Chapter 20: Q. 44 (page 568)

a. Find an expression for the $v_{rms}$ of gas molecules in terms of $p$ , $V$ and the total mass of the gas $M$ .
b. A gas cylinder has a piston at one end that is moving outward at speed $v_{piston}$ during an isobaric expansion of the gas. Find an expression for the rate at which is changing in terms of $v_{piston}$ , the instantaneous value of $v_{rms}$ , and the instantaneous value $L$ of the length of the cylinder.
c. A cylindrical sample chamber has a piston moving outward at $0.50 m / s$ during an isobaric expansion. The rms speed of the gas molecules is localid="1648640672000" $450 m / s$ at the instant the chamber length is localid="1648640676590" $1.5 m$ . At what rate is localid="1648640708264" $v_{rms}$ changing?

Short Answer

Expert verified

a.Expression for $v_{rms}$ is $\sqrt{\frac{3 pV}{M}}$ .

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

c.Rate of change of $v_{rms}$ is $75 m / s^{2}$ .

Step by step solution

Expression forvrms(part a)

The expression for the rms speed is,

where $m$ is the mass of the particle.

Ideal gas law:

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

This results in,

$v_{rms} = \sqrt{\frac{3 pV}{mN}}$

$M = mN$

Then,

$v_{rms} = \sqrt{\frac{3 pV}{M}}$

Expression forvrmsinterms of vpiston (part b)

The speed of the piston is $v_{p}$ and not $v_{piston}$ ,

The time derivative of the rms speed is ,

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

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

$V = A L$

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

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

$A = V / L$

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

$\frac{d}{dt} v_{rms} = \frac{v_{rms} (t) v_{p}}{2 L}$

Calculation for rate of change of vrms (part c)

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

$\frac{d}{dt} v_{rms} = \frac{450 \cdot 0.5}{2 \cdot 1.5} = 75 m / s^{2}$

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Short Answer

Step by step solution

Expression forvrms(part a)

Expression forvrmsinterms of vpiston (part b)

Calculation for rate of change of vrms (part c)

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