is roortional to Δ U Figure 9 9 IDENTIFY and SET UP: Use information about the ressure and volume in the ideal gas law to determine the sign of Δ T, and from that the sign of Q EXEUTE: For constant, Q= n Δ T Since the gas is ideal, = nrt and for constant, Δ = nrδ T Δ nr R Q= n = Δ Since the gas exands, Δ > 0 and therefore Q > 0 Q > 0 means heat goes into gas EALUATE: Heat flows into the gas, is ositive and the internal energy increases It must be that Q> 9 IDENTIFY: Δ U = Q For an ideal gas, Δ U = Δ T, and at constant ressure, = Δ = nrδ T SET UP: = R for a monatomic gas EXEUTE: Δ U = n( R) Δ T = Δ = Then Q= Δ U + =,so Q= EALUATE: For diatomic or olyatomic gases, is a different multile of R and the fraction of Q that is used for exansion work is different 96 IDENTIFY: For an ideal gas, Δ U = Δ T, and at constant ressure, Δ = nrδ T SET UP: = R for a monatomic gas EXEUTE: U n ( R ) T Δ = Δ = Δ = (00 0 Pa)(800 0 m 00 0 m ) = 60 J EALUATE: = nrδ T = Δ U = 0 J Q= n ( ) 600 J Δ T = n R Δ T = Δ U = 600 J of heat energy flows into the gas 0 J leaves as exansion work and 60 J remains in the gas as an increase in internal energy 97 IDENTIFY: For a constant volume rocess, Q= n Δ T For a constant ressure rocess, Q= nδ T For any rocess of an ideal gas, Δ U = n Δ T SET UP: From Table 9, for N, = 076 J/mol K and = 907 J/mol K Heat is added, so Q is ositive and Q =+ 7 J Q 7 J EXEUTE: (a) Δ T = = =+ 0 K n (00 mol)(076 J/mol K) Q 7 J (b) Δ T = = =+ 79 K n (00 mol)(907 J/mol K) (c) Δ U = n ΔT for either rocess, so ΔU is larger when ΔT is larger The final internal energy is larger for the constant volume rocess in (a) EALUATE: For constant volume = 0 and all the energy added as heat stays in the gas as internal energy For the constant ressure rocess the gas exands and > 0 Part of the energy added as heat leaves the gas as exansion work done by the gas

The First Law of Thermodynamics 9-7 98 IDENTIFY: Aly = nrt to calculate T For this constant ressure rocess, = Δ Q= n Δ T Use Δ U = Q to relate Q, and Δ U SET UP: 0 atm = 0 Pa For a monatomic ideal gas, = 7 J/mol K and = 078 J/mol K ( 0 Pa)(0 0 m ) EXEUTE: (a) T = = = K nr (00 mol)(8 J/mol K) T ( 0 Pa)(0 0 m ) = = = nr (00 mol)(8 J/mol K) 6 K (b) = Δ = ( 0 Pa)(0 0 m 0 0 m ) = 9 0 J (c) Q n T = Δ = = (d) Δ U = Q = 88 0 J (00 mol)(078 J/mol K)(6 K K) 87 0 J EALUATE: e could also calculate ΔU as Δ U = n Δ T = (00 mol)(7 J/mol K)(6 K K) = 90 0 J, which agrees with the value we calculated in art (d) 99 IDENTIFY: alculate and Δ U and then use the first law to calculate Q (a) SET UP: = d = nrt so = nrt / ( / ) / ln( / ) (work done during an isothermal rocess) = nrt d = nrt d = nrt EXEUTE: = (00 mol)(8 J/mol K)(0 K) ln(0 / ) = (6 J) ln(0) = 60 J EALUATE: for the gas is negative, since the volume decreases (b) EXEUTE: Δ U = n Δ T for any ideal gas rocess Δ T = 0 (isothermal) so Δ U = 0 EALUATE: Δ U = 0 for any ideal gas rocess in which T doesn t change (c) EXEUTE: Δ U = Q Δ U = 0 so Q= = 60 J (Q is negative; the gas liberates 60 J of heat to the surroundings) EALUATE: Q= n Δ T is only for a constant volume rocess so doesn t aly here Q= nδ T is only for a constant ressure rocess so doesn t aly here 90 IDENTIFY: = + R and γ = SET UP: R = 8 J/mol K EXEUTE: = + R R = + R= 78 J/mol K R 8 J/mol K = = = 6 J/mol K Then γ 07 EALUATE: The value of is about twice the values for the olyatomic gases in Table 9 A roane molecule has more atoms and hence more internal degrees of freedom than the olyatomic gases in the table 9 IDENTIFY: Δ U = Q Aly Q= nδt to calculate Aly Δ U = n ΔT to calculate γ = / SET UP: Δ T = 0 = 0 K Since heat is added, Q = + 970 J EXEUTE: (a) Δ U = Q =+ 970 J J = 77 J (b) Q 970 J = = = 70 J/mol K nδt (7 mol)(0 K) ΔU 77 J = = = 8 J/mol K nδt (7 mol)(0 K) 70 J/mol K γ = = = 0 8 J/mol K EALUATE: The value of γ we calculated is similar to the values given in Tables 9 for olyatomic gases 9 IDENTIFY and SET UP: For an ideal gas Δ U = n Δ T The sign of Δ U is the same as the sign of Δ T ombine Eq(9) and the ideal gas law to obtain an equation relating T and, and use it to determine the sign of Δ T EXEUTE: T = T and = nrt/ so, T γ γ = T and T = T ( / ) γ γ γ γ γ γ γ < and γ is ositive so T < T T Δ is negative so Δ U is negative; the energy of the gas decreases EALUATE: Eq(9) shows that the volume increases for this rocess, so it is an adiabatic exansion In an adiabatic exansion the temerature decreases

