Title 1. P2C1 Thermodynamics_With Solve.pdf ✅

ZvcMwZwe` ̈v  Mastery Practice Sheet 1 ZvcMwZwe` ̈v Thermodynamics cÖ_g Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| GKwU Kv‡b©v BwÄb hLb 27 C ZvcgvÎvq Zvc MÖvn‡K _v‡K ZLb Gi Kg©`ÿZv 50%| G‡K 60% Kg©`ÿ Ki‡Z n‡j Gi Dr‡mi ZvcgvÎv KZ evov‡Z n‡e? [BUET 21-22] mgvavb: 1 = 1 – T2 T1  0.5 = 1 – 300 T1  T1 = 600 K 2 = 1 – T2 T1  0.6 = 1– 300 T1  T1 = 750 K  T1 = 150 K (Ans.) 2| GKwU B‡jKwUaK †KUwji mvnv‡h ̈ 2 kg cvwbi ZvcgvÎv 25C †_‡K 80C G DbœxZ Ki‡j GbUawci cwieZ©b †ei K‡iv| [BUET 20-21] mgvavb: GbUawci cwieZ©b, s = ms ln T2 T1 = 2 × 4200 ln 80 + 273 25 + 273 = 1422.746392 JK–1 (Ans.) 3| mgvb f‡ii wZbwU wfbœ Zij c`v_© A, B, C Gi ZvcgvÎv h_vμ‡g 12, 19C Ges 28C| A †K hw` B Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv nq 16C| B †K hw` C Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv 23C| A †K hw` C Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv KZ n‡e? [BUET 20-21] mgvavb: QA = QB  msAA = msBB  sA × (16 – 12) = sB(19 – 16)  sA sB = 3 4  sA = 3 4 sB .... (i) Avevi, QB = QC  msBB = msCC  sB × (23 – 19) = sC × (28 – 23)  sB sC = 5 4  sC = 3 4 sB .... (ii) QA = QC  msAA = msCC  sA ( – 12) = sC (28 – )  sA sC = 28 –   – 12  3 4 sB 4 5 sB = 28 –   – 12  3 4 × 5 4 = 28 –   – 12   = 20.258 C (Ans.) 4| GKwU cÖZ ̈veZ©x BwÄb Zv‡ci 1/6th Ask‡K Kv‡R iƒcvšÍi K‡i| hLb Dr‡mi ZvcgvÎv wVK †i‡L MÖvn‡Ki ZvcgvÎv 62C Kgv‡bv nq, ZLb Bwćbi `ÿZv wØ ̧Y nq| MÖvnK Gi ZvcgvÎv wbY©q K‡iv| [BUET 19-20] mgvavb: W = Q × 1 6  W Q = 1 6 =   = 1 – T2 T1 = 1 6  T2 T1 = 5 6 ..... (i) Avevi, 2 = 2 6 = 1 – T2 – 62 T1  1 – 1 3 = T2 T1 – 62 T1  1 3 = 5 6 – 62 T1  T1 = 372 k  T2 = 310 k = 37C  MÖvn‡Ki ZvcgvÎv 37C (Ans.) 5| T1 Avw` ZvcgvÎvi GKwU Av`k© Avw` AvqZb 2 m3 | iæ×Zvcxq cÖmvi‡Yi d‡j Gi AvqZb 4 m3 nq Zvici m‡gvò cÖwμqvq cÖmvwiZ Kivq AvqZb 10 m3 nq, cieZ©x av‡c iæ×Zvcxq cÖwμqvq ms‡KvP‡bi d‡j Gi ZvcgvÎv cybivq T1 nq| Gi P‚ovšÍ AvqZb KZ? [BUET 18-19] mgvavb: 1g av‡c, T1V 1  – 1 = T2V 2  – 1  T1 T2 =     4 2  – 1 = 2( – 1) 2q av‡c, m‡gvò cÖmvi‡Y †h‡nZz ZvcgvÎvi cwieZ©b nq bv ZvB ZvcgvÎv؇qi cv_©K ̈ GKB _vK‡e|  V3 = 10 m3 3q av‡c, iæ×Zvcxq ms‡KvP‡b, T2V 3  – 1 = T1V 4  – 1  T 2 1  – 1 V3 = T 1 1  – 1 V4  V4 =     T2 T1 1  – 1 × V3
