Question

A sample of an ideal gas is slowly compressed to one-half its original volume with no change in pressure. If the original root-mean-square speed (thermal speed) of the gas molecules was V, the new speed is _____.(A) V/2.(B) V.(C) V/2(D)–√ 2V.(E) 2–√V.

Answer 1

Answer:

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$

Explanation:

From the kinetic theory model of gases we know that the velocity rms (speed of gas molecules) is given by:

$V_{rms}= \sqrt{\frac{3RT}{M}}$  (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 $PV= nRT$

If we solve for T we got: $T = \frac{PV}{nR}$

For the initial state we can replace T into the equation (1) and we got:

$V_{rms_i}= \sqrt{\frac{3R (\frac{PV}{nR})}{M}} = \sqrt{\frac{3PV}{M}}$

For the final state we know that :$V_f = \frac{V}{2}$ And the pressure not change , so then the final velocity would be:

$V_{rms_f}= \sqrt{\frac{3R (\frac{P(V/2)}{nR})}{M}} = \sqrt{\frac{3P(V/2)}{M}}$

$V_{rms_f}=\sqrt{\frac{3PV}{2nM}} = \sqrt{\frac{1}{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} \sqrt{\frac{3PV}{M}} = \frac{1}{\sqrt{2}} V_{rms_i}$

So then the final answer on this case would be:

$\frac{V_{rms_i}}{\sqrt{2}}$