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 (c_rms)²

substituting into the equation:

pNm/ρ = ¹/₃ Nm (c_rms)²

cancelling, we get:

p/ρ = ¹/₃ (c_rms)²

(c_rms)² = 3p/ρ

c_rms = √(3p/ρ)

c_rms = √(3.0 x 10⁵/1.3 )

c_rms = 480 ms^-1

 

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

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 (c_rms)²

substituting into the equation:

pNm/ρ = ¹/₃ Nm (c_rms)²

cancelling, we get:

p/ρ = ¹/₃ (c_rms)²

ρ = 3p/ (c_rms)²

c_rms = 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 127^oC is 600 ms^-1. Calculate the RMS speed at 1127^oC.

½m (c_rms)² = ³/₂kT

so (c_rms)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 (c_rms)²

substituting into the equation:

pNm/ρ = ¹/₃ Nm (c_rms)²

cancelling, we get:

p/ρ = ¹/₃ (c_rms)²

(c_rms)² = 3p/ρ

At STP: (c_rms)² = 3.03 x 10⁵/1.25

Now:

½m (c_rms)² = ³/₂kT

so (c_rms)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 (c_rms)² = ³/₂kT

so (c_rms)² ∝ T

But we have to deal in kelvin temperatures!

T₁ = 27 + 273 = 300K

(c_rms)²/300 = (2c_rms)²/T₂

(c_rms)²/300 = 4(c_rms)²/T₂

1/300 = 4/T₂

T₂ = 1200 K

T₂ = 1200 - 273 = 927^oC