2  Physics 2nd Paper Chapter-01 =     1 2  – 1  – 1 × V3 = 1 2 × 10 = 5 m3 (Ans.) 6| GbUawc ej‡Z Kx eyS? 100C ZvcgvÎvi ev‡®ú cwiYZ Kiv nj| GbUawci e„w× †ei Ki| [BUET 17-18] mgvavb: GbUawc n‡”Q e ̄‘i wek„•LjZvi cwigvc Ges e ̄‘i †fŠZ ag© hv iæ×Zvcxq cÖwμqvq w ̄’i _v‡K| s = m1(v) T = 4 × 2.26 × 106 373 = 24.2359 × 103 Jk–1 (Ans.) 7| GKwU wmwjÛv‡i 300 K ZvcgvÎvq I 106 Pa Pv‡c 0.001 m3 M ̈vm Av‡Q| M ̈vmwU‡K cÖ_‡g m‡gvò cÖmviY Kiv nj Ges c‡i iæ×Zvcxq cÖwμqvq AveviI cÖmviY Kiv nj, cÖwZ †ÿ‡ÎB cÖmvi‡Yi AbycvZ 1 : 2| cÖmvi‡Y †gvU Kv‡Ri cwigvY †ei K‡iv|[BUET 17-18] mgvavb: 1g av‡c, W1 = nRT ln V2 V1 = P1V1 ln V2 V1 [∵ PV = nRT] = 106 × 0.001 ln2 = 693.147 J iæ×Zvcxq cÖmvi‡Y, T1V 1  – 1 = T2V 2  – 1  T2 =     1 2 1.4 – 1 × 300 = 227.357 k PV = nRT  n = PV RT = 106 × 0.001 8.314 × 300 = 0.4 mole W2 = nR 1 –  [T2 – T1] = 0.4 × 8.314 1 – 1.4 [227.357 – 300] = 603.95 J Wnet = W1 + W2 = 1297.100902 J (Ans.) 8| 10 g IR‡bi GKwU †jvnvi †c‡iK‡K wKQzÿY GKwU evb©vi wkLvq DËß Kiv nj| DËß †c‡iKwU‡K 10C ZvcgvÎvq 100 g cvwb‡Z Wzev‡bv nj| G‡Z cvwbi ZvcgvÎv e„w× †c‡q 20C nj| cvwb‡Z Wzev‡bvi c~‡e© †c‡i‡Ki ZvcgvÎv wbY©q K‡iv| [†jvnvi Av‡cwÿK ZvcgvÎv = 0.11kcal/kgC] [BUET 16-17] mgvavb: Final Temperature  n‡j, M„nxZ Zvc = ewR©Z Zvc  10 × 0.11( – 20) = 100 × 1 × (20 – 10)   = 929.09C (Ans.) 9| GK cigvYywewkó GKwU Av`k© M ̈vm‡K 17C ZvcgvÎvq nVvr Gi g~j Avq‡Z‡bi GK-`kgvsk AvqZ‡b msKzwPZ Kiv nj| ms‡KvP‡bi ci ZvcgvÎv KZ n‡e? [BUTET 16-17] mgvavb: T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =       1 1 10 1.67 – 1 × (17 + 273) = 1356.43191 k = 1083.432 C (Ans.) 10| 1.2 atm Pvc Ges 310 K ZvcgvÎvq †Kvb M ̈v‡mi AvqZb 4.3 L| iæ×Zvcxq cÖwμqvq M ̈vm‡K msKzwPZ K‡i AvqZb 0.76 L Kiv nj| M ̈vmwUi (K) P‚ovšÍ Pvc Ges (L) P‚ovšÍ ZvcgvÎv wbY©q Ki| [M ̈vmwU‡K Av`k© M ̈vm wnmv‡e we‡ePbv Kiv hvq hvi  = 1.4] [BUET 14-15] mgvavb: (K) P1V 1  = P2V2   P2 =     