9-8 hater 9 9 IDENTIFY: For an adiabatic rocess of an ideal gas, γ γ, SET UP: For a monatomic ideal gas γ = / γ = = ( ) γ / 00800 m EXEUTE: (a) = = ( 0 0 Pa) = 76 0 Pa 0000 m (b) This result may be substituted into Eq(96), or, substituting the above form for, ( ( ) ) ( )( ) / 00800 / γ = = 0 0 Pa 00800 m = 06 0 J γ 0000 (c) From Eq(9), ( T T ) ( ) γ ( ) and T γ γ = T = = 00800 0000 = 9, and since the final temerature is higher than the initial temerature, the gas is heated EALUATE: In an adiabatic comression < 0 since Δ < 0 Q = 0so Δ U = Δ U > 0and the temerature increases 9 IDENTIFY and SET UP: (a) In the rocess the ressure increases and the volume decreases The -diagram is sketched in Figure 9 (b) For an adiabatic rocess for an ideal gas γ γ T = T γ γ, = and = nrt, Figure 9 EXEUTE: From the first equation, T = T( / ) γ = (9 K)( / 00900 ) 0 T = (9 K)() = 768 K = 9 (Note: In the equation T = imlies γ γ = = γ γ (00 atm)() 9 atm = T the temerature must be in kelvins) = ( / ) = (00 atm)( / 00900 ) γ EALUATE: Alternatively, we can use = nrt to calculate : n, R constant imlies / T = nr = constant so / T= / T = ( / )( T / T) = (00 atm)( / 00900 )(768 K/9 K) = 9 atm, which checks γ 9 IDENTIFY: For an adiabatic rocess of an ideal gas, = ( ) SET UP: γ = 0 for an ideal diatomic gas atm = 0 0 Pa and EXEUTE: Q U 0 and γ γ = L = 0 m γ =Δ + = for an adiabatic rocess, so Δ U = = ( ) ( ) = / γ = ( 0 Pa)() = 68 0 Pa = 0 Pa = ([68 0 Pa][0 0 m ] [ 0 Pa][0 0 m ]) = 0 0 J The internal energy 00 increases because work is done on the gas ( Δ U > 0) and Q = 0 The temerature increases because the internal energy has increased EALUATE: In an adiabatic comression < 0 since Δ < 0 Q = 0so Δ U = Δ U > 0and the temerature increases γ γ 96 IDENTIFY: Assume the exansion is adiabatic T = T relates and T Assume the air behaves as an ideal gas, so Δ U = n Δ T Use = nrt to calculate n SET UP: For air, = 976 J/mol K and γ = 0 = 0800 T = 9 K shere, = π r = 06 0 Pa For a

γ 00 The First Law of Thermodynamics 9-9 EXEUTE: (a) T = T = (9 K) = 0 K = 7 0800 π (b) π r (09 m) 7 0 = = = m (06 0 Pa)(7 0 m ) n = = = 09 mol RT (8 J/mol K)(9 K) Δ U = n Δ T = (09 mol)(076 J/mol K)( K 9 K) = J EALUATE: e could also use Δ U = = ( ) to calculate Δ U, if we first found from = nrt γ 97 (a) IDENTIFY and SET UP: In the exansion the ressure decreases and the volume increases The -diagram is sketched in Figure 97 Figure 97 (b) Adiabatic means Q = 0 Then Δ U = Q gives = Δ U = n Δ T = n( T T) (Eq9) = 7 J/mol K (Table 9) EXEUTE: = (00 mol)(7 J/mol K)(00 00 ) =+ J ositive for Δ > 0 (exansion) (c) Δ U = = J EALUATE: There is no heat energy inut The energy for doing the exansion work comes from the internal energy of the gas, which therefore decreases For an ideal gas, when T decreases, U decreases γ γ 98 IDENTIFY: = nrt For an adiabatic rocess, T = T SET UP: For an ideal monatomic gas, γ = / ( 00 0 Pa) (0 0 m ) EXEUTE: (a) T = = = 0 K nr (0 mol) (8 J mol K ) (b) (i) Isothermal: If the exansion is isothermal, the rocess occurs at constant temerature and the final temerature is the same as the initial temerature, namely 0 K = ( / ) = = 00 0 Pa (ii) Isobaric: Δ = 0 so = 00 0 Pa T = T( / ) = T = 60 K γ 067 T (0 K)( ) 067 T = = = (0 K) = 89 K γ 067 ( ) (iii) Adiabatic: Using Equation (9), ( ) EALUATE: In an isobaric exansion, T increases In an adiabatic exansion, T decreases γ γ 99 IDENTIFY: ombine T = T with = nrt to obtain an exression relating T and for an adiabatic rocess of an ideal gas SET UP: T = 99 K γ γ γ γ nrt nrt nrt T T EXEUTE: = so T = T and γ = γ ( γ )/ γ 0 / 080 0 Pa T = T = (99 K) 88 K 6 = = 0 0 Pa EALUATE: For an adiabatic rocess of an ideal gas, when the ressure decreases the temerature decreases 90 IDENTIFY: Aly Δ U = Q For any rocess of an ideal gas, Δ U = n Δ T For an isothermal exansion, = nrt = nrt ln ln SET UP: T = 88 K = = 00 EXEUTE: (a) Δ U = 0since Δ T = 0 (b) = (0 mol)(8 J/mol K)(88 K) ln(00) = 9 0 J > 0and work is done by the gas Since Δ U = 0, Q= =+ 9 0 J Q > 0so heat flows into the gas EALUATE: hen the volume increases, is ositive

9-0 hater 9 9 IDENTIFY and SET UP: For an ideal gas, = nrt The work done is the area under the ath in the -diagram EXEUTE: (a) The roduct increases and this indicates a temerature increase (b) The work is the area in the lane bounded by the blue line reresenting the rocess and the verticals at and The area of this traezoid is a b ( )( ) (0 0 Pa)(0000 m ) 800 J b + a b a = = EALUATE: The work done is the average ressure, ( ), + times the volume increase 9 IDENTIFY: Use = nrt to calculate T is the area under the rocess in the -diagram Use Δ U = n ΔT and Δ U = Q to calculate Q SET UP: In state c, EXEUTE: (a) c = 0 0 Pa and c = 0000 m In state a, c c (0 0 Pa)(0000 m ) Tc = = = 9 K nr (000 mol)(8 J/mol K) a = 0 0 Pa and (b) = (0 0 Pa + 0 0 Pa)(0000 m 0000 m ) + (0 0 Pa)(0000 m 0000 m ) =+ 00 J 00 J of work is done by the gas = a 0000 m a a (0 0 Pa)(0000 m ) (c) Ta = = = 9 K For the rocess, Δ T = 0, so Δ U = 0 and Q= =+ 00 J nr (000 mol)(8 J/mol K) 00 J of heat enters the system EALUATE: The work done by the gas is ositive since the volume increases 9 IDENTIFY: Use Δ U = Q and the fact that Δ U is ath indeendent > 0 when the volume increases, < 0 when the volume decreases, and = 0 when the volume is constant Q > 0 if heat flows into the system SET UP: The aths are sketched in Figure 9 Q =+ 900 J (ositive since heat flows in) acb =+ 600 J (ositive since Δ > 0) acb Figure 9 EXEUTE: (a) Δ U = Q Δ U is ath indeendent; Q and deend on the ath Δ U = U U b a This can be calculated for any ath from a to b, in articular for ath acb: Δ U = Q = 900 J 600 J = 00 J Now aly Δ U = Q to ath adb; Δ U = 00 J for this ath also =+ 0 J (ositive since Δ > 0) adb Δ U = Q so Q =Δ U + = 00 J + 0 J =+ 0 J a b adb adb acb a b adb (b) Aly Δ U = Q to ath ba: Δ Ub a = Qba ba = 0 J (negative since Δ < 0) ba Δ U = U U = ( U U ) = Δ U = 00 J b a a b b a a b Then Q =Δ U + = 00 J 0 J = 60 J ba b a ba ( Q ba < 0; the system liberates heat) (c) U a = 0, U d = 80 J Δ U = U U =+ 00 J, so U =+ 00 J a b b a rocess a d Δ U = Q a d ad ad Δ U = U U =+ 80 J a d d a b a b acb acb adb =+ 0 J and adb = ad + db But the work db for the rocess d b is zero since Δ = 0 for that rocess Therefore = =+ 0 J ad adb Then Q =Δ U + =+ 80 J + 0 J =+ 0 J (ositive imlies heat absorbed) ad a d ad

rocess d b Δ U = Q d b db db = 0, as already noted db Δ U = U U = 00 J 80 J =+ 0 J d b b d Then Q =Δ U + =+ 0 J (ositive; heat absorbed) db d b db EALUATE: The signs of our calculated these rocesses 9 IDENTIFY: Δ U = Q The First Law of Thermodynamics 9- Q ad and Q db agree with the roblem statement that heat is absorbed in SET UP: = 0 when Δ = 0 EXEUTE: For each rocess, Q=Δ U + No work is done in the rocesses ab and dc, and so = = 0 J and = = 0 J The heat flow for each rocess is: for ab, Q = 90 J For bc, Q = 0 J + 0 J = 890 J For ad adc ad, Q = 80 J + 0 J = 00 J For dc, Q = 0 J Heat is absorbed in each rocess Note that the arrows reresenting the rocesses all oint in the direction of increasing temerature (increasing U ) EALUATE: ΔU is ath indeendent so is the same for aths adc and abc Q = 00 J + 0 J = 60 J Q = 90 J + 890 J = 980 J Q and are ath deendent and are different for these two aths abc 9 IDENTIFY: Use = nrt to calculate Tc/ T a alculate ΔU and and use Δ U = Q to obtain Q SET UP: For ath ac, the work done is the area under the line reresenting the rocess in the -diagram Tc c c (0 0 J)(0060 m ) EXEUTE: (a) = = = 00 T c = Ta T (0 0 J)(000 m ) a a a (b) Since T = T, Δ U = 0for rocess abc For ab, Δ = 0 and = 0 For bc, is constant and bc c a = Δ = (0 0 Pa)(000 m ) = 0 0 J Therefore, Q= =+ 0 0 J 0 0 J of heat flows into the gas during rocess abc ab adc abc =+ 0 0 J Since Δ U = 0, (c) = (0 0 Pa + 0 0 Pa)(000 m ) =+ 80 0 J Q 80 0 J ac = ac =+ EALUATE: The work done is ath deendent and is greater for rocess ac than for rocess abc, even though the initial and final states are the same 96 IDENTIFY: For a cycle, Δ U = 0 and Q= alculate SET UP: The magnitude of the work done by the gas during the cycle equals the area enclosed by the cycle in the -diagram EXEUTE: (a) The cycle is sketched in Figure 96 (b) = (0 0 Pa 0 0 Pa)(00 m 0080 m ) =+ 0 J More negative work is done for cd than ositive work for ab and the net work is negative = 0 J (c) Q= = 0 J Since Q < 0, the net heat flow is out of the gas EALUATE: During each constant ressure rocess = Δ and during the constant volume rocess = 0 bc abc Figure 96 97 IDENTIFY: Use the st law to relate Q tot to tot for the cycle alculate ab and bc and use what we know about tot to deduce ca

9- hater 9 (a) SET UP: e aren t told whether the ressure increases or decreases in rocess bc The two ossibilities for the cycle are sketched in Figure 97 Figure 97 In cycle I, the total work is negative and in cycle II the total work is ositive For a cycle, Δ U = 0, so Qtot = tot The net heat flow for the cycle is out of the gas, so heat Q tot < 0 and tot < 0 Sketch I is correct (b) EXEUTE: tot = Qtot = 800 J tot = ab + bc + ca = 0 since Δ = 0 bc ab = Δ since is constant But since it is an ideal gas, Δ = nrδ T = nr( T T ) = 660 J ab b a ca = tot ab = 800 J 660 J = 60 J EALUATE: In rocess ca the volume decreases and the work is negative 98 IDENTIFY: Aly the aroriate exression for for each tye of rocess = nrt and = + R SET UP: R = 8 J/mol K EXEUTE: Path ac has constant ressure, so = Δ = nrδ T, and ac = ( c a) = ( mol)(8 J mol K)(9 K 00 K) = 789 0 J nr T T Path cb is adiabatic ( Q= 0), so = Q Δ U = Δ U = n Δ T, and using = R, cb ac cb = ( )( b c) = ( mol)(9 J mol K 8 J mol K)(600 K 9 K) = 67 0 J n R T T Path ba has constant volume, so ba = 0 So the total work done is EALUATE: 0 = ac + cb + ba = + = Δ > < 0when Δ < 0 and = 0 when Δ = 0 > when 0, 789 0 J 67 0 J 0 9 0 J 99 IDENTIFY: Use Q= n ΔT to calculate the temerature change in the constant volume rocess and use = nrt to calculate the temerature change in the constant ressure rocess The work done in the constant volume rocess is zero and the work done in the constant ressure rocess is = Δ Use Q= nδt to calculate the heat flow in the constant ressure rocess Δ U = n Δ T, or Δ U = Q SET UP: For N, = 076 J/mol K and = 907 J/mol K Q 0 J EXEUTE: (a) For rocess ab, Δ T = = = 9 K n (0 mol)(076 J/mol K) T = 9 K, so T = 86 K = nrt says T doubles when doubles and is constant, so T = (86 K) = 7 K = 899 (b) For rocess ab, ab = 0 For rocess bc, = Δ = Δ = = bc nr T (c) For rocess bc, (0 mol)(8 J/mol K)(7 K 86 K) 0 J = Δ = = Q n T c a = + = (0 mol)(907 J/mol K)(7 K 86 K) 6 0 J (d) Δ U = n Δ T = (0 mol)(076 J/mol K)(7 K 9 K) = 6 0 J EALUATE: The total Q is 0 J + 6 0 J = 78 0 J Δ U = Q = 78 0 J 0 J = 6 0 J, which agrees with our results in art (d) ab bc b 0 J 90 IDENTIFY: For a constant ressure rocess, Q= n Δ T Δ U = Q Δ U = n ΔT for any ideal gas rocess SET UP: For N, EXEUTE: (a) = 076 J/mol K and = 907 J/mol K Q < 0 if heat comes out of the gas Q + 0 J n = = = mol ΔT (907 J mol K)(00 K) (b) Δ U = n Δ T = Q( / ) = ( 0 J)(076/907) = 79 0 J

