Tiêu đề আদর্শ গ্যাস(Ideal Gas)_ With Solve.pdf ✅

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Varsity Practice Content 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 wjwLZ cÖkœmg~n 1| †Kv‡bv e× cv‡Î iwÿZ Av`k© M ̈v‡mi ZvcgvÎv 1C e„wׇZ Pvc 0.5% e„w× cvq| P‚ovšÍ ZvcgvÎv wbY©q K‡iv| [Medium] mgvavb: P2 = (1 + 0.005)P1 = 1.005 P1 T2 = T1 + 1  P1 T1 = P2 T2  P1 T1 = 1.005P1 T1 + 1  T1 + 1 = 1.005 T1  0.005 T1 = 1  T1 = 200 K  P‚ovšÍ ZvcgvÎv = 201 K= –72C (Ans.) 2| 27C ZvcgvÎvq 0.02 m 3 AvqZ‡bi cv‡Î 28 g wbqb I AvM©b M ̈vm wgkÖ‡Yi Pvc 1  105 Nm–2 | wbqb I AvM©‡bi MÖvg AvYweK fi h_vμ‡g 20 g/mol I 40 g/ml n‡j M ̈vm؇qi cwigvY wbY©q Ki| [Medium] mgvavb: awi wbqb M ̈vm Av‡Q xg  AvM©b M ̈vm Av‡Q (28 – x) g wbqb M ̈v‡mi Rb ̈ Pvc, P1 = n1RT V = xRT 20V AvM©b M ̈v‡mi Rb ̈ Pvc, P2 = n2RT V = (28 – x) RT 40V Wvë‡bi AvswkK Pvc m~Îvbymv‡i, P = P1 + P2  P = xRT 20V + (28 – x) RT 40V  PV RT = x 20 + 28 – x 40  1  105 0.02 8.314  300 = x + 28 40  0.8  40 = x + 28  x = 32 – 28 = 4 g  wgkÖ‡Y 4 g wbqb M ̈vm I 24 g AvM©b M ̈vm Av‡Q| (Ans.) 3| 0C ZvcgvÎv I 0.76 atm Pv‡ci evZv‡mi Nb‡Z¡i mv‡_ 27C ZvcgvÎv I 0.75 atm Pv‡c Nb‡Z¡i Zzjbv Ki| [Easy] mgvavb: P1 1T1 = P2 2T2  0.76 1  273 = 0.75 2  (27 + 273)  1 2 = 0.76  300 0.75  273  1 2 = 304 273  1 : 2 = 304 : 273 (Ans.) 4| 3  10–4 m 3 evqy‡K 0C n‡Z 50C ZvcgvÎvq DbœxZ Ki‡j †`Lv hvq †h, evqyi AvqZb 3.54  10–4 m 3 n‡q‡Q| evqyi AvqZb cÖmviY ̧Yv1⁄4 KZ? [Medium] mgvavb: V = V0 + V0   = V – V0 V0 = 3.54  10–4 – 3  10–4 3  10–4  50 = 0.54  10–4 3  10–4  50 = 9 2500 = 3.6  10–3 C –1 (Ans.) 5| 760 mm Hg Pv‡c wKQz gv‡e©jmn GKwU M ̈v‡mi AvqZb 100 Nb †mw›UwgUvi| Pvc 1000 mm Hg ch©šÍ evov‡j †gvU AvqZb 80 Nb‡mw›UwgUvi nq| gv‡e©‡ji AvqZb KZ? [Easy] mgvavb: awi, gv‡e©‡ji AvqZb x cm3  P1V1 =P2V2  760(100 – x) = 1000(80 – x)  0.76(100 – x) = 80 – x  76 – 0.76x = 80 – x  4 = 0.24x  x = 50 3 cm3 = 16.67 cm3 (Ans.) 6| 0C ZvcgvÎvq I GK evqygÐjxq Pv‡c 1 L evqyi fi 1.3 g| 120 evqygÐjxq Pv‡c Ges 30C ZvcgvÎvq 1 m 3 evqyi fi KZ? [Medium] mgvavb: 1 = m V1 = 1.3  10–3 1  10–3 kgm–3 = 1.3 kgm–3 GLb, P1 1T1 = P2 2T2  2 = P2 P1  T1 T2  1 = 120  273 1  303  1.3 = 140.55 kgm–3  m2 = 2V2 = 140.55  1 kg = 140.55 kg (Ans.)