V1 V2  × P1 = 1.2 ×     4.3 0.76 1.4 = 13.5799 atm (L) T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =     4.3 0.76 1.4 – 1 × 310 = 620.0456 k (Ans.) 11| 0C ZvcgvÎvq 1 kg eid‡K 100C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z GbUawci e„w× wbY©q K‡iv| [BUET 13-14] mgvavb: s = s1 + s2 = ml(f) T + ms ln T2 T1 = 1 × 3.36 × 106 273 + 1 × 4200 ln 373 273 = 2541.617 Jk–1 (Ans.) 12| 10C ZvcgvÎvq 1 kg eid‡K 100C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z GbUawci e„w× wbY©q K‡iv| [BUTE 13-14] mgvavb: s = ms ln T2 T1 = 5 × 4200 ln 373 283 = 5798.76 Jk–1 (Ans.) 13| GKwU cvigvYweK †evgv we‡ùvwiZ n‡j m„ó Av ̧‡bi †Mvj‡Ki e ̈vmva© nq 100 m Ges Gi ZvcgvÎv 105 K| hw` †MvjKwU iƒ×Zvc c×wZ‡Z 1000 m e ̈vmva© ewa©Z nq Z‡e Gi m¤¢ve ̈ nq Z‡e Gi m¤¢ve ̈ ZvcgvÎv KZ n‡e?    Av‡cwÿK Zvc؇qi AbycvZ   CP CV = 1.66 [BUET 13-14,CUET 05-06] mgvavb: T1V 1  – 1 = T2V 2  – 1
ZvcMwZwe` ̈v  Mastery Practice Sheet 3  T2 =     V1 V2  – 1 × T1 =     r1 r2 3 – 3 × T1    V =  4 3 r 3  V  r 3 = 105 ×     100 1000 3 × 1.66 – 3 = 1047.128548 k (Ans.) 14| GKwU Kv‡b©v Pμ cÖv_wgK 327C ZvcgvÎvq KvR m¤úbœ K‡i| cÖwZwU av‡c ms‡KvPb ev cÖmvi‡Yi AbycvZ 1 : 6 n‡j Kv‡b©v P‡μi me©wb¤œ ZvcgvÎv Ges `ÿZv wbY©q K‡iv| [ = 1.4] [BUET 07-08] mgvavb: (P1,V1, T1) A B (P2,V2, T1) (P4,V4, T2) (P3,V3, T2) D V2 V1 = 6 ; V3 V2 = 6 BC  iæ×Zvcxq cÖmviY| myZivs T1V 2  – 1 = T2V 3  – 1  T2 T1 =     V2 V3  – 1 =     1 6 1.4 – 1 = 0.488   = 1 – T2 T1 = 51.2% (Ans.) 15| GK †ivMxi †`‡ni ZvcgvÎv GKwU ÎæwUc~Y© _v‡g©vwgUv‡ii mvnv‡h ̈ †g‡c 45C cvIqv †Mj| hw` GB _v‡g©vwgUv‡i eid we›`y Ges ev®úwe›`y h_vμ‡g 3C Ges 107C †Z cvIqv hvq, Zvn‡j †ivMxi †`‡ni cÖK...Z ZvcgvÎv dv‡ibnvBU † ̄‹‡j †ei K‡iv| [BUET 04-05] mgvavb: x – 3 107 – 3 = F – 32 212 – 32  45 – 3 104 = F – 32 180  F = 104.6923077C (Ans.) 16| GKwU Kv‡b©v Bwćbi `ÿZv 1 6 Zvc MÖvn‡Ki ZvcgvÎv 65C Kgv‡j `ÿZv 1 3 nq| Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv wbY©q K‡iv| [BUET 02-03] mgvavb: 1 = 1 – T2 T1 = 1 6  T2 T1 = 5 6 ..... (i) 2 = 1 – T2 – 65 T1 = 1 3  1 – T2 T1 + 65 T1 = 