Answer:
the ratio is [tex]\frac{V_2}{V_1}=\sqrt{2}[/tex]
Explanation:
Given
[tex]Initial Temperature T_1=387 KFinal Temperature T_2=774 K[/tex]
The RMS velocity of molecules in a gas is given by
[tex]V_{rms}=\sqrt{\dfrac{3k_bT}{m}}[/tex]
where T=temperature
[tex]k_b=constant[/tex]
For T = 387K
[tex]V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1[/tex]
For T = 774
[tex]V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)[/tex]
dividing eqn 1 and eqn 2
[tex]\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}[/tex]
[tex]\frac{V_2}{V_1}=\sqrt{2}[/tex]
Thus,the ratio is [tex]\frac{V_2}{V_1}=\sqrt{2}[/tex]
Answer:
Ratio v₂/v₁ = √2
Explanation:
We are given that ;
Initial temperature (T1) = 387K
Final temperature (T2) = 774K
Now, according to kinetic theory of gases, the average speed is proportional to the square root of the absolute temperature. The speed is given by:
v = √( 3∙R∙T/M)
Where;
R = universal gas constant,
T = absolute temperature,
M = molar mass of the gas molecules
Thus, v₁ = √(3∙R∙T₁/M)
Since it's same gas,
v₂ = √( 3∙R∙T₂/M)
Hence,
Ratio v₂/v₁ = √( 3∙R∙T₂/M)/√( 3∙R∙T₁/M) = √(T₂/T₁) = √(774K/387K) = √2
