At 47^@"C"47∘C, what is the average kinetic energy of "1 g"1 g of "O"_2(g)O2(g)?
1 Answer
Since we are at ordinary temperatures, oxygen gas might rotate and vibrate, but we should check. Thus, it has translational, rotational, and maybe vibrational degrees of freedom.
When we denote degrees of freedom as
K_(avg) = N/2 nRTKavg=N2nRT where:
nn is the mols of gas.R ="8.314472 J/mol"cdot"K"R=8.314472 J/mol⋅K is the universal gas constant.T = "47 + 273.15 K"T=47 + 273.15 K is the temperature in"K"K .
To check whether we are at high enough temperatures (the classical limit), you should ask yourself, is:
Theta_"vib" = tildeomega/k_B " << " T ? This is usually NOT true at ordinary temperatures.Theta_"rot" = tildeB/k_B " << " T ? This is usually true at ordinary temperatures.
where
tildeomega is the fundamental vibrational frequency andtildeB is the rotational constant.k_b = "0.695 cm"^(-1)//"K" is the Boltzmann constant.
For
Theta_"vib" = "1580.19 cm"^(-1)//"0.695 cm"^(-1)cdot"K" = ul("2273.7 K" " >> " T) color(red)(xx)
Theta_"rot" = "1.4377 cm"^(-1)//"0.695 cm"^(-1)cdot"K" = ul("2.0686 K" " << " T) color(green)(sqrt"")
In the classical limit (i.e. the temperature is much higher than
In practice, it is better to ignore the vibrational degrees of freedom because we are NOT in the high temperature limit for vibration (meaning we do NOT have
This means the total number of degrees of freedom is:
N = overbrace(3)^"Translational" + overbrace(2)^"Rotational" + overbrace(~~0)^"Vibrational" = 5
And so, the average kinetic energy for
color(blue)(K_(avg)) = 5/2 nRT
= 2.5("1 g" xx "1 mol"/"31.998 g")("8.314472 J/mol"cdot"K")("47 + 273.15 K")
= color(blue)"207.97 J"
