c b N/m 2 (0.120 m m 3 ), = J. W total = W a b + W b c 2.00

Transcription

1 Chter 19, exmle rolems: (19.06) A gs undergoes two roesses. First: onstnt m 3, isohori. Pressure inreses from P to P. Seond: Constnt P, isori. olume omressed from m 3 to m 3. () Show oth roesses in digrm: (10 5 P) () W = 0, W = Δ (for isori roesses) = N/m 2 (0.120 m m 3 ), = J. W totl = W + W 2.00 = J. (Note: Δ = finl initil.) (m 3 ) (19.08) Biyle tire um. Nozzle losed off. Slowly deress lunger until /2. Assume ir is n idel gs. Temerture T = onstnt. () Work ositive or negtive? olume redued while ressure inresed (t onstnt T). W = Û d is negtive. Tht is, to omress the gs isothermlly, work must e done to the gs. () Het flow ositive or negtive? Q = ΔU + W. Here ΔU = 0 sine T is not hnged. So Q = W is lso negtive. Tht is, for isotherml omression, het must flow out of the system (into therml reservoir, whih is so lrge tht it n sor quite n mount of het without rising its temerture reily). If het is not llowed to flow out (y using some therml insultion), then it is lled n diti omression. ΔU must then e ositive, so tht Q n e zero. The temerture of the system must therefore go u. The ressure of the system must lso rise more thn tht of the orresonding isotherml omression y the sme Δ. () Reltive mgnitudes of the het flow nd the work: They re equl, euse ΔU = 0 when T does not hnge. (For n idel gs, U deends on T only). (19.12) A gs in ylinder. onstnt ressure N/m 2. Cooled nd Comressed from 1.70 m 3 to 1.20 m 3. Internl energy dereses y J. () Work done W = N/m 2 (1.20 m m 3 ) = J. () Q = ΔU + W = ( J) + ( J) = J. Tht is, J of het must flow out of the system to the environment. [Note: For Q, flowing in is ositive. For W, oming out (doing work to the environment) is ositive.] () The gs does not hve to e idel, sine the idel gs lw hs not een used, nd we lso did not use the ft tht U is funtion of T lone.

2 (19.22) Cylinder ontins mol of helium t T = 27.0 C. () To rise the temerture to 67.0 C, keeing the volume onstnt, the het needed is Q = nc ΔT. C = (3/2) R for montomi helium gs. Therefore Q = mol (3/2) J/mol K 40 K = J. [Note: ΔT in K is the sme s ΔT in C.] () To rise the temerture to 67.0 C, keeing the ressure onstnt, the het needed is Q = nc ΔT. C = (5/2) R for montomi helium gs. Therefore Q = mol (5/2) J/mol K 40 K = J. () The differene J J = J is due to W = Δ (for isori roesses) = ( finl initil ) = nr(t finl T initil ) = mol J/mol K 40 K = J. More het is required in the se when the ressure is ket onstnt, sine then the volume will inrese s the temerture is rised, nd therefore the system will do work to the environment. (d) The gs is idel. The hnge in U is the sme s the Q t onstnt volume, nd is equl to J, euse W = 0 in this se. Beuse ΔU is th (roess) - indeendent, so the sme hnge of U lso hens in the se when the ressure is ket onstnt. This is euse ΔU = U finl U initil deends only on the initil nd finl temertures. (For n idel gs, U deends only on T.) If you lulte ΔU using the formul ΔU = Q W for the isori roess, you get the sme nswer, nmely J J = J. You n lso use the formul U = (3/2) RT to lulte ΔU for montomi gs, nd get the sme nswer. (19.38) Cylinder ontining mol of n idel montomi gs. Initilly, = P, nd = m 3. () Initil solute temerture T = /nr = N/m m 3 / mol J/mol K = K. () 2. (i) Isotherml. T finl = T initil = K. finl = initil / 2 = P. (ii) (iii) Isori. T finl = 2T initil = K. finl = initil = P. Aditi. initil initil = finl finl. = 5/3 for montomi gs. finl = initil ( initil / finl ) = P (1/2) 5/3 = P. T finl = /nr = P m 3 / mol J/mol K = K. (You n lso lulte T finl using the formul T initil -1 initil = T finl -1 finl, nd otin T finl = T initil ( initil / finl ) -1 = K (1/2) 2/3 = K.) (19.42) 0.5 mol of idel gs tken from stte to stte : () T finl = T = N/m m 3 / mol J/mol K = K. () W totl = W + W = N/m m N/m m 3 = 500 J. () T initil = T = N/m m 3 / (10 5 P) (m 3 )

