
Calculate the Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. How close do they come to the experimental values from Table 1.5?

Interpretation:
The Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. is to be calculated. And is to be compared to the experimental values from Table 1.5.
Concept introduction:
The Boyle temperature is defined as the temperature for which the second virial coefficient, becomes zero. In other words, the Boyle temperature is the temperature at which a non-ideal gas behaves like an ideal gas. In this temperature the attractive and repulsive forces acting on the gas molecules balance each out.
Answer to Problem 1.38E
The Boyle temperature of gas are as follows;
Entry | Gas | TB (K)(calculated values) | TB (K)(experimental values) | a (atm L2 / mol2) | B (L / mol) |
1 | Carbon dioxide | 1026 | 713 | 3.592 | 0.04267 |
2 | Oxygen | 521 | 405 | 1.360 | 0.03183 |
3 | Nitrogen | 433 | 327 | 1.390 | 0.03913 |
The experimental values of Boyle temperature (TB) are compared with calculated values and found to be in low.
Explanation of Solution
The Boyle temperature (TB) is defined as the temperature for which the second virial coefficient, becomes zero. In other words, the Boyle temperature is the temperature at which a non-ideal gas behaves like an ideal gas. In this temperature the attractive and repulsive forces acting on the gas molecules balance each out. Thus ‘TB’ is given by the expression
‘a’ and ‘b’ are called as van der Waals constants and ‘a’ represents the pressure correction and it is related to the magnitude and strength of the interactions between gas particles. Similarly, ‘b’ describes the volume correction and it is having relationship to the size of the gas particles.
Since, the Boyle temperature is the temperature for which second virial coefficient becomes 0. Moreover, it is at this temperature that the forces of attraction and repulsion acting on the gas molecules balance out. Higher order be virial coefficients are smaller than the second virial coefficient, the gas likes to behave as an ideal gas over a entire range of pressures. At low pressures,
the equation transforms as,
Where, Z = Compressibility factor.
TB of carbon dioxide
Knowing a and b values of CO2 from table 1.6 TB of CO2 is calculated as follows;
R = 0.0823 L. atm / K. mol
TB of oxygen:
Knowing a and b values of oxygen from table 1.6 TB of oxygen is calculated as follows;
TB of nitrogen:
Knowing a and b values of nitrogen from table 1.6 TB of nitrogen is calculated as follows;
Moreover, the experimental Boyle temperature (table 1.5) values are compared with calculated values and the experimental values are found to be low.
Thus, the Boyle temperatures for carbon dioxide, oxygen, and nitrogen using the van der Waals constants in Table 1.6. is calculated. And compared to the experimental values from Table 1.5.
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- 6.arrow_forward0/5 alekscgi/x/sl.exe/1o_u-IgNglkr7j8P3jH-IQs_pBaHhvlTCeeBZbufuBYTi0Hz7m7D3ZcSLEFovsXaorzoFtUs | AbtAURtkqzol 1HRAS286, O States of Matter Sketching a described thermodynamic change on a phase diagram The pressure on a sample of pure X held at 47. °C and 0.88 atm is increased until the sample condenses. The pressure is then held constant and the temperature is decreased by 82. °C. On the phase diagram below draw a path that shows this set of changes. 3 pressure (atm) + 0- 0 5+ 200 temperature (K) 400 Explanation Check X 0+ F3 F4 F5 F6 F7 S 2025 McGraw Hill LLC All Rights Reserved. Terms of Use Privacy Center Accessibility Q Search LUCR + F8 F9 F10 F11 F12 * % & ( 5 6 7 8 9 Y'S Dele Insert PrtSc + Backsarrow_forward5.arrow_forward
