Tiêu đề 10. P1C10_আদর্শ গ্যাস ও গ্যাসের গতিতত্ত্ব_(With Solve).pdf ✅

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Engineering Practice Sheet................................................................................................. 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. P = 2 atm, V = 2 L, T = 30C n‡j Gi g‡a ̈ 21% O2 AYy we` ̈gvb| O2 AYyi msL ̈v KZ? [BUET 22-23] mgvavb: n = PV RT = 2 × 2 0.0821 × 303 = 0.16 mol  21% n = N NA  N = 21% × 0.16 × 6.023 × 1023  N = 2.023 × 1022 wU (Ans.) 2. GKwU wmwjÛv‡i iwÿZ Aw·‡Rb M ̈v‡mi AvqZb 1  10–2 m 3 , ZvcgvÎv 300 K Ges Pvc 2.5  105 Nm–2 | Pvc w ̄’i †i‡L wKQz Aw·‡Rb M ̈vm †ei K‡i †bIqv n‡j ZvcgvÎv †e‡o 400 K n‡jv| e ̈eüZ Aw·‡R‡bi fi KZ? [BUET 21-22] mgvavb: w = Mn = M     PV RT1 – PV RT2 = MPV R     1 T1 – 1 T2 = 32  2.5  105  1  10–2 8.314     1 300 – 1 400 g = 8.018 g (Ans.) 3. †Kv‡bv ̄’v‡bi evqyi ZvcgvÎv 26C Ges Av‡cwÿK Av`a©Zv 70%| hw` †m ̄’v‡bi ZvcgvÎv K‡g 18C nq, Z‡e evqyw ̄’Z Rjxqev‡®úi KZ kZvsk Nbxf‚Z n‡q Zij cvwb n‡e? [26C Ges 18C G m¤ú„3 Rjxqev‡®úi Pvc h_vμ‡g 25.21 mm Ges 15.48 mm cvi` ͇̄¤¢i mgvb|] [BUET 17-18] mgvavb: 26C ZvcgvÎvq, R = f F  70% = f 25.21  f = 17.647 mm (Hg)  Nbxf‚Z n‡e = 17.647 – 15.48 17.647  100% = 12.28% (Ans.) 4. KZ wWwMÖ †mjwmqvm ZvcgvÎvq Aw·‡Rb AYyi g~j Mo eM©‡eM –100C ZvcgvÎvi nvB‡Wav‡Rb AYyi g~j Mo eM©‡e‡Mi mgvb n‡e? [BUET 17-18, 00-01] mgvavb: c 2 1 = c 2 2  T1 M1 = T2 M2  T1 32 = – 100 + 273 2  T1 = 2768 K  T1 = 2495C (Ans.) 5. 0C ZvcgvÎv Ges 1.0  105 Nm–2 Pv‡c Kve©b WvB-A·vBW M ̈v‡mi NbZ¡ 1.98 kgm–3 | mgPv‡c 0C I 30C ZvcgvÎvq D3 M ̈vm AYyi g~j Mo eM©‡eM †ei Ki| [BUET 06-07] mgvavb: c1 = 3P 1 = 3  105 1.98 ms–1  c1 = 389.25 ms–1 Avevi, 1T1 = 2T2  2 = 1.98  273 303 kgm–3  2 = 1.784 kgm–3  c2 = 3P 2 = 3  105 1.784 ms–1  c2 = 410.075 ms–1 (Ans.) 6. GKRb e ̈w3 k¦vm-cÖk¦v‡m 1.12 litre evqy †meb Ki‡j (i) †m †gvU KZ ̧‡jv AYy †meb K‡i? (ii) 27C ZvcgvÎvq H AYy ̧‡jvi Mo MwZkw3 KZ? [mve©Rbxb M ̈vm aaæeK = 8.314 Jmole–1K –1 ] [BUET 02-03] mgvavb: (i) AYyi msL ̈v, N = V 22.4  NA  N = 1.12 22.4  6.023  1023  N = 3.0115  1022 wU (Ans.) (ii) GKwU AYyi Mo MwZkw3, E = 5 2 kT  E = 5 2  1.38  10–23  300 J/molecule  E = 1.035  10–20 J/molecule  N wU AYyi Mo MwZkw3 = N  1.035  10–20 = 3.0115  1022  1.035  10–20 = 311.69 J (Ans.)
