Question

Calculate the most probable speed, the mean speed, and the mean relative speed of 'COs molecules at 20°C

Answer 1

`C_(mp)=332.8046\ m/s`

`C_(avg)=375.4542\ m/s`

`C_(rms)=407.6007\ m/s`

Step-by-step explanation:

The expression for the most probable speed is:

`C_(mp)=\sqrt {\frac {2RT}{M}}`

R is Gas constant having value = 8.314 J / K mol

M is the molar mass of gas

Given that : The gas is Carbon dioxide (Corrected and assumed)

Molar mass of
`CO_2` = 44.01 g/mol

Also, 1 g = 0.001 kg

So, Molar mass of
`CO_2` = 0.04401 kg/mol

Temperature = 20°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (20 + 273.15) K = 293.15 K

T = 293.15 K

Thus,

`C_(mp)=\sqrt {(2* 8.314* 293.15)/(0.04401)}`

`C_(mp)=\sqrt{(4874.4982)/(0.04401)}`

`C_(mp)=√(110758.87752)`

`C_(mp)=332.8046\ m/s`

The expression for the mean speed is:

`C_(avg)=\sqrt {(8RT)/(\pi M)}`

R is Gas constant having value = 8.314 J / K mol

M is the molar mass of gas

So, Molar mass of
`CO_2` = 0.04401 kg/mol

T = 293.15 K

Thus,

`C_(avg)=\sqrt{(8* 8.314* 293.15)/((22)/(7)* 0.04401)}`

`C_(avg)=\sqrt{(19497.9928)/(0.13831)}`

`C_(avg)=375.4542\ m/s`

The expression for the root mean square speed is:

`C_(rms)=\sqrt {\frac {3RT}{M}}`

R is Gas constant having value = 8.314 J / K mol

M is the molar mass of gas

Molar mass of
`CO_2` = 0.04401 kg/mol

Temperature 293.15 K

Thus,

`C_(rms)=\sqrt{(3* 8.314* 293.15)/(0.04401)}`

`C_(rms)=\sqrt{(7311.7473)/(0.04401)}`

`C_(rms)=407.6007\ m/s`