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:
![PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]](https://tex.z-dn.net/?f=PV_%7Bm%7D%20%3D%20RT%5B1%2B%20%5Cfrac%7BB%7D%7BV_%7Bm%7D%20%7D%2B%20%5Cfrac%7BC%7D%7BV_%7Bm%7D%20%5E%7B2%7D%20%7D%20%2B%20...%5D)
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.
![b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%7BC%7D%20%3D%20%5Csqrt%7B1200%7D%20%3D%2034.64%5Btex%5Dcm%5E%7B3%7D%2Fmol)
[/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
The buoyancy of an object is dictated by its density. So let us calculate for density, where:density = mass / volume
Calculate the volume first of a solid cube:volume = (6 cm)^3 = 216 cm^3 = 216 mL
Therefore density is:density = 270 g / 216 mLdensity = 1.25 g / mL
Therefore this object will float in the layer in which the density is more than 1.25 g / mL.
When a specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state, this emitted energy can
<span>be used to determine the "identity of the element".</span>
Answer:
The ocean currents are too strong by the Amazon River to form deltas.
Explanation:
The Atlantic has sufficient wave and tidal energy to carry most of the Amazon's sediments out to sea, thus the Amazon does not form a true delta. The great deltas of the world are all in relatively protected bodies of water, while the Amazon empties directly into the turbulent Atlantic.
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