3 0.5 mol J/mol K = K. Thus ΔU = 0 etween the stte nd the stte, nd therefore Q = W = 500 J for this omosite roess. This het Q is ositive, nd is therefore flowing into the system. (Note tht we n not lulte Q or Q here, sine we do not know wht tye of idel gs is involved here (montomi, or ditomi, or olytomi), nd the roess is not ny of the known roess. But we do know tht Q = Q + Q. If we were given tht the idel gs is montomi or ditomi, we would e le to lulte Q. Then Q would follow from Q Q. But we still nnot lulte Q diretly.) (19.48) n = 3 mol. Idel gs. Cyle : C = 29.1 J/mol K, : isori exnsion. : isohori redution of ressure. : diti omression. T = 300 K. T = 492 K. T = 600 K. W yle = W + W + W. W = 0. W = ( ) = nr (T T ) = 3 mol J/mol K (492 K 300 K) = 4789 J. W = (U U ) = n C (T T ) = = n (C R) (T T ) = 3 mol (29.1 J/mol K J/mol K) (600 K 492 K) = 6735J, (where we hve used Q = ΔU + W nd C = C + R). W yle = 4789 J 6735J = 1946 J. (Another wy to do W is to use the formuls = = for n diti roess, nd otin W = Û d = Û d / = ( ) / ( + 1) = ( ) / ( 1) = nr (T T ) / ( 1) = 3 mol J/mol K (492 K 600 K) / ( ) = 6734 J. (Tht we did not get 6735 J s ove is euse the given C = 29.1 J/mol K does not hve four-signifint-digits ury. Sine this seond roh does not need C it is tully more urte if n nd the temertures given re ext.) [Note tht we hve used = C / C = C / (C R).]

4 (19.54) Thermodynmi roess in solid. Cu ue, 2.00 m on side. Susended y string. (dt on Tle 14.1, 17.2, 17.3) Heted from 20.0 C to 90.0 C. = 1 tm = N/m 2. () Δ =? β = ( C) 1. Δ = β ΔT 0 = ( C) 1 (90.0 C 20.0 C) (0.02 m) 3 = m 3. () W = Δ (for isori roesses) = N/m m 3 = J. Q = m ΔT = (ρ 0 ) ΔT = [ kg/ m 3 (0.02 m) 3 ] 390 J/kg C 70 C = J. () ΔU = Q J (d) This result shows tht nd re rtilly the sme for solid euse the volume hnge under onstnt ressure s temerture is hnged is too smll to e imortnt (euse β is in generl very smll). Thus isohori nd isori roesses rtilly involve the sme Q if the temerture hnge is the sme. This is very different from the ehvior of gs, for whih C = C + R is true, whih is equivlent to = + R / M. Note tht in this reset the ehviors of liquid nd solid re similr. (19.62) System with iston ontining mol of O N/m 2 nd 355 K. Tret O 2 s idel gs. First exnds isorilly to 2 0 (from 0 ). It is then omressed isothermlly k to 0. Finlly it is ooled isohorilly k to 0. () Show ll roesses in digrm: Isori exnsion: olume inresed from 0 to 2 0, Temerture inresed. T > T (= 355 K). Isotherml omression: olume redued from 2 0 k to 0, Temerture stys onstnt. T = T > T. Isohori ooling: Temerture dros k to T = 355 K. () T = 2 T = 710 K. T = T =710 K. () mx = = 2 = N/m 2. or P. (d) W yle = W + W (W = 0.) W = 0 (2 0 0 ) = = nrt = mol J/mol K 355 K = J W = Û ( ) d = nrt Û (1/) d = nrt ln ( / ) = mol J/mol K 710 K ln (1/2) = J W yle = J J = J.

5 (19.64) System with iston ontining mol of N N/m 2 nd 300 K. Tret N 2 s idel gs. First omressed isorilly to (1/2) 0 (from 0 ). It then exnds ditilly to 0. Finlly, it is heted isohorilly to originl ressure 0. () Show ll roesses in digrm: Isori omression: olume droed from 0 to (1/2) 0, Temerture droed. Aditi exnsion: olume exnded from (1/2) 0 to 0, Temerture droed further, Isohori heting: Temerture rised k to T = 300 K. () onstnt. T = 150 K. T 1 = T 1. = 7/5 for ditomi gses. T = T ( / ) 1 (1/2) 0 0 = 150 K (1/2) 0.4 = K. () min =. But =. = ( / ) = N/m 2 (1/2) 7/5 = N/m 2 or P. Note: n lso e omuted y using = nrt, or /T = nr/ = nr/ = /T = N/m 2 / 300 K. = K ( N/m 2 / 300 K) = N/m 2. The slight differene in the fourth signifint digit is euse T is more urtely K. The error is one rt in out 6800! This ours in the fifth signifint digit in T. Whih nswer is more urte? The first one whih used =, so it did not use the lulted T. Of ourse, if one uses the mesured vlue for, insted of the theoretil vlue 7/5, it would e even more urte.