قيمة الثابت العام للغازات
)(157 ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ :ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ إﻋﺪاد د /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﻟﻌﺎﻡ ﻟﻠﻐﺎﺯﺍﺕ Value of the Gas Constant ﻻ ﺗﺘﻐﯿﺮ اﻟﻘﯿﻤﺔ اﻟﻌﺪدﯾﺔ ﻟﻠﺜﺎﺑﺖ ) (Rإﻻ ﺑﺎﺧﺘﻼف اﻟﻮﺣﺪات اﻟﺘﻲ ﯾﻘﺎس ﺑﮭﺎ ﻛﻞ ﻣﻦ اﻟﻀﻐﻂ ) (Pواﻟﺤﺠﻢ ﻓﻘﻂ ) .(Vوﻋﻨﺪ اﻟﺘﻌﺎﻣﻞ ﻣﻊ ﻗﻮاﻧﯿﻦ اﻟﻐﺎزات ﯾﺘﻢ اﻟﺘﻌﻮﯾﺾ ﺑﻘﯿﻤﺔ R اﻟﻤﻨﺎﺳﺒﺔ ﻓﻲ ﺿﻮء اﻟﻮﺣﺪات اﻟﻤﺘﺎﺣﺔ. وﻣﻦ اﻟﺴﮭﻞ اﺷﺘﻘﺎق وﺣﺪات ﺛﺎﺑﺖ اﻟﻐﺎزات ﻋﻠﻰ اﻟﻨﺤﻮ اﻟﺘﺎﻟﻲ : PV pressure × (length)3 =R = nT degrees × moles force × (length)-2 × (length)3 degrees × moles force × length energy =R = degrees × moles degrees × moles =R R = J K -1mol-1 R = Nm K -1mol-1 وﻓﯿﻤﺎ ﯾﻠﻲ ﺳﻨﺄﺧﺬ ﺑﻌﺾ ھﺬه اﻟﻘﯿﻢ اﻟﻌﺪدﯾﺔ ﻟﻠﺜﺎﺑﺖ : R (١ﺍﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ) (Rﻟﻠﻐﺎﺯﺍﺕ ﺑﻮﺣﺪﺍﺕ atm L /mol K ﻣﻦ ﻗﺎﻧﻮن أﻓﻮﺟﺎدرو أن اﻟﻤﻮل اﻟﻮاﺣﺪ ﻣﻦ أي ﻏﺎز ﯾﺸﻐﻞ ﺣﺠﻤﺎً ﻗﺪره ) (22.414 Lﻋﻨﺪ ﻣﻌﺪل اﻟﻀﻐﻂ ودرﺟﺔ اﻟﺤﺮارة )ﻋﻨﺪ اﻟﻈﺮوف اﻟﻘﯿﺎﺳﯿﺔ (S.T.P = 273 K, 1 atm وﺑﺎﻟﺘﻌﻮﯾﺾ ﻓﻲ اﻟﻌﻼﻗﺔ : P V 1 atm × 22.4136 L = n T 1 mol × 273.15 K =R ⇒ R = 0.082056 atm.L/mol. K ≈ 0.0821 atm. L /mol K )(157 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (158) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د : ( ﺗﺴﺎويR) ( ﻟﻜﻦ ﻓﻲ ھﺬه اﻟﺤﺎﻟﺔ ﻓﺈن ﻗﯿﻤﺔml) وﯾﻤﻜﻦ اﺳﺘﺨﺪام وﺣﺪة اﻟﺤﺠﻢ 0.0821 L.atm 0.0821 L × 103 ml L-1 .atm R= = mol.K mol . K ⇒ R = 82.1 ml.atm/mol.K Fig. 62 : a) Decreasing the volume of the gas at constant n and T increases the frequency of collisions with the container walls and therefore increases the pressure (Boyle's law). b) Increasing the temperature (kinetic energy) at constant n and P increases the volume of the gas (Charles's law). c) Increasing the amount of gas at constant T and P increases the volume (Avogadro's law). d) Changing the identity of some molecules at constant T and V has no effect on the pressure (Dalton's law). (158) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (159) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د Fig. 63 : Schematic illustrations of Boyle's law, Charles's law, and Avogadro's law (159) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com )(160 ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ :ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ إﻋﺪاد د /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي (٢ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺍﺕ bar. L/mol K ﻋﻨﺪ اﺳﺘﺨﺪام اﻟﺒﺎر ﻛﻮﺣﺪة ﻟﻠﻀﻐﻂ ﺑﺪﻻً ﻣﻦ : atm ﻓﺈن : R )( 1 atm = 1.01325 bar PV )(1.01325 bar) × (22.4136 L = nT ) (1 mol ) × ( 273.15 K =R R = 0.08314 L bar/mol K ⇒ R = 0.08314 L bar/mol K (٣ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺍﺕ Pa . dm3/mol K اﻟﻌﻼﻗﺔ ﺑﯿﻦ اﻟﺒﺎﺳﻜﺎل ) (Paﻛﻮﺣﺪة ﻟﻠﻀﻐﻂ اﻟﺠﻮي ووﺣﺪة ) (atmھﻲ : )(1 atm = 101325 Pa وﺑﺎﻟﺘﺎﻟﻲ ﻹﯾﺠﺎد ﻗﯿﻤﺔ ) (Rﺑﻮﺣﺪة Pa . dm3/mol . K : PV nT ) (101325 Pa) × (22.4136 dm 3 =R 1mol × 273.15 K =R ⇒ R= 8314 Pa . dm 3/mol . K ⇒ R = 8.314 kPa . dm 3/mol . K وﯾﻤﻜ ﻦ اﻟﺤ ﺼﻮل ﻋﻠ ﻰ ھ ﺬه اﻟﻘﯿﻤ ﺔ ﻣ ﻦ ﻣﻌﺮﻓ ﺔ ﻗﯿﻤ ﺔ Rﺑﻮﺣ ﺪة (R = 8.314 atm : L/mol K وﻣﻦ ﻣﻌﺮﻓﺔ اﻟﻌﻼﻗﺔ ﺑﯿﻦ : )(160 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (161) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د 1 atm = 101325 Pa 1 L = 1 dm3 0.082056 atm . L × 101325 Pa/atm . dm 3 /L ) ( 0.08206 atm. L R= = mol.K mol . K 3 R = 8314.32 Pa dm / mol K R = 8.314 kPa dm3 / mol K Pa . m3/mol K ( ﺑﻮﺣﺪﺍﺕR) ( ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ٤ PV nT (101325 Pa) × (0.0224136 m3 ) R= 1mol × 273.15 K R = 8.314 Pa. m 3 /mol .K R= ⇒ R = 8.314 Pa . m 3 / mol K : ﻣﻠﺤﻮﻇﺔ : ( ﺣﯿﺚ أنJoul) ( اﻟﻰ وﺣﺪةPa . m3) وﯾﻤﻜﻦ ﺗﺤﻮﯾﻞ وﺣﺪة 1 Pa m 3 = 1 J R = 8.314 Pa . m 3 /mol . K ⇒ R = 8.314 J/ mol K ⇒ R = 8.314 J/ mol K dyne. cm /mol K ( ﺑﻮﺣﺪﺍﺕR) ( ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ٥ ( )ﺣﯿﺚ1 atm) وھﻲ ﺗﺮادف ﺿﻐﻄﺎً ﻗﺪره واﺣﺪ ﺟﻮdyne/cm2 ھﻨﺎك وﺣﺪة ﺗﺴﻤﻰ 1 dyne = 1 g . cm . s-2 : اﻟﺪاﯾﻦ وﺣﺪة ﻟﻠﻘﻮة وھﻲ ﺗﺴﺎوي (161) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com )(162 ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ :ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ إﻋﺪاد د /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي وﻟﺘﺤﻮﯾﻞ وﺣﺪة اﻟﻀﻐﻂ اﻟﺠﻮي )) (1 atmواﻟﺘﻲ ﺗﻜﺎﻓﻲء ﻋﻤﻮداً ﻣﻦ اﻟﺰﺋﺒﻖ ﻃﻮﻟﮫ (76 ) cmاﻟﻰ وﺣﺪة dyne/cm2ﻧﺴﺘﺨﺪم ﻧﻔﺲ اﻟﻌﻼﻗﺔ اﻟﺴﺎﺑﻘﺔ : P=hdg g cm ) × (980.6667 ) cm3 s2 )( d )( a 1atm = (76.00 cm) × (13.5951 1 atm = 1013251.9 g cm-1 s-2 10132252 g cm s-2 = 1 atm = 1013252 dyne/cm2 2 cm 1 atm = 1.013252 × 106 dyne/cm2 وھ ﺬه اﻟﻘﯿﻤ ﺔ ﻟﻠ ﻀﻐﻂ ﺗﻌ ﺎدل ) (1 atmوﺑﺎﻟﺘ ﺎﻟﻲ ﻟﺘﺤﻮﯾ ﻞ اﻟ ﻀﻐﻂ اﻟ ﻰ وﺣ ﺪات : dyne. cm/mol K ) P( dyne/cm 2 ) × (cm3 = N . m mol-1 K -1 mol K ) (1.013252 × 106 dyne/cm 2 ) × (22413.6 cm3 =R 1 mol × 273.15 K R = 83143419.47 dyne . cm/ mol K =R ⇒ R = 8.314 × 107 dyne . cm/ mol K (٦ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺓ ﺍﻹﺭﺝ erg/mol K اﻹرج ﻋﺒﺎرة ﻋﻦ اﻟﺸﻐﻞ اﻟﻤﺒﺬول ﻋﻨﺪﻣﺎ ﺗﺆﺛﺮ ﻗﻮة ﻗﺪرھﺎ واﺣﺪ داﯾﻦ) (1 dyneﻋﻠﻰ ﻣﺴﺎﻓﺔ ﻗﺪرھﺎ ) (1 cmوھﻮ ﻣﻦ وﺣﺪات اﻟﻄﺎﻗﺔ .وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن : 1 erg = 1 dyne. cm )(162 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com )(163 ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ :ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ إﻋﺪاد د /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي ﻭﳊﺴﺎﺏ ﻗﻴﻤﺔ Rﺑﻮﺣﺪﺓ :erg ) PV (1.013252 × 106 dyne/cm2 ) × (22413.6 cm3 =R = RT )(1 mol) × (273.15 K R = 83143419.47 dyne. cm/mol.K R = 83143419.47 erg/mol.K ⇒ R = 8.31 × 10 7 erg/mol.K وﻧﻈﺮاً ﻷن اﻹرج ﻛﻤﯿﺔ ﺻﻐﯿﺮة ﻣﻦ اﻟﻄﺎﻗﺔ ،ﻓﺈﻧﮫ ﯾﻌﺒﺮ ﺑﻮﺣﺪة أﻛﺜﺮ ﻣﻼءﻣﺔ وھﻲ اﻟﺠﻮل ﺣﯿﺚ أن : 1 Joul = 107 erg ⇒ R = 8.314 × 107 erg/mol.K R = 8.314 Joul/K.mol (٧ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺍﺕ N .m /mol K ﻓﻲ ھﺬه اﻟﺤﺎﻟﺔ ﺳﻨﺴﺘﺨﺪم وﺣﺪة اﻟﻀﻐﻂ ) (N/m2ووﺣﺪة اﻟﺤﺠﻢ ) (m3أي ﺗﻜﻮن وﺣﺪة : R ) P(N/m 2 ) × (m 3 =R = N . m mol -1K -1 mol K ﺗﻌﺮﻳﻒ ﺍﻟﻨﻴﻮﺗﻦ : وھﻮ اﻟﻘﻮة اﻟﺘﻲ ﺗﻜﺴﺐ ﻛﺘﻠﺔ ﻗﺪرھﺎ ) (1 Kgإﺳﺮاﻋﺎً ﻗﺪره ).(1 m/s2 واﻟﻨﯿﻮﺗﻦ ) (Nوﺣﺪة اﻟﻘﻮة ﻓﻲ ﻧﻈﺎم )(m . Kg . s وﻟﻠﺘﻌﺒﯿﺮ ﻋﻦ اﻟﻀﻐﻂ اﻟﺠﻮي اﻟﻘﯿﺎﺳﻲ ) (Standard Atmosphereﺑﻮﺣﺪات ):(m . Kg .s )(163 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (164) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د ﻧﻀﺮب ﻛﺘﻠﺔ اﻟﺰﺋﺒﻖ )ﺑﻮﺣﺪة اﻟﻜﯿﻠﻮ ﺟﺮام( اﻟﺬي ﯾﺆﺛﺮ ﻋﻠﻰ اﻟﻤﺘﺮ اﻟﻤﺮﺑﻊ ﺑﻌﺠﻠﺔ اﻟﺠﺎذﺑﯿﺔ : اﻷرﺿﯿﺔ P=h.d.g P = h (m) . d (Kg/m 3 ) . g (m/s 2 ) = Kg m -1 s -2 = Kg m s -2 /m 2 = N/m 2 Newton ( ﻓﺈن ھﺬهh = 76 cm, d = 13.596 g/cm3, g = 980.7 cm/s2) ﻓﺈذا ﻋﻠﻤﻨﺎ أن : ( ﻛﻤﺎ ﯾﻠﻲKg, m) اﻟﻮﺣﺪات ﯾﺠﺐ ﺗﺤﻮﯾﻠﮭﺎ إﻟﻰ h Hg = 76.00 cm = 0.7600 m 13.5951 g 13.5951 g × 10-3 Kg g -1 = = 13595.1 Kg m-3 3 3 -6 3 -3 cm 1 cm × 10 m cm 980.6665 cm 980.6665 cm × 10-2 m cm -1 g= = = 9.806665 ms-2 2 2 s s -3 -2 P = h (m) . d (kg m ) . g (ms ) d Hg = P = (0.7600 m) × (13595.1 kg m- 3 ) × (9.806665 m s- 2 ) P = 101325.1694 kg m-1 s-2 = 1.01325 × 105 kg m-1s-2 P = 1.01325 × 105 Kg m s-2 /m2 (where N = Kg m s-2 ) N P = 1.01325 × 10 N/m 2 5 N. m/mol ( وﺑﺎﻟﺘ ﺎﻟﻲ ﻟﺘﺤﻮﯾ ﻞ اﻟ ﻀﻐﻂ اﻟ ﻰ وﺣ ﺪات1 atm) وھﺬه اﻟﻘﯿﻤﺔ ﻟﻠﻀﻐﻂ ﺗﻌﺎدل :K P (N/m2 ) × (m3 ) = N . m mol-1 K-1 mol K (1.01325 × 105 N/m2 ) × (0.0224136 m3 ) R= 1 mol × 273.15 K R= ⇒ R = 8.314 N . m/ mol K (164) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com )(165 ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ :ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ إﻋﺪاد د /ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي (٨ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺍﺕ J/mol K أوﺟﺪﻧﺎ ﻓﻲ اﻟﻔﻘﺮة اﻟﺴﺎﺑﻘﺔ ﻗﯿﻤﺔ اﻟﺜﺎﺑﺖ ). (R = 8.314 N. m/mol K ﻭﺍﳉﻮﻝ ھﻮ ﻗﻮة ﻗﺪرھﺎ ﻧﯿﻮﺗﻦ ﻣﻀﺮوﺑﺔ ﻓﻲ ﻣﺴﺎﻓﺔ ﻗﺪرھﺎ ):(1 m 1 Joule = N . m وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن : J = N. m R = 8.314 N. m/mol K ⇒ R = 8.314 J/mol. K وﺑﺎﻟﺘ ﺎﻟﻲ ﻋﻨ ﺪ اﻟﺘﻄﺒﯿ ﻖ ﻓ ﻲ ﻣﻌﺎدﻟ ﺔ اﻟﻐ ﺎز اﻟﻤﺜ ﺎﻟﻲ ﺑﻘﯿﻤ ﺔ ) (8.314 J/mol Kﻓﯿﺠ ﺐ أن ﻧﺴﺘﺨﺪم ﻟﻠﻀﻐﻂ وﺣﺪة ﺑﺎﺳﻜﺎل ) (Paوﻟﻠﺤﺠﻢ وﺣﺪة ﻣﺘﺮ ﻣﻜﻌﺐ ).