Question

A box contains about 6.07 x 1021 hydrogen

atoms at room temperature 21°C.

Find the thermal energy of these atoms.

Answer in units of J.

01. 31.3714942932129

2.31.7979030609131

O 3.43.3718528747559

4. 44.6510772705078

5.38.376781463623

6.35.391918182373

7. 42.2753715515137

O 8. 36.9757232666016

9. 32.1633949279785

10. 32.5898056030273

Answer 1

Answer:

E_T = 36.95J

6.  35.391918182373

Explanation:

To find the thermal energy of the total amount of molecules you use the following formula:

$E_T=\frac{3}{2}nk_BT$   ( 1 )

n: number of molecules = 6.07*10^21

T: temperature = 21°C = 294.15K

k_B: Boltzman's constant = 1.38*10^-23 J/K

You replace the values of the parameters in the equation (1):

$E_T=\frac{3}{2}(6.07*10^{21}( 1.38*10^{-23} J/K)(294.15K)\\\\E_T=36.95J$

hence, the thermal energy is 36.95J

6.  35.391918182373