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HACTEHA [7]
3 years ago
10

A balloon (approximate as a sphere) is inflated such that its

Chemistry
1 answer:
Likurg_2 [28]3 years ago
8 0

Answer: 1436 mL

Explanation: 4/3 • 3.14 • 7^3 <— calculate

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0.05 is the answer to your question
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3 years ago
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At high pressures how does the volume of a real gas compare with the volume of an ideal gas under the same conditions and why
malfutka [58]
Answer: No, a<span>t high pressures, volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

Reason: 
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, </span>volume of a real gas does not  compare with the volume of an ideal gas under the same conditions.

3 0
3 years ago
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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
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

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:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     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[/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

3 0
3 years ago
what's the density of an element if a sample has a mass of 43.2 g and a volume of 96.5 mL. (0.45 g/mL)
lisov135 [29]
To find the answer you need to use the formula that will help you to find the density. Density = mass/volume
d = 43.2g/96.5mL = 0.45g/mL
7 0
3 years ago
The answer to the question
VMariaS [17]
The correct answer should be Letter B
6 0
3 years ago
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