One of such properties is heat capacity, which is the amount of heat energy required to raise the temperature of a substance by one degree Kelvin. Mathematically, it is ? = q? T , where q is the amount of heat absorbed by a substance and ? T is the change in temperature measured. When substances absorb heat, their molecules translate, rotate and vibrate due to the rise in temperature . As a result of the motion of movement of molecules in these modes, there is a contribution of energy towards determining the heat capacity of that substance.
The heat capacity is, however, defined through constant volume (Cv) or constant pressure (Cp) with a relationship, Cp = Cv + nR and CpCv , the heat capacity ratio for ideal gases which is further determined by obtaining the pressure difference with atomospheric pressure in adiabatic conditions. That is lnp1-lnp2lnp1-lnp3. The energy contribution through the modes of movement of molecules is the total of their, translational, rotational and vibrational energies. For ideal gases, this can be calculated theoretically as a result of their classes, Monatomic, Diatomic and Linear polyatomic.
Monatomic gases such as Helium, move in translation with the energy 32RT. Diatomic gases such as Nitrogen, move in all 3 modes with the energy 72RT. And the linear polyatomic gases such as CO2 move with the energy 132RT. The constant volume heat capacity for these ideal gases can be determined as a result of its relationship with these energies as the energy U = nRT and Cv is the derivative with respect to volume. i. e Cv = ? U? Tv . This leads to the following Cv for the 3 classes of gases; 12. 5 Jmol*K for monatomic, 29.
1 Jmol*K for diatomic, and 54. 0 Jmol*K for linear polyatomic. Data Room Temperature = 16. 2 oC ± 0. oC p2 = Room Pressure Room Pressure = 760. 84 mmHg ± 0. 22 mmHg Helium Trial| P1 (mmHg)(±0. 3)| P3 (mmHg)(±0. 3)| 1| 300. 4| 75. 6| 2| 275. 7| 69. 0| 3| 281. 9| 74. 8| Carbon Dioxide Trial| P1 (mmHg)(±0. 3)| P3 (mmHg)(±0. 3)| 1| 290. 3| 34. 1| 2| 277. 8| 25. 3| 3| 283. 1| 40. 1| The values for Helium and Carbon dioxide were gotten from the other group who performed the experiment. Nitrogen Trial| P1 (mmHg)(±0. 3)| P3 (mmHg)(±0. 3)| 1| 278. 7| 63. 7| 2| 286. 6| 89. 7| 3| 270. 5| 58. 9| 4| 294. 2| 85. 0| 5| 285. 5| 89. 7| 6| 291. 4| 70. 0| 7| 268. 1| 54. 1| 8| 289. 0| 64. 8| 9| 281. 5| 65. 8| 10| 265. 3| 59. 7|
Values in bold are the 3 best trial obtained. Answers to Questions 1) C, mathematical defined as C = q? T , is the heat capacity, the amount of energy required to raise the temperature of a substance by one degree Kelvin. Cv, is the heat capacity per unit volume while, Cp , is the heat capacity per unit pressure. Both are related mathematically by the equation Cp = Cv + nR. The expected heat capacity for the three classes of gases are as follows Monatomic = 3R2=12. 5 Jmol. K Diatomic = 7R2=29. 1 Jmol. K Linear triatomic = 13R2=54. 0 Jmol. K The equations leading to the heat capacity ratio, ? , is summarized by CpCv= ln(p1p2)ln? p1p3) The vibrational contribution to Cv can be determined once the vibrational frequencies of the molecule is known. That is Rx2e-x where is x>5 . x = (NA hRT)v Where NA = Avogadro's number, h = Planck's constant and v = vibration frequency.
