Answer:
![PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%20%2B%20%28b-%5Cfrac%7Ba%7D%7BRT%7D%29%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV%5E%7B2%7D%20_%7Bm%7D%20%7D%20%2B%20...%5D)
B = b -a/RT
C = b^2
a = 1.263 atm*L^2/mol^2
b = 0.03464 L/mol
Explanation:
In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:
Using the van deer Waals equation of state:

With further simplification, we have:
![P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D-b%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%5E%7B2%7D%20%7D%5D)
Then, we have:
![P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BRT%7D%7BV_%7Bm%7D%20%7D%20%5B%5Cfrac%7B1%7D%7B1-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%20%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Therefore,
![PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B%281-%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%29%20%5E%7B-1%7D%20-%20%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Using the expansion:

Therefore,
![PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%5Cfrac%7Bb%7D%7BV_%7Bm%7D%20%7D%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%20-%5Cfrac%7Ba%7D%7BRTV_%7Bm%7D%20%7D%5D)
Thus:
equation (1)
Using the virial equation of state:
![P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]](https://tex.z-dn.net/?f=P%20%3D%20RT%5B%5Cfrac%7B1%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%5E%7B2%7D%7D%2B%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B3%7D%20%7D%2B%20...%5D)
Thus:
equation (2)
Comparing equations (1) and (2), we have:
B = b -a/RT
C = b^2
Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.
[/tex] = 0.03464 L/mol
a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2
Answer:
62.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2C₈H₁₈ + 25O₂ —> 16CO₂ + 18H₂O
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Finally, we shall determine the number of mole of O₂ needed to react with 5 moles of C₈H₁₈. This can be obtained as shown below:
From the balanced equation above,
2 moles of C₈H₁₈ reacted with 25 moles of O₂.
Therefore, 5 moles of C₈H₁₈ will react with = (5 × 25) / 2 = 62.5 moles of O₂.
Thus, 62.5 moles of O₂ is needed for the reaction.
Answer:
Quantity A is weight and Quantity B is mass
Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object
I have posted the answetr in this picture.
Answer: By looking at the periodic table, you will see that Fluorine has 9 protons. Since the number of electrons equal the number of protons, Fluorine has 9 electrons as well. Meanwhile, it's mass number of 19, minus 10 neutrons, gives you 9 protons or electrons. Hence, the atom would be Fluorine.
Explanation: