Tiêu đề ideal gas solve.pdf ✅

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  HSC Preparation 1 `kg Aa ̈vq Av`k© M ̈vm I M ̈v‡mi MwZZË¡ Ideal Gas and Kinetics of Gases ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| w ̄’i ZvcgvÎvq 1.5  105 Nm–2 Pv‡c wbw`©ó wKQz M ̈v‡mi AvqZb 0.003 m2 ; 9  105 Nm–2 Pv‡c M ̈vmwUi AvqZb KZ? mgvavb: P1V1 = P1V2  1.5  105  0.003 = 5  105  V2  V2 = 9  10–4 m 3 (Ans.) 2| w ̄’i ZvcgvÎvq 0.75 m (cvi`) Pv‡c 50  10–6 m 3 nvB‡Wav‡Rb M ̈vm ivLv Av‡Q| hw` Pvc e„w×i d‡j M ̈v‡mi AvqZb 38  10–6 m 3 nq Z‡e Pvc e„w×i cwigvY KZ? mgvavb: P1V1 = P1V2  hg  V1 = P2V2  0.76  13600  9.8  50  10–6 = P2  38  10–6  P2 = 133280 Pa  Pvc e„w× = 133280 – (0.76  9.8  13600) Pa = 31987.2 Pa (Ans.) 3| GKwU cv‡Î 0C ZvcgvÎvq wKQz M ̈vm iwÿZ Av‡Q| KZ ZvcgvÎvq M ̈v‡mi Pvc 0C ZvcgvÎvi Pv‡ci GK Z...Zxqvsl n‡e? (aiv hvK, cv‡Îi AvqZb AcwiewZ©Z _v‡K? mgvavb: P1 T1 = P2 T2  P 273 = P 3 T2  T2 = 91 k 4| GKwU d¬v‡· 30C ZvcgvÎv Ges 1 evqygÛjxq Pv‡c wKQz evZvm Ave× Av‡Q| GLb d¬v‡·i gyL wQwc w`‡q AvUwK‡q G‡K DËß Kiv ïiæ n‡jv| wQwcwU Lyj‡Z hw` 4 evqygÐjxq Pv‡ci cÖ‡qvRb nq Z‡e KZ ZvcgvÎv ch©šÍ G‡K DËß Ki‡j wQwcwU Ly‡j hv‡e? mgvavb: P1 T1 = P2 T2  1 (30 + 273) = 4 T2  T2 = 1212 k (Ans.) 5| wQwc AvUv †evZ‡j ̄^vfvweK Pv‡c 27C ZvcgvÎv wKQz M ̈vm Av‡Q| evZv‡mi ZvcgvÎv 50C-G DbœxZ Ki‡j M ̈v‡mi Pvc KZ n‡e? mgvavb: P1 T1 = P2 T2  101325 (27 + 273) = P2 (50 + 273)  P2 = 109093.25 Pa (Ans.) 6| w ̄’i Pv‡c KZ ZvcgvÎvq 0C ZvcgvÎvi M ̈v‡mi AvqZb wØ ̧Y n‡e? mgvavb: V1 T1 = V2 T2  v 273 = 2v T2  T2 = 546 k (Ans.) 7| w ̄’i Pv‡c 27C ZvcgvÎvq 100 m3 M ̈v‡mi ZvcgvÎv KZ n‡j AvqZb 150 m3 n‡e? mgvavb: V1 T1 = V2 T2  100 (27 + 273) = 150 T2  T2 = 450 k (Ans.) 8| 27C ZvcgvÎvq 0.76m cvi` Pv‡c GKwU M ̈v‡mi AvqZb 45 wjUvi| KZ Pv‡c 77C ZvcgvÎvq H M ̈v‡mi AvqZb 30 wjUvi n‡e? mgvavb: P1V1 T1 = P2V2 T2  0.76  45  10–3 (27 + 273) = P2  30  10–3 (77 + 273)  P2 = 1.33m (Hg) (Ans.) 9| 10C wWwMÖ ZvcgvÎvq 1 L evqy‡K Zvc †`Iqv n‡jv| KZ ZvcZvÎvq Zvi AvqZb I Pvc wØ ̧Y n‡e? mgvavb: P1V1 T1 = P2V2 T2  P  1 (10 + 273) = 2P  2  1 T2  T2 = 1132 k (Ans.) 10| GKwU wmwjÛv‡i iwÿZ M ̈v‡mi AvqZb 1  10–2 m 3 , ZvcgvÎv 300 K Ges Pvc 2.5  105 Nm–2 | ZvcgvÎv w ̄’i †i‡L wKQz Aw·‡Rb †ei K‡i †bIqvi ci