- 9arrow_forwardalekscgi/x/lsl.exe/1o_u-IgNslkr7j8P3jH-IQs_pBanHhvlTCeeBZbufu BYTI0Hz7m7D3ZS18w-nDB10538ZsAtmorZoFusYj2Xu9b78gZo- O States of Matter Sketching a described thermodynamic change on a phase diagram 0/5 The pressure on a sample of pure X held at 47. °C and 0.88 atm is increased until the sample condenses. The pressure is then held constant and the temperature is decreased by 82. °C. On the phase diagram below draw a path that shows this set of changes. pressure (atm) 3- 200 temperature (K) Explanation Chick Q Sowncharrow_forward0+ aleksog/x/lsl.exe/1ou-lgNgkr7j8P3H-IQs pBaHhviTCeeBZbufuBYTOHz7m7D3ZStEPTBSB3u9bsp3Da pl19qomOXLhvWbH9wmXW5zm O States of Matter Sketching a described thermodynamic change on a phase diagram 0/5 Gab The temperature on a sample of pure X held at 0.75 atm and -229. °C is increased until the sample sublimes. The temperature is then held constant and the pressure is decreased by 0.50 atm. On the phase diagram below draw a path that shows this set of changes. F3 pressure (atm) 0- 0 200 Explanation temperature (K) Check F4 F5 ☀+ Q Search Chill Will an 9 ENG F6 F7 F8 F9 8 Delete F10 F11 F12 Insert PrtSc 114 d Ararrow_forward
- x + LEKS: Using a phase diagram a X n/alekscgi/x/lsl.exe/10_u-IgNsikr7j8P3jH-IQs_pBan HhvlTCeeBZbufu BYTI0Hz7m7D3ZcHYUt80XL-5alyVpw ○ States of Matter Using a phase diagram to find a phase transition temperature or pressure Use the phase diagram of Substance X below to find the melting point of X when the pressure above the solid is 1.1 atm. pressure (atm) 16 08- solid liquid- 0 200 400 gas 600 temperature (K) Note: your answer must be within 25 °C of the exact answer to be graded correct. × 5arrow_forwardS: Using a phase diagram leksogi/x/sl.exe/1ou-IgNs kr 7j8P3jH-IQs_pBan HhvTCeeBZbufuBYTI0Hz7m7D3ZdHYU+80XL-5alyVp O States of Matter Using a phase diagram to find a phase transition temperature or pressure se the phase diagram of Substance X below to find the boiling point of X when the pressure on the liquid is 1.6 atm. pressure (atm) 32- 16- solid liquid 0. gas 100 200 temperature (K) 300 Note: your answer must be within 12.5 °C of the exact answer to be graded correct. 10 Explanation Check § Q Search J 2025 McGraw Hill LLC. All Rights Researrow_forward151.2 254.8 85.9 199.6 241.4 87.6 242.5 186.4 155.8 257.1 242.9 253.3 256.0 216.6 108.7 239.0 149.7 236.4 152.1 222.7 148.7 278.2 268.7 234.4 262.7 283.2 143.6 QUESTION: Using this group of data on salt reduced tomato sauce concentration readings answer the following questions: 1. 95% Cl Confidence Interval (mmol/L) 2. [Na+] (mg/100 mL) 3. 95% Na+ Confidence Interval (mg/100 mL)arrow_forward
- Results Search Results Best Free Coursehero Unloc xb Success Confirmation of Q x O Google Pas alekscgi/x/lsl.exe/1o_u-IgNslkr 7j8P3jH-IQs_pBanHhvlTCeeBZbufu BYTI0Hz7m7D3ZcHYUt80XL-5alyVpwDXM TEZayFYCavJ17dZtpxbFD0Qggd1J O States of Matter Using a phase diagram to find a phase transition temperature or pressure Gabr 3/5 he pressure above a pure sample of solid Substance X at 101. °C is lowered. At what pressure will the sample sublime? Use the phase diagram of X below to nd your answer. pressure (atm) 24- 12 solid liquid gas 200 400 temperature (K) 600 ote: your answer must be within 0.15 atm of the exact answer to be graded correct. atm Thanation Check © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center I Q Search L³ ملةarrow_forward301.7 348.9 193.7 308.6 339.5 160.6 337.7 464.7 223.5 370.5 326.6 327.5 336.1 317.9 203.8 329.8 221.9 331.7 211.7 309.6 223.4 353.7 334.6 305.6 340.0 304.3 244.7 QUESTION: Using this group of data on regular tomato sauce concentration readings answer the following questions: 1. 95% Cl Confidence Interval (mmol/L) 2. [Na+] (mg/100 mL) 3. 95% Na+ Confidence Interval (mg/100 mL)arrow_forwardSearch Results Search Results Best Free Coursehero Unlo x b Success Confirmation of Q aleks.com/alekscgi/x/sl.exe/10_u-lgNslkr7j8P3jH-IQs_pBan HhvlTCeeBZbufu BYTIOHz7m7D3ZcHYUt80XL-5alyVpwDXM TEZayFYCav States of Matter Using a phase diagram to find a phase transition temperature or pressure Use the phase diagram of Substance X below to find the temperature at which X turns to a gas, if the pressure above the solid is 3.7 atm. pressure (atm) 0. 32- 16 solid liquid gas 200 temperature (K) Note: your answer must be within 20 °C of the exact answer to be graded correct. Дос Xarrow_forward
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