2  Physics 1st Paper Chapter-10 7| 30C ZvcgvÎvi 20 g He M ̈v‡mi ZvcgvÎv 60C G e„w× Ki‡Z KZUzKz AwZwi3 kw3 mieivn Ki‡Z n‡e? [Easy] mgvavb: cÖ‡qvRbxq kw3 = EK = 3 2 nRT = 3 2  20 4  8.314  (273 + 60 – 273 – 30) = 3 2  5  8.314  30 = 225  8.314 = 1870.65 J (Ans.) 8| GKwU cv‡Îi AvqZb 360 m3 Ges ZvcgvÎv 25C| ZvcgvÎv 30C G DbœxZ n‡j evqyi kZKiv KZ Ask †ei n‡q hv‡e? [Easy] mgvavb: GLv‡b, n1RT1 = n2RT2  n2 n1 = T1 T2  n1 – n2 n1 = T2 – T1 T2  n1 – n2 n1  100% = 5 300  100% = 1.67% (Ans.) 9| M ̈v‡mi MwZm~Î n‡Z cÖgvY K‡iv, 1 mol Av`k© M ̈v‡mi MwZkw3 E = 3 2 RT [Hard] mgvavb: PV = 1 3 Mc 2 Avevi, PV = RT [n = 1 mol]  1 3 Mc2 = RT  2 3  1 2 Mc 2 = RT  E = 3 2 RT (Proved) 10| M ̈v‡mi MwZZ‡Ë¡i mvnv‡h ̈ Pvj©m Gi m~Î cÖwZcv`b K‡iv| [Hard] mgvavb: 1 †gvj M ̈v‡mi Rb ̈, PV = RT Ges PV = 1 3 mNc 2 ......... (i)  RT = 1 3 mNc 2  mc2 = 3  R N  T = 3KT (i) n‡Z, PV = 1 3  3KT  N  PV = NKT  Pvc w ̄’i _vK‡j, AvqZb (V)  ZvcgvÎv (T)  V  T 11| M ̈v‡mi MwZZ‡Ë¡i mvnv‡h ̈ e‡qj Gi m~Î cÖwZcv`b K‡iv| [Hard] mgvavb: 1 †gvj M ̈v‡mi Rb ̈, PV = RT Ges PV = 1 3 mNc 2 .......... (i)  RT = 1 3 mNc 2  mc2 = 3  R N  T = 3KT (i) n‡Z, PV = 1 3  3KT  N  PV = NKT  w ̄’i ZvcgvÎvq NKT = aaæeK  PV = aaæeK  P  1 V (Proved) 12| M ̈v‡mi MwZZ‡Ë¡i mvnv‡h ̈ Pvc Gi m~Î cÖwZcv`b K‡iv| [Hard] mgvavb: 1 †gvj M ̈v‡mi Rb ̈, PV = RT Ges PV = 1 3 mNc 2 ........... (i)  RT = 1 3 mNc 2  mc2 = 3 R N T = 3KT (i) n‡Z, PV= 1 3  3KT  N  PV = NKT  AvqZb w ̄’i _vK‡j, Pvc (P)  ZvcgvÎv (T)  P  T (Ans.) 13| GKwU N‡ii AvqZb 150 m3 | mKv‡j NiwUi ZvcgvÎv wQj 27C| `ycy‡i ZvcgvÎv †e‡o 42C n‡j, Ni †_‡K cÖv_wgK AvqZ‡bi KZ kZvsk evqy †ewi‡q hv‡e? [Medium] mgvavb: V2 V1 = T2 T1  V2 – V1 V1 = T2 – T1 T1  V2 – V1 V1  100 = 42 – 27 27 + 273  100 = 15 300  100 = 5% (Ans.) 14| cÖgvY Ki †h, w ̄’i Pv‡c Ges w ̄’i AvqZ‡b M ̈v‡mi Av‡cwÿK Zv‡ci AbycvZ = 1 + 2 f (†hLv‡b, f = M ̈vmwUi ̄^vaxbZvi gvÎv)| [Medium] mgvavb: GK †gvj M ̈v‡mi †gvU ̄^vaxbZvi gvÎv = NAf  GK †gvj M ̈v‡mi †gvU kw3, E = f 2 NAKT  E = f 2 RT    K =  R NA
Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Varsity Practice Content 3 GLb, CV =     dE dT V = f 2 R  CP = R + CV = R + f 2 R =     1 + f 2 R  CP CV = 1 + f 2 f 2  CP Cv = 2 + f f  CP Cv = 1 + 2 f  CP Cv = 1 + 2 f (Proved) 15| GKwU Mvmxq wgkÖ‡Y 1 mol wnwjqvg (CP = 2.5R, CV = 1.5R) Ges 1 mol nvB‡Wav‡Rb (CP = 3.5R, CV = 2.5R) M ̈vm i‡q‡Q| wgkÖYwUi CP,CV I  Gi gvb wbY©q K‡iv| [Medium] mgvavb: wgkÖ‡Yi Rb ̈ CPm , CVm I m awi,  CPm = n1CP1 + n2CP2 n1 + n2  CPm = 1  2.5R + 1  3.5R 1 + 1  CPm = 3R  CVm = n1CV1 + n2CV2 n1 + n2  CVm = 1  1.5R + 1  2.5R 1 + 1  CVm = 2R  m = CPm CVm = 3R 2R = 3 2 (Ans.) 16| 100 mL AvqZ‡bi GKwU cv‡Î M ̈v‡mi Pvc 200 kPa| hw` cÖwZwU AYyi Mo •iwLK MwZkw3 6  10–21 J nq Z‡e cv‡Î Dcw ̄’Z M ̈v‡mi AYymsL ̈v wbY©q K‡iv| [Medium] mgvavb: hw` cv‡Î N msL ̈K AYy _v‡K, †gvU kw3, NE = 3 2 RT = 3 2 PV  N = 3 2 PV E  N = 3 2  200  103  0.1  10–3 6  10–21  N = 3 2  1 3  1022 = 0.5  1022 wU  N = 5  1021 wU (Ans.) 17| GKwU 2 mL AvqZ‡bi cv‡Î 50 mg M ̈vm 100 kPa Pv‡c i‡q‡Q| cÖwZwU M ̈vm AYyi fi 8  10–26 kg n‡j cÖwZwU AYyi Mo •iwLK MwZkw3 wbY©q K‡iv| [Medium] mgvavb: cv‡Î Dcw ̄’Z AYyi msL ̈v, N = 50  10–6 8  10–26  N = 5  0.125  1021  N = 6.25  1020 wU  †gvU kw3, E = 3 2 PV = 3 2  100  103  2  10–6 = 3  10–1  cÖwZwU AYyi Rb ̈, E = E N = 3  10–1 6.25 1020  E  3 6  10–21  E  5  10–22 J (Ans.) 18| 2 atm Pv‡c 5 L wnwjqvg M ̈v‡mi MwZkw3 E n‡j, 3 atm Pv‡c 15 L Aw·‡Rb M ̈v‡mi AYy ̧‡jvi MwZkw3‡K E Gi gva ̈‡g cÖKvk K‡iv| [Easy] mgvavb: E = 3 2 PV = 3 2  2  5 = 15 atmL E1 = 5 2 PV = 5 2  3  15 = 15 2  15 atmL = 7.5E (Ans.) 19| 300 K ZvcgvÎvq 1 †gvj cwigvY †Kv‡bv GKcigvYyK M ̈vm‡K w ̄’i Pv‡c AvqZb wØ ̧Y Kiv n‡j, M ̈v‡mi Af ̈šÍixY kw3i cwieZ©b wbY©q K‡iv| [Medium] mgvavb: w ̄’i Pv‡c, V  T Avevi E  T  V2 = 2V1 n‡j, E2 = 2E1  kw3i cwieZ©b, E = E2 – E1 = 2E1 – E1 = E1 = 3 2 nRT = 3 2  1  R  300 = 450R (Ans.)