(c) = Q Δ U = 7 0 J (d) ΔU is the same for both rocesses, and if Δ = 0, = 0 and Q=Δ U = 79 0 J The First Law of Thermodynamics 9- EALUATE: For a given Δ T, Q is larger when the ressure is constant than when the volume is constant 9 IDENTIFY and SET UP: Use the first law to calculate and then use = Δ for the constant ressure rocess to calculate Δ EXEUTE: Δ U = Q Q = 0 J (negative since heat energy goes out of the system) Δ U = 0 so = Q= 0 J onstant ressure, so = d = ( ) = Δ 0 J Then Δ = = = 06 m 90 0 Pa EALUATE: Positive work is done on the system by its surroundings; this inuts to the system the energy that then leaves the system as heat Both Eq(9) and (9) aly to all rocesses for any system, not just to an ideal gas 9 IDENTIFY: = nrt For an isothermal rocess = nrtln( / ) For a constant ressure rocess, = Δ SET UP: L = 0 m EXEUTE: (a) The -diagram is sketched in Figure 9 00 0 Pa (b) At constant temerature, the roduct is constant, so = ( / ) = ( L) 600 L = The 0 0 Pa final ressure is given as being the same as = = The final volume is the same as the initial 0 Pa volume, so T = T( ) = 70 K (c) Treating the gas as ideal, the work done in the first rocess is = nrtln( ) = ln( ) 00 0 Pa = (00 0 Pa)( 0 m ) ln = 08 J 0 0 Pa For the second rocess, = ( ) = ( ) = ( ( )) 00 0 Pa = (0 0 Pa)( 0 m ) = J 0 0 Pa The total work done is 08 J J = 9 J (d) Heat at constant volume No work would be done by the gas or on the gas during this rocess EALUATE: hen the volume increases, > 0 hen the volume decreases, < 0 Figure 9 9 IDENTIFY: Δ = 0 βδ T = Δ since the force alied to the iston is constant Q= mc Δ T Δ U = Q SET UP: m= ρ EXEUTE: (a) The fractional change in volume is Δ = 0 βδ T = (0 0 m )(0 0 K )(00 K) = 0 m (b) = Δ = ( F A) Δ = ((00 0 N) (0000 m ))( 0 m ) = 68 J (c) Q= mc Δ T = 0 ρc Δ T = (0 0 m )(79 kg m )( 0 J kg K)(00 K) Q = 7 0 J

9- hater 9 (d) Δ U = Q = 7 0 J to three figures (e) Under these conditions is much less than Q and there is no substantial difference between c and c EALUATE: Δ U = Q is valid for any material For liquids the exansion work is much less than Q 9 IDENTIFY: Δ = β0 Δ T = Δ since the alied ressure (air ressure) is constant Q= mc Δ T Δ U = Q SET UP: For coer, β = 0 ( ), c = 90 J/kg K and ρ = 890 0 kg/m EXEUTE: (a) (b) = Δ = 88 0 J Δ = Δ = = 8 β T 0 ( 0 ( ) )(700 )(00 0 m) 86 0 m 6 (c) Q= mcδ T = ρ0 cδ T = (89 0 kg m )(800 0 m )(90 J kg K)(700 ) = 9 J (d) To three figures, Δ U = Q= 90 J (e) Under these conditions, the difference is not substantial, since is much less than Q EALUATE: Δ U = Q alies to any material For solids the exansion work is much less than Q 9 IDENTIFY and SET UP: The heat roduced from the reaction is Q = ml, where reaction reaction L reaction is the heat of reaction of the chemicals Q = +Δ U reaction sray EXEUTE: For a mass m of sray, = mv = m(9 m/s) = (80 J/kg) m and Δ Usray = Qsray = mcδ T = m(90 J/kg K)(00 0 ) = (, 00 J/kg) m Then Qreaction = (80 J/kg +, 00 J/kg) m= (,80 J/kg) m and Qreaction = mlreaction imlies mlreaction = (,80 J/kg) m The mass m divides out and L = 0 J/kg reaction EALUATE: The amount of energy converted to work is negligible for the two significant figures to which the answer should be exressed Almost all of the energy roduced in the reaction goes into heating the comound γ γ 96 IDENTIFY: The rocess is adiabatic Aly = and = nrt Q = 0so Δ U = = ( ) γ SET UP: For helium, γ = 67 = 00 atm = 0 0 Pa = 0900 atm = 97 0 Pa T = 88 K = 00 0 m EXEUTE: (a) / γ /67 γ = γ 00 atm = = (00 0 m ) = 0 m 0900 atm T T (b) = nrt gives = 0900 atm 0 m 00 atm 00 0 m T = T = (88 K) = 76 K = 0 7 (c) Δ U = ([0 0 Pa)(00 0 m )] [97 0 Pa)( 0 m )] = 0 J 067 EALUATE: The internal energy decreases when the temerature decreases γ γ 97 IDENTIFY: For an adiabatic rocess of an ideal gas, T = T = nrt 7 SET UP: For air, γ = 0 = EXEUTE: (a) As the air moves to lower altitude its density increases; under an adiabatic comression, the temerature rises If the wind is fast-moving, Q is not as likely to be significant, and modeling the rocess as adiabatic (no heat loss to the surroundings) is more accurate nrt γ γ γ γ γ γ (b) =, so T = T gives T = T The temerature at the higher ressure is ( ( ) )/ ( ) /7 T = T / γ γ = (8 K) [8 0 Pa]/[60 0 Pa] = 87 K = 0 so the temerature would rise by 9 EALUATE: In an adiabatic comression, Q = 0 but the temerature rises because of the work done on the gas

98 IDENTIFY: For constant ressure, γ γ = + R SET UP: γ = = = + EXEUTE: (a) The -diagram is sketched in Figure 98 The First Law of Thermodynamics 9- = Δ For an adiabatic rocess of an ideal gas, ( ) (b) The work done is = 0( 0 0) + ( 0( 0) ( 0)) = 0(0 0) γ and so R ( γ ) 0 0 = and R = + R Note that 0 is the absolute ressure (c) The most direct way to find the temerature is to find the ratio of the final ressure and volume to the original and treat the air as an ideal gas = =, since = Then γ γ = 0 = 0 = 0 = 0 γ ( ) T T T T T 0 0 0 0 (d) Since ( )( ) Q > 0 0 0 γ γ n=, Q= + R T0 T0 = 0 0 + This amount of heat flows into the gas, since RT RT R EALUATE: In the isobaric exansion the temerature doubles and in the adiabatic exansion the temerature 7 decreases If the gas is diatomic, with γ = γ = and T = 0 T0, = 0 0and Q= 0 0 0 Δ U = 9 Δ U > 0and this is consistent with an increase in temerature 0 0, Figure 98 99 IDENTIFY: Assume that the gas is ideal and that the rocess is adiabatic Aly Eqs(9) and (9) to relate ressure and volume and temerature and volume The distance the iston moves is related to the volume of the gas Use Eq(9) to calculate (a) SET UP: γ = / = ( + R)/ = + R/ = 0 The two ositions of the iston are shown in Figure 99 Figure 99 γ EXEUTE: adiabatic rocess: = h A = h A γ γ γ γ / γ /0 0 0 Pa h 0 Pa γ = 0 0 Pa = 0 0 Pa + = 0 Pa 6 air = ha = h A h = = (00 m) = 0077 m The iston has moved a distance h h = 00 m 0077 m = 07 m

9-6 hater 9 (b) T = T γ γ Th A = Th A γ γ γ γ γ 00 h 00 m T T h 0077 m = = 00 K = 797 K = 07 (c) = n ( T T) (Eq9) = (00 mol)(08 J/mol K)(00 K 797 K) = 77 0 J EALUATE: In an adiabatic comression of an ideal gas the temerature increases In any comression the work is negative 960 IDENTIFY: M γ γ m= ρ The density of air is given by ρ = For an adiabatic rocess, T = T RT = nrt nrt γ γ γ γ SET UP: Using = in T = T gives T = T EXEUTE: (a) The -diagram is sketched in Figure 960 (b) The final temerature is the same as the initial temerature, and the density is roortional to the absolute ressure The mass needed to fill the cylinder is then 0 Pa m= = ( kg/m )(7 0 m ) = 0 0 kg 6 0 air 0 0 Pa ithout the turbocharger or intercooler the mass of air at 0 707 0 kg T = 0 and = 0 0 Pa in a cylinder is m= ρ = The increase in ower is roortional to the increase in mass of air in the cylinder; the 0 0 kg ercentage increase is = 0 = % 707 0 kg ( γ )/ γ (c) The temerature after the adiabatic rocess is T = T The density becomes ( γ ) / γ / γ T ρ = ρ0 = ρ0 = ρ0 T The mass of air in the cylinder is 0 6 0 Pa m = ( kg/m )(7 0 m ) = 96 0 kg, 0 0 Pa 96 0 kg = 00 = 0% The ercentage increase in ower is 707 0 kg EALUATE: The turbocharger and intercooler each have an areciable effect on the engine ower Figure 960 96 IDENTIFY: In each case calculate either Δ U or Q for the secific tye of rocess and then aly the first law (a) SET UP: isothermal ( Δ T = 0) Δ U = Q ; = + 00 J For any rocess of an ideal gas, Δ U = n Δ T EXEUTE: Therefore, for an ideal gas, if Δ T = 0 then Δ U = 0 and Q= =+ 00 J (b) SET UP: adiabatic ( Q = 0) Δ U = Q ; =+ 00 J EXEUTE: 0 Q = says Δ U = = 00 J