2 .......................................................................................................................................  Physics 1st Paper Chapter-10 weMZ mv‡j KUET-G Avmv cÖkœvejx 1. w ̄’i Pv‡c KZ wWwMÖ †mjwmqvm ZvcgvÎvq †Kvb M ̈v‡mi AYyi g~j Mo eM©‡eM cÖgvY Pvc I ZvcgvÎvi g~j Mo eM©‡eM Gi A‡a©K n‡e? [KUET 19-20] mgvavb: c = 3RT M  c  T  T2 T1 =     c2 c1 2  T2 =       1 2 c1 c1 2  273 = 68.25 K  T2 = –204.75C (Ans.) 2. †Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU evqy ey`ey‡`i e ̈vm wØ ̧Y nq| n«‡`i c„‡ô evqygЇji Pvc ̄^vfvweK evqygÐjxq Pv‡ci mgvb Ges n«‡`i cvwbi DòZv aaæeK n‡j n«‡`i MfxiZv KZ? [RUET 15-16, 09-10; CUET 13-14; KUET 04-05] mgvavb: P1V1 = P2V2  P1 P2 = V2 V1 =     d2 d1 3 = 23 = 8  hg + Patm Patm = 8  h  103  9.8 + 101325 101325 = 8  h = 72.375 m (Ans.) 3. GKwU cyKz‡i cvwbi MfxiZv 6 m| evqygЇji ZvcgvÎv 27C Ges cvwbi g‡a ̈ Dnv cÖwZ wgUvi MfxiZvi Rb ̈ 0.5C K‡g| cvwbi Nb‡Z¡i cwieZ©b D‡cÿv K‡i cyKz‡ii Zj‡`‡k Drcbœ GKwU gvk© M ̈v‡mi ey`ey` Dnvi DcwiZ‡j †cu.Qv‡j AvqZ‡bi cwieZ©‡bi kZKiv nvi wbY©q K‡iv| [KUET 05-06; RUET 05-06] mgvavb: P1V1 T1 = P2V2 T2  V2 V1 = P1 P2  T2 T1  V2 V1 = hg + Patm Patm  300 300 – 0.5h  V2 V1 = 6  103  9.8 + 101325 101325  300 300 – 0.5h  V2 V1 = 1.596  V2 – V1 V1  100% = (1.596 – 1)  100% = 59.62% (e„w×) (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 1. 27C ZvcgvÎvi M ̈vm‡K KZ ZvcgvÎvq †bIqv n‡j Mo‡eM wØ ̧Y n‡e? [RUET 19-20] mgvavb: c –  T  T2 T1 =      c  – 2 c – 1 2  T2 =      2c  – 1 c – 1 2  300 K = 1200 K (Ans.) 2. hw` 0C DòZvi Ges 106 dyne cm–2 Pv‡c 1 gm nvB‡Wav‡Rb M ̈v‡mi AvqZb 11.2 litre nq Z‡e †gvjvi aaæeK R-Gi gvb KZ n‡e? [RUET 06-07] mgvavb: PV = w M RT  R = MPV wT = 2  106  10–5 10–4  11.2  10–3 1  273 Jmol–1K –1  R = 8.205 Jmol–1K –1 (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 1. GKwU wmwjÛv‡i iwÿZ Aw·‡Rb 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 †bqv nj| d‡j Pvc K‡g 1.3  105 Nm–2 nj| e ̈eüZ Aw·‡R‡bi fi wbY©q K‡iv| [CUET 13-14, 07-08] mgvavb: PV = nRT  n = PV RT  n1 – n2 = (P1 – P2)  V RT  n = (2.5  105 – 1.3  105 )  10–2 8.314  300  n = 0.4811 mol  w = M  n = 32  0.4811  w = 15.3952 g (Ans.) 2. GKwU 500 m3 AvqZ‡bi N‡ii evZv‡mi ZvcgvÎv 37C| Gqvi Kzjvi e ̈envi Kivi Rb ̈ evZv‡mi ZvcgvÎv K‡g 23C nj| hw` N‡ii evqyPvc mgvb _v‡K, Z‡e kZKiv KZfvM evZvm N‡ii g‡a ̈ Avm‡e/evwni n‡q hv‡e? [CUET 04-05] mgvavb: evqyPvc mgvb _vK‡j, V2 V1 = T2 T1  V2 V1 = 23 + 273 310 = 0.954  V1 – V2 V1  100% = (1 – 0.954)  100% = 4.6% (wfZ‡i Avm‡e) (Ans.)
Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Engineering Practice Sheet................................................................................................ 3 3. GKwU Aw·‡Rb wmwjÛv‡ii AvqZb 5  105 cm3 Ges G‡Z 300 evqygÐjxq Pv‡c Aw·‡Rb fwZ© Av‡Q| wKQzUv e ̈env‡ii ci †`Lv †Mj †h Pvc 100 evqygÐjxq Pv‡c †b‡g †M‡Q| †h cwigvY Aw·‡Rb e ̈eüZ n‡q‡Q Zvi AvqZb KZ? [CUET 03-04] mgvavb: P1V1 = P2V2  V2 = 300  5  105 100 = 1.5  106 cm3  Aw·‡Rb M ̈vm e ̈eüZ n‡q‡Q = 1.5  106 – 5  105 cm3 = 106 cm3 (Ans.) weMZ mv‡j BUTex-G Avmv cÖkœvejx 1. †evjR&g ̈vb aaæe‡Ki gvb KZ? [BUTex 10-11] mgvavb: †evjRg ̈vb aaæe‡Ki gvb, k = 1.38  10–23 Jmolecule–1K –1 (Ans.) 2. m¤ú„3 ev®ú e‡qj I Pvj©‡mi m~Î †g‡b P‡j Kx? [BUTex 10-11] mgvavb: m¤ú„3 ev®ú e‡qj I Pvj©m Gi m~Î †g‡b P‡j bv| (Ans.) 3. 29C ZvcgvÎvq 3 g bvB‡Uav‡R‡bi †gvU MwZkw3 wbY©q K‡iv| [bvB‡Uav‡R‡bi MÖvg AvYweK fi 28 g] [BUTex 09-10] mgvavb: †gvU MwZkw3, E = f 2 nRT  E = 5 2  3 28  8.314  302 J  E = 672.543 J (Ans.) 4. wbw`©ó †Kv‡bv w`‡b wkwkivsK 8.5C Ges evqyi ZvcgvÎv 18.4C| Av‡cwÿK Av`a©Zv wbY©q K‡iv| 8C, 9C, 18C I 19C ZvcgvÎvq me©vwaK ev®úPvc h_vμ‡g 8.04  10–3 m, 8.61  10–3 m, 15.46  10–3 m I 16.46  10–3 m cvi`| [BUTex 03-04] mgvavb: 8.5C ZvcgvÎvq m¤ú„3 ev®úPvc = 8.04 + 8.61 – 8.04 1  0.5 = 8.325 mm (Hg) 18.4C ZvcgvÎvq m¤ú„3 ev®úPvc = 15.46 + 16.46 – 15.46 1  0.4 = 15.86 mm (Hg)  Av‡cwÿK Av`a©Zv, R = 8.325 15.86  100%  R = 52.49% (Ans.) MCQ weMZ mv‡j BUET-G Avmv cÖkœvejx 1. n«‡`i Zj‡`‡k Ae ̄’vbiZ GKwU †ejy‡bi AvqZb f‚-c„‡ô 8 ̧Y nq| n«‡`i MfxiZv KZ? [evZv‡mi Pvc Pa = 1.013  105 Pa] [BUET Preli 22-23] 63 m 72 m 81 m 91 m DËi: 72 m e ̈vL ̈v: P1V1 = P2V2  Patm  8V2 = (Patm + hg)  V2  1.013  105  8 = (1.013  105 + h  103  9.8)  h = 72.36 m 2. AvqZvKvi iæ‡gi gv‡S 3 mole Aw·‡Rb M ̈vm 27 C ZvcgvÎvq Ae ̄’vb Ki‡Q| iæ‡gi gvÎv 5  3  2 [m GK‡K]| AvqZvKvi iægwU‡Z Aw·‡Rb M ̈v‡mi Pvc KZ? [BUET Preli 22-23] 250 Pa 350 Pa 450 Pa 150 Pa DËi: 250 Pa e ̈vL ̈v: V = (5 × 3 × 2) m3 = 30 m3 T = 300 K n = 3 mole P = nRT V = 249.42 Pa  250 Pa 3. SI Gi GK‡K R Gi gvb KZ? [BUET Preli 21-22; RUET 12-13] 8.314 JK–1mol 8.314 J–1K –1mol–1 8.314 JKmol–1 8.314 Jmol–1K –1 DËi: 8.314 Jmol–1K –1 4. 7 g N2 M ̈v‡mi 30C G MwZkw3 KZ? [BUET Preli 21-22] 944.68 J 954.86 J 944.96 J None DËi: 944.68 J e ̈vL ̈v: Ek = 3 2 nRT  Ek = 3 2  7 28  8.314  303 J  Ek = 944.68 J 5. †evëRg ̈v‡bi Mo gy3 c‡_i mgxKiY †KvbwU? [BUET Preli 21-22] 1 2n 2 3 4n 2 1 n 2 4 3n 2 DËi: 3 4n 2
