Question

a. Find an expression for the${\mathrm{v}}_{\mathrm{rms}}$of gas molecules in terms of$\mathrm{p}$,$\mathrm{V}$and the total mass of the gas $\mathrm{M}$.

b. A gas cylinder has a piston at one end that is moving outward at speed $v_{\text{piston}}$during an isobaric expansion of the gas. Find an expression for the rate at which is changing in terms of ${\mathrm{v}}_{\text{piston}}$, the instantaneous value of ${\mathrm{v}}_{\text{rms}}$, and the instantaneous value $\mathrm{L}$of the length of the cylinder.

c. A cylindrical sample chamber has a piston moving outward at $0.50\mathrm{m}/\mathrm{s}$during an isobaric expansion. The rms speed of the gas molecules is localid="1648640672000" $450\mathrm{m}/\mathrm{s}$at the instant the chamber length is localid="1648640676590" $1.5\mathrm{m}$. At what rate is localid="1648640708264" ${\mathrm{v}}_{\mathrm{rms}}$changing?

Short answer

a.Expression for${\mathrm{v}}_{\mathrm{rms}}$is$\sqrt{\frac{3\mathrm{pV}}{\mathrm{M}}}$.

b.Expression for the rate at which${\mathrm{v}}_{\mathrm{rms}}$is changing in terms of ${\mathrm{v}}_{\mathrm{piston}}$is $\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{Tms}}=\frac{{\mathrm{v}}_{\mathrm{rms}}\left(\mathrm{t}\right){\mathrm{v}}_{\mathrm{p}}}{2\mathrm{L}}$.

c.Rate of change of${\mathrm{v}}_{\mathrm{rms}}$is$75\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Expression forvrms(part a)

a.

The expression for the rms speed is,

where $\mathrm{m}$is the mass of the particle.

Ideal gas law:

$\mathrm{pV}={\mathrm{Nk}}_{\mathrm{B}}\mathrm{T}\Rightarrow {\mathrm{k}}_{\mathrm{B}}\mathrm{T}=\frac{\mathrm{pV}}{\mathrm{N}}$

This results in,

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{pV}}{\mathrm{mN}}}$

$\mathrm{M}=\mathrm{mN}$

Then,

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{pV}}{\mathrm{M}}}$

Expression forvrmsinterms of vpiston (part b)

b.

The speed of the piston is${\mathrm{v}}_{\mathrm{p}}$and not ${\mathrm{v}}_{\text{piston}}$,

The time derivative of the rms speed is ,

$\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{p}}{\mathrm{M}}}\frac{\mathrm{d}}{\mathrm{dt}}\sqrt{\mathrm{V}}=\frac{1}{2\sqrt{\mathrm{V}}}\sqrt{\frac{3\mathrm{p}}{\mathrm{M}}}\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{V}$

Substitute the volume as the product of the cross section area $\mathrm{A}$of the piston times the instantaneous value$\mathrm{L}$of its position.

$\mathrm{V}=\mathrm{A}\mathrm{L}$

$\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{V}=\mathrm{A}\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{L}={\mathrm{Av}}_{\mathrm{p}}$

$\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\frac{1}{2\sqrt{\mathrm{V}}}\sqrt{\frac{3\mathrm{p}}{\mathrm{M}}}{\mathrm{Av}}_{\mathrm{p}}$

$\mathrm{A}=\mathrm{V}/\mathrm{L}$

$\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\frac{1}{2\sqrt{\mathrm{V}}}\sqrt{\frac{3\mathrm{p}}{\mathrm{M}}}\frac{\mathrm{V}}{\mathrm{L}}{\mathrm{v}}_{\mathrm{p}}=\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\frac{1}{2}\sqrt{\frac{3\mathrm{pV}}{\mathrm{M}}}{\mathrm{Lv}}_{\mathrm{p}}$

$\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\frac{{\mathrm{v}}_{\mathrm{rms}}\left(\mathrm{t}\right){\mathrm{v}}_{\mathrm{p}}}{2\mathrm{L}}$

Calculation for rate of change of vrms (part c)

c.

Taking the conclusion obtained in part (b) and applying it numerically, we have

$\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{v}}_{\mathrm{rms}}=\frac{450·0.5}{2·1.5}=75\mathrm{m}/{\mathrm{s}}^2$