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(a) What is the average kinetic energy in joules of hydrogen atoms on the 5500ºC surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is $6.00 \times 10^{5} \mathrm{K} ?$

(a)$1.2 \times 10^{-19} \mathrm{~J}$(b) $1.25 \times 0^{-17} \mathrm{~J}$

Physics 101 Mechanics

Chapter 13

Temperature, Kinetic Theory, and the Gas Laws

Temperature and Heat

Cornell University

Simon Fraser University

Hope College

Lectures

04:16

In mathematics, a proof is…

04:48

In mathematics, algebra is…

01:42

What is the average kineti…

01:53

What are (a) the average k…

02:12

(I) What is the average ki…

02:32

The product of the pressu…

00:33

01:19

05:05

(a) How many atoms of heli…

04:19

04:33

The Fermi energy of sodium…

03:26

(I) $(a)$ What is the aver…

06:17

A helium-filled balloon ha…

07:52

01:30

01:07

Much of the gas near the S…

02:54

01:04

What is the average transl…

01:02

Oxygen atoms are an import…

00:55

01:54

The temperature near the c…

00:28

The average power generate…

So, for part a we know that the temperature is 5500 degrees celsius, which means it is 5500 plus 273 kelvins, 723 kelvin, and we want to find out the energy of a hydrogen molecules. It'S very simple energy, if you remember of our not having a particle, is 3 by 2 k. So that's 3 by 2 times, if you remember woldsman constant, is r by avocation number. So that's 8.314 joule per mole kelvin divided by 6 into 10 part 23 per mole times 5723, which turns out to be 12 into 10 power. Minus 20 joules 1.210 power minus 19 for part b. We are. Temperature is 6 into 10 power 5 instead and notice that if you add 273 to it, it basically remains the same, but you don't need to because it's already in kelvin now the energy again is 3 by 2 k t so that's the again 3 by 2 Into 8.314 by 6 into 10 power, 23 into 6 times 10 power 5. It stands out to be 12.5 into 10 power. Minus 18 joules, which is 1.25 into 10 power. Minus 17 joulet.

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