- #1

BoanviaFx

- 30

- 0

## Homework Statement

Can someone confirm if I answered correctly? Thanks for support and help!

a) In the Brownian motion experiment, small particles of matter are seen moving randomly. Give an explanation for this motion.

b) Boyle's Law is fundamental when discussing the gas laws.

i) State Boyle's law.

ii) Draw a labelled diagram of the apparatus used in an experiment to verify this law.

iii) Sketch the graph which is obtained from such an experiment

c) The pressure exerted on a surface by gas molecules is given by P=1/3nm² where n is the number of molecules per unit volume, m is the mass of a molecule and c² is the mean square velocity of the molecules.

i) state four assumptions used to derive this equation

ii) Derive the above equation for the pressure exerted by gas molecules

d) A quantity of an ideal mono-atomic gas of density 1.2kg/m

^{3}occupies a volume of 1.2x10

^{-2}m

^{-3}at a pressure of 1x10

^{5}Pa and a temperature of 255K.

i) How many moles of gas are present?

ii) What is the internal energy of this gas?

iii) Calculate the root mean square speed of the molecules.

## Homework Equations

Molar gas constant, R=8.31Jmol

^{-1}K

^{-1}

## The Attempt at a Solution

a) The small particles of matter are seen moving randomly due to the bombardment of air particles traveling in random directions.

bi) For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.

ii)

iii)

ci) 1) The molecules are considered as hard, identical spheres undergoing rapid, random motion, their size being much smaller than their separations.

2) Their kinetic energy is assumed constant as all collisions inside the container are taken to be perfectly elastic in nature.

3) Intermolecular forces are assumed negligible except during collisions but then the time of collisions is assumed to be of insignificant magnitude.

4) Newtonian mechanics is perfectly applicable to the motion of the molecules.

ii) P=1/3ρc²

di) PV=nRT

1x10

^{5}*1.2x10

^{-2}=n*8.31Jmol

^{-1}K

^{-1}*255

n=0.566

ii) E=3/2nRT

E=3/2*0.566*8.31*255 =1800J

iii) Crms =

**√**(3RT/M)

ρ=m/v

m= 1.2*0.012 = 0.0144kg

Now to find Molar Mass

M = m/n

M = 0.0144/0.566

M = 0.0254

Crms =

**√**(3*8.31*255/0.0254)

Crms = 499m/s

Last edited: