Question

A cubic box with sides of 20.0 cm contains 2.00 × 1023 molecules of helium with a root-mean-square speed (thermal speed) of 200 m/s. The mass of a helium molecule is 3.40 × 10-27 kg. What is the average pressure exerted by the molecules on the walls of the container? (The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.314 J/mol•K .) (12 pts.)

Answer 1

Answer:

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Explanation:

Side of the cubic box = s = 20.0 cm

Volume of the box ,V= $s^3$

$V=(20.0 cm)^3=8000 cm^3=8\times 10^{-3} m^3$

Root mean square speed of the of helium molecule : 200m/s

The formula used for root mean square speed is:

$\mu=\sqrt{\frac{3kN_AT}{M}}$

where,

= root mean square speed

k = Boltzmann’s constant = $1.38\times 10^{-23}J/K$

T = temperature = 370 K

M = mass helium = $3.40\times 10^{-27}kg/mole$

$N_A$ = Avogadro’s number = $6.022\times 10^{23}mol^{-1}$

$T=\frac{\mu _{rms}^2\times M}{3kN_A}$

Moles of helium gas = n

Number of helium molecules = N =$2.00\times 10^{23}$

N = $N_A\times n$

Ideal gas equation:

PV = nRT

Substitution of values of T and n from above :

$PV=\frac{N}{N_A}\times R\times \frac{\mu _{rms}^2\times M}{3kN_A}$

$PV=\frac{N\times R\times \mu ^2\times M}{3k\times (N_A)^2}$

$R=k\times N_A$

$PV=\frac{N\times \mu ^2\times M}{3}$

$P=\frac{2.00\times 10^{23}\times (200 m/s)^2\times 3.40\times 10^{-27} kg/mol}{3\times 8\times 10^{-3} m^3}$

$P=1133.33 Pa =1.133 kPa$

(1 Pa = 0.001 kPa)

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.