2) Data obtained in the experiment is presented in the data section above. 3) Sample error calculation 2300. 42*0. 32+760. 842*0. 222 =817. 9962127 2817. 99621271061. 242+0. 22760. 842 = 0. 770793 275. 62*0. 32+760. 842*0. 222 =168. 9143383 2817. 99621271061. 242+168. 9143383836. 442 = 0. 79681 0. 7707931. 39482677 = 0. 55261 0. 796811. 26875807 = 0. 6280336 20. 552610. 327702282+0. 62803361. 268758072 = 3. 12 This was applied to calculate all errors in the heat capacity ratios below. Helium Trial| ? = ln(p1p2)ln? (p1p3)| Error| 1| ln(300. 4+760. 84760. 84)ln? (300. 4+760. 8475. 6+760. 84) = 1. 40| ± 3. 12| 2| ln(275. 7+760. 84760. 84)ln? (275. 7+760. 8469. 0+760. 84) = 1. 39| ± 3. 12| 3| ln(281. 9+760. 84760. 84)ln? (281. 9+760. 8474. 8+760. 84) = 1. 43| ± 3. 12| Carbon Dioxide Trial| ? = ln(p1p2)ln? (p1p3)| Error| 1| ln(290. 3+760. 84760. 84)ln? (290. 3+760. 8434. 1+760. 84) = 1. 16| ±3. 08| 2| ln(277. 8+760. 84760. 84)ln? (277. 8+760. 8425. 3+760. 84) = 1. 12| ± 3. 07| 3| ln(283. +760. 84760. 84)ln? (283. 1+760. 8440. 1+760. 84) = 1. 19| ±3. 08| Nitrogen Trial| ? = ln(p1p2)ln? (p1p3)|
Error| 1| ln(278. 7+760. 84760. 84)ln? (278. 7+760. 8463. 7+760. 84) = 1. 35| ± 3. 11| 2| ln(289. 0+760. 84760. 84)ln? (289. 0+760. 8464. 8+760. 84) = 1. 34| ± 3. 11| 3| ln(265. 3+760. 84760. 84)ln? (265. 3+760. 8459. 7+760. 84) = 1. 34| ± 3. 11| 4) Theoretical Cv for CO2 Translation = 3R2 = 3*8. 3142 = 12. 471 Jmol*K Rotational =22 R = 8. 314 Jmol*K Vibrational v1 = 4. 02 x 1013 s x = NA hRTv = 6. 02 x 1023*6. 63 x 10-348. 314*2984. 02 x 1013 = 6. 48 Therefore contribution = 8. 314(6. 48)2 * e-6. 48 =0. 54 Jmol*K 3 = 7. 05 x 1013 s x = NA hRTv = 6. 02 x 1023*6. 63 x 10-348. 314*298 7. 05 x 1013 = 11. 36 therefore contribution = 8. 314(11. 36)2 * e-11. 36 = 0. 013 Jmol*K v2 = v4 = 2. 00 x 1013 x = NA hRTv = 6. 02 x 1023*6. 63 x 10-348. 314*298 2. 00 x 1013 = 3. 22 This is less than 5. therefore contribution = 8. 314* 3. 222 *e3. 22e3. 22 -12 = 3. 74
Jmol*K Cv for CO2 = 12. 471 +8. 314 +0. 54 +0. 013 + 2(3. 74) = 28. 818 = 29. 0 Jmol*K 5) Cp,m = Cv,m + R so Cv,m = RCp,mCv,m- 1 Average experimental ? CO2 = 1. 16+1. 12+1. 193 = 1. 16 Therefore Experimental Cv,m = 8. 3141. 16 - 1 = 51. 96 = 52. 0 JK While Theoretical Cv,m = 8. 141. 29-1 = 28. 67 = 29. 0 JK Percentage error = 29 -5229*100 = 79% 6) Experimental ratio were precise but not accurate to the theoretical values as calculated Gas| Average ratio| Percentage error (%)| Helium| 1. 40+1. 39+1. 433 = 1. 41| 1. 67-1. 411. 67*100 = 15. 57| Nitrogen| 1. 34+1. 34+1. 353 = 1. 34| 1. 40-1. 341. 40*100 = 4. 29| Carbon dioxide| 1. 16+1. 12+1. 193 =1. 16| 1. 29-1. 161. 29*100 = 10. 08| Sources of experimental errors would include; leakage through the hose connecting the gas cylinder to the adiabatic vessel and the speed with which the brass cover plate is replaced after the gas expansion.
The vibrational contribution to Cv is very much dependent on the temperature. At low temperature, the contribution is zero. As the temperature increases, the lowest vibrational energy is comparable to RT and therefore some contribution to the constant volume heat capacity. While at high temperatures the contribution is at its highest. Conclusion The experiment was successful as the heat capacity ratios were achieved to minimal errors from the theoretical values. Reference 1. Thomas Engel, Physical Chemistry, 2nd Edition, Prentice Hall, 2010, pg 21-22, 806 - 807. 2. Lab manual for Chem 2103, experiment 1. 3. Tip for Experiment 1 on CUlearn.
