Hc Varma Solutions

CHAPTER 24 KINETIC THEORY OF GASES 1. Volume of 1 groyne of boast RT 0.082 ? 273 3 2 3 PV = nRT ? V = = = 22.38 ? 22.4 L = 22.4 × 10 = 2.24 × 10 m P 1 2. 3. 1 ? 1? 10 ?3 10 ?3 PV 1 = = = RT 0.082 ? 273 22.4 22400 1 23 19 No of molecules = 6.023 × 10 × = 2.688 × 10 22400 3 5 V = 1 cm , T = 0°C, P = 10 mm of Hg n= n= 1.36 ? 980 ? 10 ?6 ? 1 PV Æ'gh ? V 13 = = = 5.874 × 10 RT RT 8.31 ? 273 23 13 11 No. of moluclues = No × n = 6.023 × 10 × 5.874 × 10 = 3.538 × 10 1 ? 1? 10 ?3 10 ?3 PV = = RT 0.082 ? 273 22.4 4. n= 5. ? 32 3 g = 1.428 × 10 g = 1.428 mg 22.4 Since crapper is same n1 = n2 = n nR ? three hundred nR ? 600 P1 = , P2 = V0 2V0 mass = ?10 ?3 ? 2V0 600 K V0 ccc K 2V0 P1 nR ? 300 1 ? = = =1:1 P2 V0 nR ? 600 1 6. V = 250 cc = 250 × 10 3 3 3 6 3 P = 10 mm = 10 × 10 m = 10 × 13600 × 10 pascal = 136 × 10 pascal T = 27°C = 300 K n= 3 7. 136 ? 10 ?3 ? 250 PV 136 ? 250 = ? 10 ? 3 = ? 10 ? 6 8.3 ? 300 RT 8.3 ? 30 0 136 ? 250 No. of molecules = ? 10 ? 6 ? 6 ? 10 23 = 81 × 1017 ? 0.8 × 1015 8.3 ? 300 5 6 P1 = 8.0 × 10 Pa, P2 = 1 × 10 Pa, T1 = 300 K, Since, V1 = V2 = V T2 = ? 8. 9. P1V1 PV 1 ? 10 6 ? V 8 ? 10 5 ? V 1? 10 6 ? 300 = 2 2 ? = ? T2 = = 375° K T1 T2 300 T2 8 ? 10 5 3 6 3 T = 300 K, P=?

m = 2 g, V = 0.02 m = 0.02 × 10 cc = 0.02 × 10 L, M = 2 g, m 2 PV = nRT ? PV = RT ? P × 20 = ? 0.082 ? 300 M 2 0.082 ? 300 5 5 ?P= = 1.23 atm = 1.23 × 10 pa ? 1.23 × 10 pa 20 nRT m RT Æ'RT P= = ? = V M V M 3 3 Æ' ? 1.25 × 10 g/cm 7 R ? 8.31 × 10 ert/deg/mole T ? 273 K ?M= 1.25 ? 10 ?3 ? 8.31 ? 10 7 ? 273 Æ'RT 4 = = 0.002796 × 10 ? 28 g/mol P 13.6 ? 980 ? 76 24.1  kinetic Theor y of Gases 10. T at Simla = 15°C = 15 + 273! = 288 K 2 P at Simla = 72 cm = 72 × 10 × 13600 × 9.8 T at Kalka = 35°C = 35 + 273 = 308 K 2 P at Kalka = 76 cm = 76 × 10 × 13600 × 9.8 PV = nRT m m PM ? PV = RT ? PM = RT ? Æ' = M V RT PSimla ? M RTKalka Æ'Simla ? = Æ'Kalka RTSimla PKalka ? M 288 ? 76 ? 10 ? 13600 ? 9.8 Æ'Kalka 1 = = 0.987 Æ'Simla...If you want to get a full essay, put together it on our website: OrderEssay.net

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