(m3 (٩ﺇﳚﺎﺩ ﻗﻴﻤﺔ ﺍﻟﺜﺎﺑﺖ ﺍﳌﻮﻻﺭﻱ ﻟﻠﻐﺎﺯﺍﺕ ) (Rﺑﻮﺣﺪﺍﺕ ﺍﻟﺴﻌﺮ Cal/mol K ﻏﺎﻟﺒﺎً ﻣﺎ ﺗﺴﺘﻌﻤﻞ ﻓﻲ ﻣﺴﺎﺋﻞ اﻟﻜﯿﻤﯿﺎء اﻟﺤﺮارﯾﺔ واﻟﺪﯾﻨﺎﻣﯿﻜﺎ اﻟﺤﺮارﯾﺔ وﺣﺪة ﻟﻠﻄﺎﻗﺔ ﺗﻌﺮف ﺑﺎﻟﺴﻌﺮ ) (calorieﺣﯿﺚ أن اﻟﻌﻼﻗﺔ ﺑﯿﻦ اﻟﺴﻌﺮ واﻟﺠﻮل: 1 Cal. = 4.184 J وﺑﺎﻟﺘﺎﻟﻲ ﻹﯾﺠﺎد ﻗﯿﻤﺔ Rﺑﻮﺣﺪة اﻟﺴﻌﺮ ﻓﺈن : 8.314 J / mol K 4.184 J / Cal =R ⇒ R = 1.987 C al/K m ol ≈ 2 C al/m ol K )(165 PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (166) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د ﺍﻟﻌﻼﻗﺔ ﺑﲔ ﺍﻟﻨﻴﻮﺗﻦ ﻭﺍﻟﺪﺍﻳﻦ 1 dyne = 10-5 N :(1 m) واﻟﺠﻮل ھﻮ ﻗﻮة ﻗﺪرھﺎ ﻧﯿﻮﺗﻦ ﻣﻀﺮوﺑﺔ ﻓﻲ ﻣﺴﺎﻓﺔ ﻗﺪرھﺎ 1 Joule = N . m N/m2 اﻟﻰdyne/cm2 وﻟﺘﺤﻮﯾﻞ وﺣﺪة اﻟﻀﻐﻂ ﻣﻦ 1dyne 1 × 10-5 N 2 P= = = 0.1 N/m cm 2 1 × 10-4 m 2 ⇒ P = 1dyne/cm2 = 0.1 N/m2 : وﺑﻤﺎ أن 1atm = 1.013 × 106 dyne/cm2 ⇒ 1dyne/cm2 = 0.1N/m2 1dyne/cm2 = 0.1 N/m2 ⇒ 1.013 × 106 dyne/cm2 = P 0.1N/m2 × 1.013 × 106 dyne/cm2 = 1.013 × 105 N/m2 P= 2 1dyne/cm (1 J = 1 N. ( ﺣﯿﺚJ/K.mol ) أوN.m/K.mol ﺑﻮﺣﺪةR وﺑﺎﻟﺘﺎﻟﻲ ﻟﺤﺴﺎب ﻗﯿﻤﺔ m) R= R= PV nT (1.013 × 105 Newton m-2 ) × ( 0.0224136 m3 ) (1 mol) × (273.15 K) ⇒ R = 8.314 J/K.mol (166) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com (167) ﻗﻮﺍﻧﲔ ﺍﻟﻐﺎﺯﺍﺕ: ﺍﻟﻔﺼﻞ ﺍﻟﺜﺎﻧﻲ ﻋﻤﺮ ﺑﻦ ﻋﺒﺪ اﷲ اﻟﮭﺰازي/إﻋﺪاد د : k ﺛﺎﺑﺖ ﺑﻮﻟﺘﺰﻣﺎﻥ ( وھﻮ ﻋﺒﺎرة ﻋﻦ ﺛﺎﺑﺖ اﻟﻐﺎز ﻟﻜﻞ ﺟﺰيءBoltzmann constant) (k) ﺛﺎﺑﺖ ﺑﻮﻟﺘﺰﻣﺎن (Gas Constant per Molecule) R 8.314 × 107 erg K -1 mol-1 k= = = 1.380 × 10-16 erg / K. molecule 23 -1 NA 6.023 × 10 molecules . mol k= R 8.314 J K -1 mol-1 = = 1.380 × 10-23 J / K . molecule 23 -1 NA 6.023 × 10 molecules . mol وﻓﻘﺎً ﻟﻮﺣﺪات اﻟﻀﻐﻂ واﻟﺤﺠﻢR ﻣﻠﺨﺺ ﻟﺒﻌﺾ ﻗﯿﻢ اﻟﺜﺎﺑﺖ:١٥ ﺟﺪول اﻟﻘﯿﻤﺔ اﻟﻌﺪدﯾﺔ R ﻟـ 0.082056 82.056 0.08314 8.314 8.314 8.314 8.314 8.314 x 107 8.314 x 107 1.987 وﺣﺪة R atm L / mol K atm ml / mol K bar L / mol K KPa. dm3/mol K Pa. m3 / mol K N. m / mol K J / mol K dyne. cm/mol K erg / mol K cal / mol K وﺣﺪة اﻟﻀﻐﻂ atm atm bar kPa Pa N/m2 Pa dyne/cm2 dyne/cm2 وﺣﺪة اﻟﺤﺠﻢ L ml L dm3 m3 m3 m3 cm3 cm3 (٥١) ﻣﺜﺎﻝ ﻋﻨﺪ درﺟﺔ4.20 g ﻣﻘﺪارھﺎF2(g) اﺣﺴﺐ ﺣﺠﻢ اﻟﻐﺎز اﻟﺬي ﺗﺸﻐﻠﮫ ﻛﻤﯿﺔ ﻣﻦ ﻏﺎز اﻟﻔﻠﻮر .(F = 19) : )ﻋﻠﻤﺎً ﺑﺄن اﻟﻜﺘﻠﺔ اﻟﺬرﯾﺔ720 torr وﺗﺤﺖ ﺿﻐﻂ42 °C (167) PDF created with FinePrint pdfFactory Pro trial version http://www.pdffactory.com
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