Question

A person is in a closed room (a racquetball court) with ????=453m3 hitting a ball (m=42.0g) around at random without any pauses. The average kinetic energy of the ball is 2.30 J. (a) What is the average value of ????2x? Does it matter which direction you take to be x? (b) Applying the methods of this chapter, find the average pressure on the walls? (c) Aside from the presence of only one "molecule" in this problem, what is the main assumption in Pressure, Temperature, and RMS Speed that does not apply here?

Answer 1

Answer:

(a) $v^{2} _{x} = 36.508 m^2/s^2$

(b)  0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of Van der Waal is greater than zero.

Explanation:

In the question, we are given the following variables:

V = 453 m^3

$m_{ball} = m = 42.0 g$

Kinetic energy (K.E) = 2.30 J

(a) The equation for kinetic energy is:

$K.E = \frac{1}{2}mv^{2}$

Since the average velocity components of the ball in the three dimensions are equal. Thus, we have:

$v_{x} ^{2} = v^{2} _{y} =v^{2} _{z}$

In addition:

$v^{2} =$$v_{x} ^{2} + v^{2} _{y} +v^{2} _{z}$ = $3v^{2} _{x}$

Therefore:

$K.E = \frac{1}{2}m[3v^{2} _{x}]$ = $\frac{3}{2}mv^{2} _{x}$

Thus:  

$v^{2} _{x} = \frac{KE}{1.5*m} = \frac{2.3}{1.5*0.042} = 36.508 m^2/s^2$

It does not matter which direction we take as x-direction because the average velocity component in every direction is the same.

(b) The average pressure on the walls can be calculated using the equation below:

$P =\frac{mv^{2} _{x} }{V}$

where V = 453 m^3

Thus: P = (0.042*36.508^2)/453 = 0.1235 Pa

(c) The volume of the ball is significant when compared with the volume of the closed room. Thus, the constant 'b' of the equation of van der Waal is greater than zero.