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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 302 K to 604 K. If v1 is the average speed of the gas molecules before the absorption of heat and v2 their average speed after the absorption of heat, what is the ratio v2/v1?

1.v2/v1 = 2.
2.v2/v1 = 4
3.v2/v1 = √2
4.v2/v1 = 2
5.v2/v1 =1/2

Respuesta :

Answer:3

Explanation:

Given

Initial Temperature [tex]T_1=302 K[/tex]

Final Temperature [tex]T_2=604 K[/tex]

root mean square velocity is given

[tex]V_{rms}=\sqrt{\frac{3k_bT}{m}}[/tex]

where [tex]T=[/tex]temperature

[tex]k_b=[/tex]constant

[tex]V_1=\sqrt{\frac{3k_b\cdot 302}{m}}**********1[/tex]

for [tex]T=604 K[/tex]

[tex]V_2=\sqrt{\frac{3k_b\cdot 604}{m}}********2[/tex]

dividing 1 & 2

[tex]\frac{V_2}{V_1}=\sqrt{\frac{604}{302}}[/tex]

[tex]\frac{V_2}{V_1}=\sqrt{2}[/tex]

adefioyelaoye

The ratio v2/v1 is √2.

What is an Ideal gas?

This is defined as a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions.

Parameters

Initial Temperature = 302K

Final Temperature = 604K

Root mean square velocity = √3KbT/m

V1 = √3kb × 302/m

V2 = √3kb × 604/m

V2/V1 = √604/302

          = √2

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