Question

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 1

T2= 1132-273= 859°C

Step-by-step explanation:

The root mean velocity of a molecule is given by

`V_(rsm)=\sqrt{(3RT)/(M_0) }`

R= gas constant

T= temperature of the gas

Mo= molecular weight of the gas

⇒V_{rsm}∝√T

⇒
`\frac{V{rsm}}{2V_(rsm)} =\sqrt{(T_0)/(T_2) }`

T_0= 273+10=283K

`\frac{V{rsm}}{2V_(rsm)} =\sqrt{(283)/(T_2) }`

solving we T2= 1132 K

therefore T2= 1132-273= 859°C

Answer 2

Step-by-step explanation:

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

rms velocity ∝ √T

V₁ / V₂ =
`\sqrt{(T_1)/(T_2) }`

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