4  Physics 1st Paper Chapter-10 20| `ywU GKcigvYyK M ̈vm, hv‡`i †gvjmsL ̈v (n1, n2) I ZvcgvÎv (T1, T2) †K wgwkÖZ Kiv n‡jv| wgkÖ‡Yi ZvcgvÎv wbY©q K‡iv| [Hard] mgvavb: wgkÖ‡Yi c~‡e© †gvU kw3 = E1 + E2 Ges wgkÖ‡Yi c‡i †gvU kw3 = E3 + E4 E1 + E2 = E3 + E4  3 2 n1RT1 + 3 2 n2RT2 = 3 2 n1RT + 3 2 n2RT  n1T1 + n2T2 = (n1 + n2) T  T = n1T1 + n2T2 n1 + n2 (Ans.) 21| 4 g nvB‡Wav‡Rb M ̈v‡mi mv‡_ STP †Z 11.2 L wnwjqvg M ̈vm‡K 20 L Gi GKwU cv‡Î wgwkÖZ Kiv n‡jv| wgkÖ‡Yi ZvcgvÎv 300 K n‡j wgkÖ‡Y M ̈v‡mi Pvc wbY©q K‡iv| [Medium] mgvavb: nH2 = W M = 4 2 = 2 mol nHe = V 22.4 = 11.2 22.4 = 0.5 mol  nH2 + nHe = ntotal  2 + 0.5 = PV RT  P = 0.08  2.5  300 20  P = 24  2.5 20  P = 6 5  5 2 = 3 atm (cÖvq) (Ans.) 22| Av`k© Ae ̄’vq †Kv‡bv M ̈v‡mi NbZ¡ 1.3 kgm–3 Ges M ̈v‡m k‡ãi †eM 330 ms–1 | M ̈vmwUi AYy ̧‡jvi ̄^vaxbZvi gvÎv wbY©q K‡iv| [Hard] mgvavb: v = P   3302 =   105 1.3  332  102  1.3  10–5 =   (900 + 9 + 180)  1.3  10–3 =   1089  1.3  10–3 =     1.3  1.4  M ̈vmwU wØcigvYyK, ̄^vaxbZvi gvÎv = 5| (Ans.) 23| 4 L AvqZ‡bi GKwU cv‡Î 8 g Aw·‡Rb, 14 g bvB‡Uav‡Rb Ges 22 g Kve©b WvB-A·vB‡Wi wgkÖY Av‡Q| wgkÖ‡Yi DòZv 27C n‡j M ̈vm wgkÖYwUi Pvc KZ? [R = 8 Jmol–1K –1 ] [Medium] mgvavb: PV = ntotalRT  P  4  10–3 =     8 32 + 14 28 + 22 44  8  300  P = 1.25  2  3  105 = 2.5  3  105  P = 5 2  3  105 = 7.5  105 Nm–2 (Ans.) 24| 1 kg wØcigvYyK M ̈v‡mi Pvc 8  104 Nm–2 , M ̈v‡mi NbZ¡ 4 kg/m 3 | Zvcxq MwZi Rb ̈ M ̈vmwUi kw3 KZ n‡e? [Medium] mgvavb: E = 5 2 nRT = 5 2 PV = 5 2 P  M d = 5 2  8  104  1 4 = 5  104 J (Ans.) 25| NTP †Z 2 g wnwjqv‡gi †gvU ̄^vaxbZvi gvÎv KZ? [Medium] mgvavb: 2 g He G AYymsL ̈v, N = W M  NA = 2 4  6.02  1023 = 3.01  1023 wU  2 g He Gi †gvU ̄^vaxbZvi gvÎv = 3  N = 3  3.01  1023 = 9.03  1023 (Ans.) 26| 5 L I 3 L AvqZ‡bi `ywU cv‡Î h_vμ‡g 3 ̧Y Ges 7 ̧Y evqygÐjxq Pv‡c evZvm Av‡Q| cvÎ `ywU‡K miæ bj w`‡q hy3 Ki‡j Dfqcv‡Îi mvaviY PvcÑ [Easy] mgvavb: n1 + n2 = ntotal  P1V1 RT + P2V2 RT = P  (V1 + V2) RT  P = P1V1 + P2V2 V1 + V2 = 5  3 + 3  7 8  P = 4.5 evqygÐjxq Pvc| (Ans.) 27| †h ZvcgvÎvq c„w_exi evqygÐjw ̄’Z bvB‡Uav‡Rb AYyi rms MwZ‡eM c„w_exi AwfKl©‡ÿÎ †_‡K gyw3‡e‡Mi mgvb n‡e Zv wbY©q K‡iv| [Medium] mgvavb: ve = crms  2gr = 3RT M  2gr = 3RT M  T = 2grM 3R  T = 2  10  6.4  106  28  10–3 3  8  T = 2  8  28 3  103  T  16  27 3  103  144  103 K (cÖvq) (Ans.)