Tiêu đề 10. P1C10_আদর্শ গ্যাস ও গ্যাসের গতিতত্ত্ব Merge_Ridoy 2.5.24.pdf ✅

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Engineering Question Bank 1 Av`k© M ̈vm I M ̈v‡mi MwZZË¡ Ideal Gas & Kinetic Theory of Gas `kg Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| †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.) 2| 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` I Dnvi DcwiZ‡j †cu.Qvi Ae ̄’vq 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 = hg + Patm Patm  300 300 – 0.5 h  V2 V1 = 6  103  9.8 + 101325 101325  300 300 – 0.5 h  V2 V1 = 1.596  V2 – V1 V1  100% = (1.596 – 1)  100% = 59.62% (e„w×) (Ans.) 3| 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.) 4| 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 cm 3  Aw·‡Rb M ̈vm e ̈eüZ n‡q‡Q = 1.5  106 – 5  105 = 106 cm 3 (Ans.) 5| 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.) 6| 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.) 7| 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.) 8| hw` 0C DòZvi Ges 106 dynecm–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?
2  Physics 1st Paper Chapter-10 [RUET 06-07] mgvavb: PV = W M RT  R = MPV WT = 2  106  10–5 10–4  11.2  10–3 1  273  R = 8.205 Jmol–1K –1 (Ans.) 9| 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.) 10| 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 =     c2 c1 2  T2 =     2c1 c1 2  300 = 1200 K (Ans.) 11| 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.) 12| 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 d1 = 3  105 1.98  c1 = 389.25 ms–1 Avevi, d1T1 = d2T2  d2 = 1.98  273 303  d2 = 1.784 kgm–3  c2 = 3P d2 = 3  105 1.784  c2 = 410.075 ms–1 (Ans.) 13| †evjR&g ̈vb aaæe‡Ki gvb KZ? [BUTex 10-11] mgvavb:†evjRg ̈vb aaæe‡Ki gvb, K = 1.38  10–23 Jmoecule–1K –1 (Ans.) 14| 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  E = 672.543 J (Ans.) 15| 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, W = V 22.4(L)  NA  N = 1.12 22.4  6.023  1023  N = 3.0115  1022 wU (ii) Mo MwZkw3, E = 5 2 kT  E = 5 2  1.38  10–23  300  E = 1.035  10–20 J/molecule (Ans.) 16| †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 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.) 17| 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| 18| 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  10–3m, 15.46  10–3 m I 16.46  10–3m 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
Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Engineering Question Bank 3 = 15.86 mm(Hg)  Av‡cwÿK Av`a©Zv, R = 8.325 15.86  100%  R = 52.49% (Ans.) weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. 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 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 2. 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  8 V2 = (Patm + hg)  V2  1.013  105  8 = (1.013  105 + h  103  9.8)  h = 72.36 m Shortcut: h = (n – 1)P g = (8 – 1)  1.013  105 103  9.8 = 72.36 m 3. One afternoon, Jamil and his friends ware gossiping beside a lake. All on a suden, Jamil noticed that a bubble from the bottom of the lake of trasparent water was coming out on the surface of the water. After coming to the surface, the bubble took a large size. Size of the bubble on the surace was 5 times, and atmos pheric pressure was 105 Nm–2 . [Density of water,  = 1000 kgm–3 ]. Calculate the depth of the lake. [IUT 21-22] 38.82 m 40.82 m 42.82 m 44.82 m DËi: 40.82 m e ̈vL ̈v: Pa , 5V1 (Pa + Pw), V1 (Pa = Pw).V1 = Pa .5V1  (Pa + hg) = 5 Pa  h = 5Pa – Pa g = 4a g = 4  105 103  9.8 = 40.82 m 4. Compared to the initial value, what is the resulting pressure for an ideal gas that is compressed isothermally to one-third of its initial volume? [IUT 20-21] Equal Three times larger Larger, but less than three times larger More than three times larger DËi: Three times larger e ̈vL ̈v: P1V1 = P2V2  P1V1 = P2  V1 3  P2 = 3P1 5. In a still water lake a fish emits a 2.0 m3 bubble at a depth of 1.5 m. What is the volume of the bubble when it reaches the surface of the lake? Atmos pheric pressure is 100 kPa. Assume that the temperature remains constant. [IUT 19-20] 4.94 mm3 4.74 mm3 4.34 mm3 4.54 mm3 DËi: Blank e ̈vL ̈v: Pa , V2 (Pa + Pw) (Pa + Pw).V1 = Pa .V2  (Pa + hg).V1 = Pa .V2  V2 = (100  103 + 1.5  103  9.8)  2 100  103  V2 = 2.294 m3 6. †Kv‡bv n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmv 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 = 651.4 m
4  Physics 1st Paper Chapter-10 Shortcut for these type of maths h = (n – 1)  P g [hw` c„‡ô ey`ey‡`i AvqZb Zj‡`‡ki Zzjbvq n ̧Y nq] h = (n 3 – 1)  P g [hw` c„‡ô ey`ey‡`i e ̈vm/e ̈vmva© Zj‡`‡ki Zzjbvq n ̧Y nq] hLb ZvcgvÎv constant _vK‡e ZLbB cÖ‡hvR ̈| 7. A bubble rises from the bottom of a lake of depth 80.0 m where the temperature is 4C. The water temperature at the surface is 18C. If the bubble’s initial diameter is 1.0 mm, what is the diameter when it reaches the surface? [IUT 16-17] 2.35 mm 2.10 mm 2.50 mm 2.40 mm DËi: 2.10 mm e ̈vL ̈v:(Pa + Pw) V1 T1 = Pa .V2 T2  (101325 + 103  80  9.8)  1 3 4 + 273 = 101325  d 3 2 18 + 273  d2 = 2.09378 mm 8. T ZvcgvÎvi GK wjUvi evqy‡K DËß Kiv n‡jv hZÿY bv evqyi Pvc I AvqZb DfqB wØ ̧Y nq| P‚ovšÍ ZvcgvÎv KZ? [BUTex 15-16] 2 T 4 T T 2 T 4 DËi: 4 T e ̈vL ̈v: P1V1 T1 = P2V2 T2  T2 = P2V2 P1V1  T1 = 2  2 T1 = 4 T1 9. GKwU 500 m3 AvqZ‡bi N‡ii evZv‡mi ZvcgvÎv 37C| Gqvi Kzjvi e ̈envi Kivi Rb ̈ evZv‡mi ZvcgvÎv K‡g 22C nj| hw` N‡i evqyPvc mgvb _v‡K, Z‡e kZKiv KZ fvM evZvm N‡ii g‡a ̈ Avm‡e/evwni n‡q hv‡e? [CUET 14-15] 4.84% 2.42% 24.2% None of them DËi: 4.84% e ̈vL ̈v: V1 T1 = V2 T2  V2 = 22 + 273 37 + 273  500 = 475.806 m3 V = 24.193 m3  V V = 24.193 500  100% = 4.8387%  †fZ‡i Avmv evZv‡mi cwigvY 4.8387%| 10. GKwU eo cv‡Îi AvqZb 480 m3 Ges ZvcgvÎv 293 K ZvcgvÎv 29815 G DbœxZ n‡j evqyi kZKiv KZ Ask †ei n‡q hv‡e? (Pvc AcwiewZ©Z Av‡Q) [CUET 14-15] 1.71% 48.71% 20.17% None of them DËi: 1.71% e ̈vL ̈v: V2 =     T2 T1  V1 =     298 293  480 m 3 = 488.1911 m3 V V = 488.1911 – 480 480  100% = 1.70647% 11. A tank of helium gas used to inflate a toy balloon is at 15.5  106 Pa pressure at 293 K. Its volume is 0.20 m3 . How large a balloon would it fill at 1.00 atmos phere at 323 K? [1 atmosphere = 101.3  103 Pa] [IUT 14-15] 3.70 m3 3.40 m3 4.70 m3 4.30 m3 DËi: Blank e ̈vL ̈v: P1V1 T1 = P2V2 T2  V2 = P1V1 P2  T2 T1 = 15.5  106  0.2  323 101.3  103  293 = 33.735 m3 12. A partially inflated balloon contains 500 m3 of Helium at 27C and 1 atm pressure. What is the volume of the Helium at an altitude of 3000 m, where the pressure is 0.5 atm and tempreture is –3C? [IUT 14-15] 900 m3 910 m3 790 m3 850 m3 DËi: 900 m3 e ̈vL ̈v: P1V1 T1 = P2V2 T2  V2 = P1V1 P2  T2 T1  V2 = 1  500  (273 – 3) (27 + 273)  0.5 = 900 m3 13. GKwU wbw`©ó f‡ii ï®.. evqyi 20C ZvcgvÎvq AvqZb 100 cc| hw` D3 ï®.. evqy‡K w ̄’i Pv‡c 50C ch©šÍ DËß Kiv nq, Z‡e AvqZb KZ n‡e? [KUET 13-14] 109 cc 115 cc 112 cc 110.2 cc 102 cc DËi: 110.2 cc e ̈vL ̈v: V2 = T2 T1  V1