enazer

28.08.2021 • 

Chemistry

Record your analysis of the information by answering the following questions. 고
• What organization is the author of the information?
• What seems to be the main purpose of the information? Is it to inform, persuade, or entertain?
Cite textual evidence from the reading to support your analysis.
• How is food irradiation performed? Describe each method.
• Using the knowledge you gained from this unit, identify parts of the text that convey similar
information as what you already know about radiation. Then, list any parts of the text that do
not match what you know about radiation.
• What are some benefits of food irradiation? Cite textual evidence from the reading, or conduct
your own research and cite those findings.
• What are some disadvantages of food irradiation? Cite textual evidence from the reading, or
conduct your own research and cite those findings.
• Do you think food irradiation should be used? Why or why not? List three facts from the reading
or your own research to support your point of view.
Works Cited
"Irradiation of Food." CDC. Centers for Disease Control and Prevention, 2009. Web.
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Ответ:

rleiphart1

01.05.2023

Chemistry

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(a) Temperature = 330 K, and mass = 0.321 kg

(b) T₂ = 171.56 K, mass = 0.32223 kg

Explanation:

For a constant temperature process we have

p₁v₁ = p₂v₂

Where p₁ = initial pressure = 10 bar = 1000000 Pa

p₂ = final pressure = 1 atm = 101325 Pa

v₁ initial volume = 0.3 m³

v₂ = final volume = unknown

From the relation we have v₂ = 2.96 m³

Therefore at constant temperature 2.93 m³ - 0.3 m³ or 2.66 m³ will be expelled from the container

Temperature = 330 K, and mass =

Also from the relation p1v1 = mRT1

We have, (1000000×0.3)/(8314×330) = 109..337 mole

For air mass

Mass = 3.171 kg

After opening we have

p2v2/(RT1) = n2 = 11.07 mol or 0.321 kg

or

(b) This is said to be adiabatic condition hence

Here

But cp = 29 (J/mol K).

and p₁v₁ = RT₁ therefore R = 1000000*0.3/330 = 909.1 J/mol·K

And For perfect gas γ = 1.4

Hence T₂ = 171.56 K

γ =cp/cv therefore cv=cp/γ = 29/1.4 = 20.714 (J/mol K). and R =cp-cv = 8.29 J/mol·K

Therefore p1v1/(RT1) = 109.66 moles and we have

p2v2/(R×T2) = 11.11 mole left

For air that is 0.32223 kg

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