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The molecules in an ideal gas at 10∘C have a root-mean-square (rms) speed vrms. At what temperature T2 (in degrees Celsius) will the molecules have twice the rms speed, 2vrms?

Answer :

Answer:

Explanation:

The rms velocity of gas molecules is proportional to square root of absolute temperature , that is

rms velocity ∝ √T

V₁ / V₂ = [tex]\sqrt{\frac{T_1}{T_2} }[/tex]

T₁ = 10°C = 10 + 273

= 283

T₂ = ?

Substituting the given values

( 1/2 )²  = 283 / T₂

T₂ = 283 X 4

= 1132 K

= 1132 - 273

= 859°c

Manetho

Answer:

T2= 1132-273= 859°C

Explanation:

The root mean velocity of a molecule is given by

[tex]V_{rsm}=\sqrt{\frac{3RT}{M_0} }[/tex]

R= gas constant

T= temperature of the gas

Mo= molecular weight of the gas

⇒V_{rsm}∝√T

⇒[tex]\frac{V{rsm}}{2V_{rsm}} =\sqrt{\frac{T_0}{T_2} }[/tex]

T_0= 273+10=283K

[tex]\frac{V{rsm}}{2V_{rsm}} =\sqrt{\frac{283}{T_2} }[/tex]

solving we T2= 1132 K

therefore T2= 1132-273= 859°C

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