4 .......................................................................................................................................  Physics 1st Paper Chapter-10 weMZ mv‡j CKRUET-G Avmv cÖkœvejx 1. GKwU Gqvi KwÛkbvi 40C ZvcgvÎv Ges 75% Av‡cwÿK Av`a©Zv wewkó evwn‡ii evqy †fZ‡i †U‡b 20C ZvcgvÎv Ges 54% Av‡cwÿK Av`a©Zvi evZv‡m cwiYZ K‡i| G‡ÿ‡Î cÖwZ NbwgUv‡i KZUv Rjxq ev®ú Nbxf‚Z n‡e? (40C I 20 C ZvcgvÎvq m¤ú„3 Rjxq ev‡®úi NbZ¡ h_vμ‡g 45 gm m–3 Ges 17 gm m–3 ) [CKRUET 23-24] 23.57 gm 25.57 gm 34.57 gm 24.57 gm 22.57 gm DËi: 24.57 gm e ̈vL ̈v: 75% = f 45  f = 33.75 gm m–3 Avevi, 54% = f 17  f = 9.18 gm m–3  Rjxqev®ú Nbxf‚Z n‡e 33.75 – 9.18 = 24.57 gm 2. GKwU Aw·‡Rb wmwjÛv‡ii AvqZb 10  105 cm3 Ges G‡Z 300 evqygÐjxq Pv‡c Aw·‡Rb fwZ© Av‡Q| wKQzUv e ̈env‡ii ci †`Lv †Mj †h, Pvc 200 evqygÐjxq Pv‡c †b‡g †M‡Q| e ̈enviK...Z Aw·‡Rb M ̈v‡mi AvqZb KZ? [CKRUET 22-23] 1000 L 500 L 1500 L 2000 L 12000 L DËi: 500 L e ̈vL ̈v: V2 = P1V1 P2 = 10  105  10–3  300 200 L = 1500 L  V = (1500 – 1000) L = 500 L 3. 27C ZvcgvÎvi M ̈vm‡K KZ ZvcgvÎvq †bIqv n‡j Mo‡eM wØ ̧Y n‡e? [CKRUET 21-22] 54C 273C 540C 1000 K 927C DËi: 927C e ̈vL ̈v: cavg  T  c2 c1 = T2 T1  T2 =     c2 c1 2  T1 = 4  (27 + 273) K = 1200 K = 927C 4. 30C ZvcgvÎvq 150 m3 AvqZ‡bi K‡ÿ GKwU cvwbi cvÎ ivLv Av‡Q| KZUzKz cvwb ev®ú nIqvi ci Aewkó cvwb I ev®ú mvg ̈ve ̄’vq _vK‡e? [30C ZvcgvÎvq m¤ú„3 ev®úPvc 31.83 mm Hg Pvc] [CKRUET 20-21] 1.13 kg 3.03 kg 4.32 kg 4.55 kg 4.55 gm DËi: 4.55 kg e ̈vL ̈v: cvwb ev®úxf‚Z n‡j ev‡®úi, Pvc, P = 31.83 mm (Hg) AvqZb, V = 150 m3 ZvcgvÎv, T = 303 K  PV = w M RT  w = MPV RT = 18  31.83 760  101325  150 8.314  303 gm = 4548.31 gm  ev‡®úi fi, w = 4.55 kg weMZ mv‡j KUET-G Avmv cÖkœvejx 1. †Kv‡bv n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j G‡j GKwU evqy ey`ey‡`i e ̈vm 4 ̧Y nq| n«‡`i c„‡ô evqygЇji Pvc ̄^vfvweK evqygЇji Pv‡ci mgvb Ges n«‡`i cvwbi DòZv aaæeK n‡j n«‡`i MfxiZv KZ? [n«‡`i c„‡ô evqyi Pvc = 101325 Pa] [KUET 18-19] 72.4 m 289.6 m 580 m 651.4 m 950 m DËi: 651.4 m e ̈vL ̈v: (Pa + Pw)  V1 = Pa  V2  (Pa + hg)  r 3 1 = Pa  r 3 2  h = 651.375 m  651.4 m Shortcut for these type of maths hw` c„‡ô ey`ey‡`i AvqZb Zj‡`‡ki Zzjbvq n ̧Y nq, h = (n – 1)  P g hw` c„‡ô ey`ey‡`i e ̈vm/e ̈vmva© Zj‡`‡ki Zzjbvq n ̧Y nq, h = (n 3 – 1)  P g [hLb ZvcgvÎv constant _vK‡e ZLbB cÖ‡hvR ̈|] 2. GKwU eÜ wmwjÛv‡i 10 gm Aw·‡Rb M ̈vm Av‡Q| 30C ZvcgvÎvq Kx cwigvY MwZkw3 jvf Ki‡e? [KUET 17-18] 1080.28 J 1108.28 J 1180.28 J 1100 J 1801.28 J DËi: 1180.28 J