1 3  T1 = 390 k  T2 = 5 6 × 390 = 325 k (Ans.) 17| GKwU †gvUi Uvqvi‡K 15C ZvcgvÎvq 2 evqygÐjxq Pv‡c cv¤ú Kivq UvqviwU nVvr †d‡U †M‡j| Gi d‡j, ZvcgvÎv KZ K‡g hv‡e Zv †ei K‡iv|  = 1.4 [BUET 02-03] mgvavb: †d‡U hvIqvi c‡i Uvqv‡ii Pvc n‡e evqygÛjxq Pv‡ci mgvb|  P2 = 1 atm T1P1 1 –   = T2P2 1 –    T2 =     P1 P2 1 –   × T1 =     2 1 1 – 1.4 1.4 × (15 + 273) = 236.26 k = – 36.74C ZvcgvÎv n«vm = 15 – (–36.74)C = 51.74C (Ans.) 18| GKwU Av`k© Kv‡b©v Bwćbi Drm Ges wm‡éi ZvcgvÎv 450 K Ges 350 K| cÖwZ mvB‡K‡j BwÄbwU hw` Drm n‡Z 1 kcal Zvc MÖnY K‡i Zvn‡j (i) cÖwZ mvB‡K‡j wm‡é ewR©Z Zvc (ii) BwÄbwUi `ÿZv Ges (iii) cÖwZ mvB‡K‡j m¤úvw`Z Kv‡Ri cwigvY wbY©q K‡iv| [J = 4.184 kJ/kcal] [BUET 01-02] mgvavb: (i) †h‡nZz Kv‡b©v BwÄb ZvB, Q  T Q2 Q1 = T2 T1  Q2 = 350 450 × 1 kcal = 0.78 kcal (Ans.) (ii)  = 1 – T2 T1 = 1 – 350 450 = 0.222 = 22.2% (Ans.) (iii) cÖwZ P‡μ m¤úvw`Z KvR, W = Q1 – Q2 = (1 – 0.78) kcal = 0.22 kcal = 0.92048 kj (Ans.) 19| ̄^vfvweK ZvcgvÎv I Pv‡c wnwjqv‡gi GK wK‡jv‡gvj AYyi AvqZb 22.42 L| w ̄’i AvqZ‡bi wnwjqv‡gi Av‡cwÿK Zvc hw` 3.0 calg–1K –1 nq Z‡e w ̄’i Pv‡c wnwjqv‡gi Av‡cwÿK Zvc wbY©q K‡iv| (†`Iqv Av‡Q, cvi‡`i NbZ¡ = 13.6  103 kgm–3 | J = 4200 J/kcal) [BUET 01-02] mgvavb: GLv‡b, R = P0V0 nT0 = 0.76  13.6  103  9.8  22.42  10–3 1 × 273 Jmol–1K –1 = 101325  22.42  10–3 273 Jmol–1K –1
4  Physics 2nd Paper Chapter-01 = 8.31 J mol–1K –1  CP = CV + R = (3  4.2 + 8.31) J mol–1K –1 = 20.91 J mol–1K –1 (Ans.) 20| 40 wWwMÖ †mjwmqvm ZvcgvÎvq Ges 76 †m. wg. evqy Pv‡c wKQz cwigvY evqy‡K iæ×Zvcxq cÖwμqv AvqZb wØ ̧Y Kiv n‡j ZvcgvÎv KZ n‡e? Avevi hw` cwiewZ©Z ZvcgvÎv †_‡K 20 wWwMÖ †mjwmqvm e„w× Kiv nq, Z‡e Gi Pvc KZ n‡e? [ = 1.4] [KUET 19-20] mgvavb: T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =     1 2 1.4 – 1 × (40 + 273) = 237.2096 k Avevi, P1V1  = P2V2   P2 =     V1 V2  × P1 =     1 2 1.4 × 76 = 28.7986 cm ZvcgvÎv Av‡iv 20C ev 20 k evov‡j, T2  = 257.2096 k T1  P 1 1 –   = T2  P 2 1 –    P2 =      T1  T2   1 –  × P1 =     237.2096 257.2096 1.4 1 – 1.4 × 28.7986 = 38.2192 cm (Ans.) 21| ÒZvc I ZvcgvÎvÓ Gi msÁv `vI| [KUET 04-05] mgvavb: Zvc nj e ̄‘i Zvcxq kw3| ZvcgvÎv nj Zvcxq Ae ̄’v, hv wba©viY K‡i H e ̄‘ Ab ̈ †Kv‡bv e ̄‘i ms ̄ú‡k© Avm‡j Zvc MÖnY Ki‡e ev eR©b Ki‡e| (Ans.) 22| m‡gvò cÖwμqv Kx? †Kv‡bv wbw`©ó f‡ii M ̈v‡mi ZvcgvÎv 40C| iƒ×Zvc cÖwμqvq Gi AvqZb wØ ̧Y Ki‡j P‚ovšÍ ZvcgvÎv KZ n‡e? ( = 1.4) [KUET 04-05] mgvavb: †h cÖwμqvq wbw`©ó ZvcgvÎvq †Kv‡bv M ̈v‡mi Pvc cwieZ©‡bi d‡j AvqZb cwiewZ©Z nq, Zv‡K m‡gvò cwieZ©b e‡j| T1V 1  – 1 = T2V 2  – 1  T2 =     1 2 1.4 – 1 × (40 + 273) = 237.2 k = – 35.8C (Ans.) 23| GKwU wmwjÛv‡i g‡a ̈ 3atm Pv‡c Ges 300 K DòZvq 10 litres evqy Av‡Q| (a) Pvc hw` nVvr K‡i wØ ̧Y nq Zvn‡j evqyi AvqZb I DòZv KZ n‡e? (b) Pvc hw` Lye ax‡i wØ ̧Y Kiv nq Zvn‡j evqyi AvqZb I DòZv KZ n‡e? [KUET 03-04] mgvavb: (a) iæ×Zvcxq cÖwμqvq,  P1V1  = P2V2   V2 =     P1 P2 1  × V1 =     1 2 1 1.4 × 10 = 6.095 Letre T1P1 1 –   = T2P2 1 –    T2 =     P1 P2 1 – 1.4 1.4 × T1 =     1 2 0.4 1.4 × 300 = 365.704 k (Ans.) (b) ax‡i ax‡i cwieZ©b  m‡gvò cÖwμqv m‡gvò cwieZ©‡bi †ÿ‡Î, P1V1= P2V2  V2 = P1 P2 × V1 = 10 2 = 5 L †h‡nZz m‡gvò cwieZ©b ZvB ZvcgvÎvi †Kv‡bv cwieZ©b n‡e bv| (Ans.) 24| 100C ZvcgvÎvq 0.02 kg Rjxqev®ú Nbxf‚Z n‡q –10C ZvcgvÎvq cwiYZ Ki‡Z KZ Zvc eR©b Ki‡Z n‡e? ev‡®úi Nbxfe‡bi Av‡cwÿK myßZvc = 2268000 Jkg–1 , ei‡di Av‡cwÿK Zvc = 2100 Jkg–1K –1 Ges ei‡di Mj‡bi Av‡cwÿK myßZvc = 336000 Jkg1 | [RUET 19-20] mgvavb: 100C Rjxqev®ú‡K 100C cvwb‡Z cwiYZ Ki‡Z cÖ‡qvRbxq Zvc, Q1 = M1(v) = 0.02 × 2.268 × 106 = 45360 J 100C cvwb‡K 0C cvwb‡Z cwiYZ Ki‡Z cÖ‡qvRbxq Zvc, Q2 = ms = 0.02 × 4200 × (100 – 0) = 8400 J 0C cvwb‡K 0C ei‡di cwiYZ Ki‡Z cÖ‡qvRbxq Zvc, Q3 = m1(f) = 0.02 × 3.36 × 105 = 6720 J (Ans.) 0C eid‡K –10C ei‡d cwiYZ Ki‡Z cÖ‡qvRbxq Zvc, Q4 = ms = 0.02 × 2100 × (10 – 0) = 420 J (Ans.)  Qnet = Q1 + Q2 + Q3 + Q4 = 60900 J (Ans.) 25| Av`k© ZvcgvÎv I Pv‡c wbw`©ó AvqZ‡bi ïé M ̈vm‡K (i) m‡gvò Ae ̄’vq, Ges (iii) iæ×Zvc Ae ̄’vq wZb ̧Y AvqZ‡b cÖmvwiZ n‡Z †`qv nj| cÖwZ‡ÿ‡Î P‚ovšÍ Pvc KZ n‡e wbY©q K‡iv| ( = 1.4) [RUET 18-19] mgvavb: (i) P1V1 = P2V2  P2 = 101325 × 1 3 Pa = 33775 Pa (Ans.)