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3 years ago

Help me please, it's so hard

Physics

2 answers:

fomenos3 years ago

8 0

Answer:

4.58×10²³ atoms

5.94×10⁻²¹ J

1340 m/s

Explanation:

Use ideal gas law to find moles of gas.

PV = nRT

(1.266 atm × 101300 Pa/atm) (4/3 π (0.15 m)³) = n (8.31451 J/mol/K) (14 + 273) K

n = 0.760 mol

Use Avogadro's number to find number of atoms.

(0.760 mol) (6.02214×10²³ atom/mol) = 4.58×10²³ atoms

Average kinetic energy per molecule is:

KE = 3/2 kT

KE = 3/2 (1.38066×10⁻²³ J/K) (14 + 273) K

KE = 5.94×10⁻²¹ J

RMS speed of each atom is:

KE = 1/2 mv²

5.94×10⁻²¹ J/atom = 1/2 (0.004 kg/mol) (1 mol / 6.02214×10²³ atom) v²

v = 1340 m/s

valentina_108 [34]3 years ago

6 0

1340 m/s is the answer and I got it how he got it^ but I’m too lazy to right it all down lol

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2 years ago

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3 years ago

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3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

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a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

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Let's calculate the maximum emission wavelength

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a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

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