Pvc K‡g 1.3  105 Nm–2 nq| e ̈eüZ Aw·‡R‡bi fi KZ? mgvavb: e ̈eüZ Aw·‡R‡bi †gvjmsL ̈v, n = n1 – n2 = P1V RT – P2V RT = 1  10–2 8.314  300  (2.5  105 – 1.3  105 ) = 0.48 mol  e ̈eüZ Aw·‡R‡bi fi = (0.48  32)g = 15.4 gm 11| 0C ZvcgvÎvq I GK evqygÛjxq Pv‡c 1  10–3 m 3 evqyi fi 1.3  10–3 kg| 120 evqygÛjxq Pv‡c Ges 30C ZvcgvÎv 1m3 evqyi fi †ei K‡iv? mgvavb: P1V1 T1 – P2V2 T2  1  1  10–3 273 = 120  v1 (30 + 273)
2  HSC Physics 1st Paper Chapter-10  v2 = 9.25  10 –6 m 3 GLv‡b, 9.25  10–6m 3 evqyi IRb 1.3  10–3 kg  1m3 evqyi IRb = 1.3  10–3 9.25  10–6 kg = 140.55 kg (Ans.) 12| 20C ZvcgvÎvq 1 wjUvi evqy‡Z Zvc †`Iqv n‡jv, †h ch©šÍ Zvi AvqZb I Pvc wØ ̧Y bv nq| evqyi ZvcgvÎv †ei K‡iv| mgvavb: P1V1 T1 – P2V2 T2  P  1 (20 + 273) = 2P  1  2 T2  T2 = 1172 k (Ans.) 13| ̄^vfvweK Pvc I ZvcgvÎvq evqyi NbZ¡ 1.29 kg m–3 30C ZvcgvÎvq Ges 75  10–2 m cvi` Pv‡c H evqyi NbZ¡ wbY©q K‡iv| mgvavb: 1T1 P1 – 2T2 P2  1.29  273 0.76 = 2  (30 + 273) 75  10–2  2 = 1.15 kgm–3 (Ans.) 14| 0C ZvcgvÎv I 0.76 m Pv‡ci evZv‡mi Nb‡Z¡i mv‡_ 27C ZvcgvÎv I 0.75m Pv‡ci Nb‡Z¡i Zzjbv K‡iv? mgvavb: 0C ZvcgvÎvq evZv‡mi NbZ¡ = 1 27C ZvcgvÎvq evZv‡mi NbZ¡ = 2  P1 1T1 = P2 2  T2  0.76 1  273 = 0.75 2  (27 + 273)  2 1 = 0.898  2 = 0.898 1 A_©vr, 2 < 1  1 †ewk = 1 – 0.898 1 1  100% = 10.197% 15| 0C ZvcgvÎvq I 1 evqygÛjxq Pv‡c 1 wjUvi evqyi fi 1.3  10–3 kg| KZ ZvcgvÎvq 130 evqygÛjxq Pv‡c 1000 wjUvi evqyi fi 140 kg n‡e? mgvavb: 140 kg fi nq 1000 L evqyi  1.3  10–3 fi nq = 1000  1.3  10–3 140 L evqy = 9.286  10–3L evqy GLb, P1V1 T1 – P2V2 T2  1  1 273 = 130  9.286  10–3 T2  T2 = 329.56 k (Ans.) 16| GKRb Wzeyix †Kvb weï× cvwbi n«‡` 20m Mfxi cvwb‡Z KvR Kivi mgq 0.000003 m3 AvqZ‡bi evZv‡mi ey`ey` Dc‡ii w`‡K cÖevwnZ nq| evqygÛ‡ji Pvc 98000 Nm–2 n‡j, cvwbi DcwiZ‡j †c.Qv‡j cÖwZwU ey`ey‡`i AvqZb KZ n‡e? (cvwbi ZvcgvÎv me©Î mgvb a‡iv)| mgvavb: P1V1 = P2V2  (hPg + Patm)  v1 = Patm  v2  (20  1000  9.8 + 98000)  0.000003 = 98000  v2  v2 = 9  10–6 m 3 (Ans.) 17| †Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU evqy ey`ey‡`i AvqZb 5 ̧Y nq| evqygÛ‡ji Pvc 105 Nm–2 n‡j n«‡`i MfxiZv KZ? mgvavb: P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (105 + h  1000  9.8)  v = 105  5v  h = 40.82 m (Ans.) 18| †Kv‡bv