Tiêu đề 10. P1C10 Ideal Gas & Kinetic Theory of Gas_With Solve.pdf ✅

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Mastery Practice Sheet 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 cm3  Aw·‡Rb M ̈vm e ̈eüZ n‡q‡Q = 1.5  106 – 5  105 = 106 cm3 (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?

Av`k© M ̈vm I M ̈v‡mi MwZZË¡  Mastery Practice Sheet 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 atmospheric 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? Atmospheric 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 Shortcut for these type of maths