(c) SET UP: isobaric Δ = 0 Use to calculate Δ T and then calculate Q EXEUTE: = Δ T = nrδ T; Δ T = / nr Q= nδ T and for a monatomic ideal gas = R The First Law of Thermodynamics 9-7 Thus Q= n RΔ T = ( Rn/)( / nr) = / =+ 70 J Δ U = n Δ T for any ideal gas rocess and = R= R Thus Δ U = / =+ 0 J EALUATE: 00 J of energy leaves the gas when it erforms exansion work In the isothermal rocess this energy is relaced by heat flow into the gas and the internal energy remains the same In the adiabatic rocess the energy used in doing the work decreases the internal energy In the isobaric rocess 70 J of heat energy enters the gas, 00 J leaves as the work done and 0 J remains in the gas as increased internal energy 96 IDENTIFY: = nrt For the isobaric rocess, = Δ = nrδ T For the isothermal rocess, f = nrtln i SET UP: R = 8 J/mol K EXEUTE: (a) The diagram for these rocesses is sketched in Figure 96 (b) Find T For rocess, n, R, and are constant so T T T = = constant = and nr T = T = ( K)() = 70 K (c) The maximum ressure is for state For rocess, n, R, and T are constant = and = = (0 0 Pa)() = 80 0 Pa (d) rocess : = Δ = nrδ T = (00 mol)(8 J/mol K)(70 K K) = 78 K rocess : = nrtln = (00 mol)(8 J/mol K)(70 K) ln = 0 J rocess : Δ = 0 and = 0 The total work done is 78 J + ( 0 J) = 8 J This is the work done by the gas The work done on the gas is 8 J EALUATE: The final ressure and volume are the same as the initial ressure and volume, so the final state is the same as the initial state For the cycle, Δ U = 0 and Q= = 8 J During the cycle, 8 J of heat energy must leave the gas Figure 96 96 IDENTIFY and SET UP: Use the ideal gas law, the first law and exressions for Q and for secific tyes of rocesses EXEUTE: (a) initial exansion (state state ) = 0 0 Pa, T = K, = 0 0 Pa, = = nrt; T/ = / nr= constant, so T/ = T/ and T = T( / ) = K( / ) = 70 K Δ = 0 so = Δ = nrδ T = (00 mol)(8 J/mol K)(70 K K) =+ 78 J Q= nδ T = (00 mol)(97 J/mol K)(70 K K) =+ 90 J Δ U = Q = 90 J 78 J = 80 J

9-8 hater 9 (b) At the beginning of the final cooling rocess (cooling at constant volume), T = 70 K The gas returns to its original volume and ressure, so also to its original temerature of K Δ = 0 so = 0 Q= n Δ T = (00 mol)(08 J/mol K)( K 70 K) = 80 J Δ U = Q = 80 J (c) For any ideal gas rocess Δ U = n Δ T For an isothermal rocess Δ T = 0, so Δ U = 0 EALUATE: The three rocesses return the gas to its initial state, so Δ U total = 0; our results agree with this 96 IDENTIFY: = nrt For an adiabatic rocess of an ideal gas, T γ γ = T SET UP: For N, γ = 0 EXEUTE: (a) The -diagram is sketched in Figure 96 (b) At constant ressure, halving the volume halves the Kelvin temerature, and the temerature at the beginning of the adiabatic exansion is 0 KThe volume doubles during the adiabatic exansion, and from Eq (9), the 00 temerature at the end of the exansion is (0 K)( ) = K (c) The minimum ressure occurs at the end of the adiabatic exansion (state ) During the final heating the volume is held constant, so the minimum ressure is roortional to the Kelvin temerature, min = (80 0 Pa)(K 00 K) = 68 0 Pa EALUATE: In the adiabatic exansion the temerature decreases Figure 96 96 IDENTIFY: Use the aroriate exressions for Q, and ΔU for each tye of rocess Δ U = Q can also be used SET UP: For N, = 076 J/mol K and = 907 J/mol K EXEUTE: (a) = Δ = nrδ T = (00 mol)(8 J mol K)( 0 K) = 87 J, Q= n Δ T = (00 mol)(907 mol K)( 0 K) = 6 J, Δ U = Q = 67 J (b) From Eq (9), using the exression for the temerature found in Problem 96, (00 mol)(8 J/mol K)(0 K)( (/ 00 = ) = J Q = 0 for an adiabatic rocess, and 00 Δ U = Q = = J (c) Δ = 0, so = 0 Using the temerature change as found in Problem 96 and art (b), Q= n Δ T = (00 mol)(076 J mol K)(00 K 7 K) = 80 J and Δ U = Q = Q= 80 J EALUATE: For each rocess we could also use Δ U = n ΔT to calculate Δ U 966 IDENTIFY: Use the aroriate exression for for each tye of rocess SET UP: For a monatomic ideal gas, γ = / and = R/ = nrt = nrt = EXEUTE: (a) ( ) (b) Q = 0 so = Δ U = n Δ T T = n T = / ( ( ) ) 0 J ln / ln () 9 0 J γ γ / = T gives T = T(/) Then (c) =, so = Δ = = nrt = 600 0 J (d) Each rocess is shown in Figure 966 The most work done is in the isobaric rocess, as the ressure is maintained at its original value The least work is done in the adiabatic rocess (e) The isobaric rocess involves the most work and the largest temerature increase, and so requires the most heat Adiabatic rocesses involve no heat transfer, and so the magnitude is zero (f) The isobaric rocess doubles the Kelvin temerature, and so has the largest change in internal energy The isothermal rocess necessarily involves no change in internal energy