n«‡`i Zj‡`k †_‡K c„‡ô Avmvi d‡j GKwU evZv‡mi ey`ey‡`i e ̈vmva© wZb ̧Y n‡q hvq| e ̈v‡ivwgUv‡i cvi` ͇̄¤¢i D”PZv 75 cm n‡j n«‡`i MfxiZv KZ? (cvi` Gi NbZ¡ 13596 kg m–3 ) mgvavb: P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (0.75  13596  9.8 + 1000  9.8)  4 3 r1 3 = 0.75  13596  9.8  4 3  (3r1) 3  h = 265.12 m (Ans.) 19| Rjvk‡qi KZ MfxiZv GKwU ey`ey‡`i AvqZb DcwiZ‡j _vKvKvjxb AvqZb A‡cÿv A‡a©K n‡e? H mgq evqygÛ‡ji Pvc 760mm cvi` Ges cvi‡`i NbZ¡ 13.6  103 kgm–3 | H mgq evqygÛ‡ji Pvc 760mm cvi` Ges cvi‡`i NbZ¡ 13.6  103 kgm–3 | mgvavb: P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (760  10–3  13.6  103  9.8 + h  1000  9.8)  v2 2 = 760  10–3  13.6  103  9.8  v2  h = 10.336 m (Ans.) 20| cÖgvY ZvcgvÎv I Pv‡c 8g Aw·‡R‡bi AvqZb wbY©q Ki| mgvavb: Pv = nRT  101325  v = 8 32  8.314  273  v = 5.6  10–3 m 3 21| 30C ZvcgvÎvq Ges 2atm Pv‡c GKwU †ejy‡bi g‡a ̈ 24g Aw·‡Rb M ̈vm Av‡Q| GK †gvj Aw·‡R‡bi fi 32gm, Aciw`‡K †Kvb GKwU cyKz‡ii Dcwi‡`‡k evqygÛ‡ji Pvc 1.5 atm, cvwbi NbZ¡ 1050 kgm–3 I MfxiZv 20m Ges Ab ̈ GKwU cyKz‡ii Dcwi‡`‡k evqygÛ‡ji Pvc 1.2 atm cvwbi NbZ¡ 1000 kgm–3 I MfxiZv 25m| [1 atm = 1.013  105 Pa, R = 8.314 Jmol–1K –1 Ges g = 9.8 m– 2 ]
Av`k© M ̈vm I M ̈v‡mi MwZZË¡  HSC Preparation 3 mgvavb: (K) †ejybwUi M ̈v‡mi AvqZb wbY©q Ki| Pv = nRT  2  1.013 105  v = 24 32  8.314  (30 + 273)  v = 9.32  10–3 m 3 (L) †Kvb cyKz‡ii Zj‡`‡k M ̈vm fwZ© †ejy‡bi AvqZb Kg n‡e? MvwYwZKfv‡e we‡kølY Ki| 1g cyKz‡ii †ÿ‡Î Zj‡`‡k †ejy‡bi AvqZb = v1m 3 n‡j, P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (1.013  105  1.5  20  9.8  1000)  v1 = 1.5  1.013  105  v2  v2 = 2.2899 v1  v1 = 0.44 v2  v1 = v1 < v1 2q cyKz‡ii Zj‡`‡k †ejy‡bi AvqZb = v1 n‡j, P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (1.2  1.013  105 + 25  103  9.8) 1.2  1.013  105  v1 = 1.5  1.013  105  v2  v1 = 0.332 v2  2q cyKz‡ii Zj‡`‡k M ̈vm fwZ© wejy‡bi AvqZb Kg n‡e| 22| hw` R = 8.31 JK–1 mol–1 nq Z‡e 72 cm cvi` Pv‡c Ges 27C ZvcgvÎvq 20 g Aw·‡R‡bi AvqZb wbY©q K‡iv| mgvavb: Pv = nRT  PV =  m RT  0.72  13.6 103  9.8  v = 20 32  8.31  (27 + 273)  v = 0.0162 m3 23| n«‡`i Zj‡`k †_‡K c„ô‡`‡k Avmvq GKwU ey`ey‡`i e ̈vm wØ ̧Y nq| n«‡`i c„‡ô evqygÛjxq Pvc 100 Nm–2 n‡j, n«‡`i MfxiZv wbY©q K‡iv| mgvavb: P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (105  h  1000  9.8)  4 3      d1 2 3 = 105  4 3      2  d1 2 3  (105 + 9800h)  1 8 = 105  h = 71.43 m 24| w ̄’iPv‡c 4  10–3 m 3 AvqZ‡bi †Kvb M ̈vm‡K 0C †_‡K 68.25C ch©šÍ DËß Kivi d‡j Gi AvqZb 1  10–3 m 3 e„w× †c‡j cig k~b ̈ ZvcgvÎvi gvb KZ? mgvavb: V2 = V1 (1 + )  (4 10–3 +1 10–3 ) = 4  10–3 (1+   68.25)   = 1 273 C –1 GLb, cig k~b ̈ ZvcgvÎv = t C n‡j, 0 = V1 (1 + t  )  t = – 1  t = – 1  = –1 1 273 = – 273C (Ans.) 25| w ̄’i AvqZb 0C ZvcgvÎvq †Kvb M ̈v‡mi Pvc 3  105 Pa n‡j 60C ZvcgvÎvq Gi Pvc KZ n‡e? mgvavb: P1 T1 = P2 T2  3  105 273 = P2 (60 + 273)  P2 = 3.659  105 Pa (Ans.) 26| 0.64 m cvi` ̄ͤ¢ Pv‡c Ges 39C ZvcgvÎv †Kvb M ̈v‡mi AvqZb 5.7  10–4 m 3 cÖgvY Pvc I ZvcgvÎvq M ̈v‡mi AvqZb KZ? mgvavb: P1V1 T1 – P2V2 T2  0.64  5.7  10–4 (39 + 273) = 0.76  v 273  v = 4.2  10–4 m 3 (Ans.) 27| w ̄’i ZvcgvÎvq 7.5  104 N.m–2 Pv‡c wbw`©ó f‡ii M ̈v‡mi AvqZb 0.006m3 | 3  105 N.m–2 Pv‡c H M ̈v‡mi AvqZb KZ n‡e? mgvavb: P1V1 = P2V2  7.5  104  0.006 = 3  105  v2  V2 = 1.5  10–3 m 3 (Ans.) 28| 20m cvwbi wb‡P GKwU ey`ey‡`i AvqZb 0.00003m3 | evqygÛ‡ji Pvc n‡j cvwbi DcwiZ‡j ey`ey`wUi AvqZb KZ n‡Z? mgvavb: P1V1 = P2V2  (Patm + hg)  v1 = Patm  v2  (105  20  1000  9.8)  0.00003 = 105  v2  v2 = 8.88  10–5 m 3 (Ans.) 29| 27C ZvcgvÎvi wbw`©ó f‡ii M ̈v‡mi AvqZb 0.0003 m 3 | Pvc w ̄’i †i‡L ZvcgvÎv 127C Kiv n‡j Gi AvqZb KZ n‡e? mgvavb: V1 T1 = V2 T2  0.0003 (27 + 273) = v2 (127 + 273)  v2 = 4  10–4 m 3 (Ans.) 30| KZ ZvcgvÎvq †Kv‡bv M ̈v‡mi Pvc I AvqZb – 73C ZvcgvÎvq Pvc I AvqZ‡bi wZb ̧Y n‡e? mgvavb: P1V1 T1 – P2V2 T2  P1V1 (–73 + 273) = 3P1  3v1 T2  T2 = 1800 k (Ans.)
4  HSC Physics 1st Paper Chapter-10 31| w ̄’i Pv‡c 0.003 m3 AvqZ‡bi †Kv‡bv M ̈vm‡K 0C †_‡K 75C ZvcgvÎvq DËß Kiv n‡j Gi AvqZb 0.003825 m3 nq| w ̄’i Pv‡c M ̈v‡mi AvqZb cÖmviY mnM wbY©q Ki| mgvavb: V2 = V1 (1 +   )  0.003825 = 0.003 (1 +   75)   3.67  10–3 (C)–1 (Ans.) 32| cÖgvY ZvcgvÎv I Pv‡c 20g Aw·‡R‡bi AvqZb wbY©q Ki| mgvavb: PV =  m RT  101325  v = 20 32  8.314  273  v = 0.014 m3 (Ans.) 33| cÖgvY ZvcgvÎv I Pv‡c GKwU †ejyb 175kg fi enb Ki‡Z cv‡i| mvg ̈ve ̄’vb †_‡K DuPz †Kv‡bv ̄’v‡b †hLv‡b e ̈v‡iv‡gwUaK Pvc 50cm cvi` Pv‡ci mgvb Ges ZvcgvÎv – 10C †mLv‡b †ejybwU KZ fvi enb Ki‡Z cvi‡e? [AvqZb AcwiewZ©Z/aaæeK a‡i] Aw·‡R‡bi AvqZb wbY©q Ki| mgvavb: 1T1 P1 – 2T2 P2  m1 v  T1 P1 = m2 v  T2 P2  m2 = M1T1P2 P1T2  m2 = 175  273  50 76  (– 10 + 273)  m2 = 119.5 kg AvqZb = v L n‡j, v 22.4 = 119.5 32  10–3  v = 83650 L (Ans.) 34| 500cc Ges 100cc AvqZ‡bi `ywU KvPbj D‡cÿYxq AvqZ‡bi bj Øviv mshy3 Av‡Q| e ̈e ̄’vwU hLb evqy‡iva Kiv nj ZLb Af ̈šÍixY ZvcgvÎv 20C Ges Pvc 70cm Hg| hw` 100cc bj‡K 20C ZvcgvÎvq Ges 500cc bj‡K 100C ZvcgvÎvq DËß Kiv nq Z‡e e ̈e ̄’vwUi Af ̈šÍixY Pvc KZ? mgvavb: mshy3 Ae ̄’vq †gvjmsL ̈v = n n‡j, n = PV RT = 70  (500 + 100) R  (273 + 20)  n = 70  600 R  293 ......... (i) 1g b‡ji Rb ̈, n1 = P1V1 RT1  n1 = 100 P1 R  293 ......... (ii) 2q b‡ji Rb ̈, n2 = P1V1 RT1  n2 = P1  500 R  373 .......... (iii) GLb, n = n1 + n2  70  600 2  273 = 100P1 R  273 + 500P1 R  373  P1 = 85.23 cm (Hg) (Ans.) 35| mgAvqZ‡bi `ywU cv‡Î h_vμ‡g 48gm I 8gm O2 I He M ̈vm ivLv Av‡Q| cvÎ؇qi ZvcgvÎv h_vμ‡g 60C I 20C| GKwU D‡cÿYxq AvqZ‡bi bj Øviv cvÎØq‡K mshy3 Ki‡j mvg ̈ve ̄’vq mgMÖ e ̈e ̄’vwUi ZvcgvÎv †ei K‡iv| mgvavb: awi, e ̈e ̄’vwUi ZvcgvÎv = T n1c1 T1 = n2 c2 T2  48 32  5 2  R  (60 + 273 – T) = 8 5  3 2  R  {T – (20 + 273)}  15 4 (333 – T) = 3(T – 29)  1248.75 + 879 =     3 + 15 4  T  T = 315.22 k (Ans.) 36| MÖx®§Kv‡j 60C ZvcgvÎvq GKwU M ̈vmc~Y© wmwjÛv‡ii wc÷‡b m kg fvi Pvcv‡j Zv f~-c„‡ô mvg ̈ve ̄’vq _v‡K| evqygЇji KvR D‡cÿv K‡i kxZKv‡j –10C ZvcgvÎvq wmwjÛviwU‡K f~-c„ô †_‡K D”PZvq ̄’vcb Ki‡j Zvi AvqZb AcwiewZ©Z _vK‡e| mgvavb: awi, h D”PZvq ̄’vcb Ki‡Z n‡e| P1 T1 = P2 T2  mg1 A mg2 A = T1 T2  g1 g2 = (60 + 273) (–10 + 273)      R + b R 2 = 333 263  3 + h R = 1.125  h = 0.125  6400 km  h = 801.52 km (Ans.) 37| 6 wjUvi m‡e©v”P aviYÿgZv m¤úbœ GKwU †ejyb 20C ZvcgvÎvq wbw`©ó cwigvY M ̈vm Øviv c~Y© Kiv nq ZLb Gi Pvc 3atm| cieZ©x‡Z †ejybwU‡K 15C ZvcgvÎvi †Lvjv gv‡V Iov‡Z †M‡j †d‡U hvq| (K) DÏxc‡Ki †ejybwUi †kl Pvc wbY©q Ki| mgvavb: P1 T1 = P2 T2  3 (20 + 27 + 3) = P2 (35 + 273)  P2 = 3.15 atm (Ans.) (L) †ejybwU‡Z 5 wjUvi M ̈vm Øviv c~Y© Kiv n‡j dvU‡e wKbv? MvwYwZKfv‡e we‡kølY Ki| mgvavb: ÔKÕ n‡Z P2 = 3.15 atm P1V1 T1 – P2V2 T2  3  5 (20 + 273) = 3.15  v2 (35 + 273)  v2 = 5.006 L (Ans.) †h‡nZz, v2 < 6 A_©vr †ejybwU dvU‡e bv|