Respuesta :
Answer:
[tex] V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}[/tex]
So then the final answer on this case would be:
[tex] \frac{V_{rms_i}}{\sqrt{2}}[/tex]
Explanation:
From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:
[tex] V_{rms}= \sqrt{\frac{3RT}{M}}[/tex] (1)
Where V represent the velocity
R the constant for ideal gases
T the temperature
M the molecular weight of the gas
We also know from the ideal gas law that [tex] PV= nRT[/tex]
If we solve for T we got: [tex] T = \frac{PV}{nR}[/tex]
For the initial state we can replace T into the equation (1) and we got:
[tex] V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}[/tex]
For the final state we know that :[tex] V_f = \frac{V}{2}[/tex] And the pressure not change , so then the final velocity would be:
[tex] V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}[/tex]
[tex] V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}[/tex]
So then the final answer on this case would be:
[tex] \frac{V_{rms_i}}{\sqrt{2}}[/tex]