The First Law of Thermodynamics 9-9 EALUATE: The work done is the area under the ath for the rocess in the -diagram Figure 966 shows that the work done is greatest in the isobaric rocess and least in the adiabatic rocess Figure 966 967 IDENTIFY: Assume the comression is adiabatic Aly T SET UP: For N, γ = 0 = 00 L, EXEUTE: (a) γ 00 = T and = nrt γ γ = 00 atm = 0 0 Pa, T = 7 K = / = 0 L = = = = = 0 T T (7 K) (7 K)() 60 K 87 = / T T T 60 K = = (00 atm) = 6 atm T / 7 K nr T 7 K (b) is constant, so = = constant and = = = (0 L) = L T T T T T 60 K EALUATE: In an adiabatic comression the temerature increases 968 IDENTIFY: At equilibrium the net uward force of the gas on the iston equals the weight of the iston hen the iston moves uward the gas exands, the ressure of the gas dros and there is a net downward force on the iston For simle harmonic motion the net force has the form F = ky, for a dislacement y from equilibrium, k and f = π m SET UP: = nrt T is constant (a) The difference between the ressure, inside and outside the cylinder, multilied by the area of the iston, must mg mg be the weight of the iston The ressure in the traed gas is 0 + = 0 + A π r (b) hen the iston is a distance h+ y above the cylinder, the ressure in the traed gas is mg h 0 + π r h+ y h y y and for values of y small comared to h, = + ~ h y h The net force, taking the ositive direction to + h mg y y be uward, is the then Fy = 0 + 0 ( πr ) mg = ( 0πr + mg ) π r h h This form shows that for ositive h, the net force is down; the traed gas is at a lower ressure than the equilibrium ressure, and so the net force tends to restore the iston to equilibrium 0π r + mg h g 0π r (c) The angular frequency of small oscillations would be given by ω = = + m h mg / 0π ω g r f = = + π π h mg y ( )

9-0 hater 9 If the dislacements are not small, the motion is not simle harmonic This can be seen be considering what haens if y ~ h; the gas is comressed to a very small volume, and the force due to the ressure of the gas would become unboundedly large for a finite dislacement, which is not characteristic of simle harmonic motion If y >> h (but not so large that the iston leaves the cylinder), the force due to the ressure of the gas becomes small, and the restoring force due to the atmoshere and the weight would tend toward a constant, and this is not characteristic of simle harmonic motion h EALUATE: The assumtion of small oscillations was made when was relaced by y/ h; this is h+ y accurate only when y/ his small 969 IDENTIFY: = d SET UP: For an isothermal rocess of an ideal gas, = nrt ( ) ln EXEUTE: (a) Solving for as a function of and T and integrating with resect to, nrt an nb = and = d nrt ln an nb = + nb hen a b 0, = nrtln, as exected = = ( ) (b) (i) Using the exression found in art (a), = ( 80 mol)( 8 J/mol K)( 00 K) ( 00 0 m ) ( 80 mol)( 68 0 m / mol) ln ( 00 0 m ) ( 80 mol)( 68 0 m / mol) + ( 0 J m mol )( 80 mol) 00 0 m 00 0 m = 80 0 J (ii) = nrtln() = 0 J (c) The work for the ideal gas is larger by about 00 J For this case, the difference due to nonzero a is more than that due to nonzero b The resence of a nonzero a indicates that the molecules are attracted to each other and so do not do as much work in the exansion EALUATE: The difference in the two results for is about 0%, which can be considered to be imortant