RMS Speed Questions

Standard Temperature and Pressure (STP) is:

pressure of 1 atmosphere = 1.01 x 105 Pa

temperature of 0°C = 273K

At STP 1 mole of ideal gas occupies 22.4 litres = 22.4 x 10-3 m3

 

Q1. Calculate the RMS speed of air molecules in a container in which the pressure is 1.0 x 105 Pa and the density of air is 1.3 kg m-3.

Density is mass/volume, and mass depends on the mass of each partile and the number of particles - Nm

so ρ = Nm/V and V = Nm/ρ

We have that:

pV = 1/3 Nm (crms)2

substituting into the equation:

pNm/ρ = 1/3 Nm (crms)2

cancelling, we get:

p/ρ = 1/3 (crms)2

(crms)2 = 3p/ρ

crms = √(3p/ρ)

crms = √(3.0 x 105/1.3 )

crms = 480 ms-1

 

Q2. The following table shows the distribution of the speeds of 20 particles:

Speed/ms-1 10 20 30 40 50 60
Number of particles 1 3 8 5 2 1
Squared Speed/m 2 s2 100 400 900 1600 2500 3600

 

Find

(a) the most probable speed,The most probable speed is the on that most particles are travelling at - the mode value:

30 m/s

(b) the average speed,

= (10 + 3 x 20 + 8 x 30 + 5 x 40 + 2 x 50 + 60)/20

= (10 + 60 + 240 + 200 + 100 + 60)/20

= 670/20 = 33.5 m/s

= 34 m/s (the answer should be given to 2 sf)

(c) the RMS speed.

First we square the speeds (see the table above)

Then we find the mean value of the square speeds:

= (100 + 3 x 400 + 8 x 900 + 5 x 1600 + 2 x 2500 x 3600)/20

= (100 + 1200 + 7200 + 8000 + 5000 + 3600)/20

= 25100/20

= 1255

Then we 'root' the number:

35.4 m/s

So, to 2 sf the answer is 35 m/s

Q3. The RMS speed of helium at STP is 1.30 km s-1. Calculate the density of helium at STP.

Density is mass/volume, and mass depends on the mass of each partile and the number of particles - Nm

so ρ = Nm/V and V = Nm/ρ

We have that:

pV = 1/3 Nm (crms)2

substituting into the equation:

pNm/ρ = 1/3 Nm (crms)2

cancelling, we get:

p/ρ = 1/3 (crms)2

ρ = 3p/ (crms)2

crms = 1.30 km s-1 = 1300 m s-1

ρ = 3 x 1.01 x 105/ (1300)2

ρ = 0.179 kg m-3

 

Q4. The RMS speed of nitrogen molecules at 127oC is 600 ms-1. Calculate the RMS speed at 1127oC.

½m (crms)2 = 3/2kT

so (crms)2 ∝ T

But we have to deal in kelvin temperatures!

T1 = 127 + 273 = 400K

T2 = 1127 + 273 = 1400K

6002/400 = x2/1400

x2 = 6002 x 1400/400 = 1 260 000

x = 1120 m/s

Q5. If the density of nitrogen at STP is 1.25 kg m-3, calculate the RMS speed of nitrogen at 227 °C.

Density is mass/volume, and mass depends on the mass of each partile and the number of particles - Nm

so ρ = Nm/V and V = Nm/ρ

We have that:

pV = 1/3 Nm (crms)2

substituting into the equation:

pNm/ρ = 1/3 Nm (crms)2

cancelling, we get:

p/ρ = 1/3 (crms)2

(crms)2 = 3p/ρ

At STP: (crms)2 = 3.03 x 105/1.25

Now:

½m (crms)2 = 3/2kT

so (crms)2 ∝ T

But we have to deal in kelvin temperatures!

T1 = 0 + 273 = 273K

T2 = 227 + 273 = 500K

3.03 x 105/1.25 x 273 = x2/500

x2 = 500 x 3.03 x 105/1.25 x 273

x = 666 m/s

Q6. Calculate the temperature in degrees Celsius at which the RMS speed of oxygen molecules is twice as great as their RMS speed at 27 °C.

½m (crms)2 = 3/2kT

so (crms)2 ∝ T

But we have to deal in kelvin temperatures!

T1 = 27 + 273 = 300K

(crms)2/300 = (2crms)2/T2

(crms)2/300 = 4(crms)2/T2

1/300 = 4/T2

T2 = 1200 K

T2 = 1200 - 273 = 927oC