Question 43: (II) (a) What is the average translational kinetic energy of a nitrogen molecule at STP? (b) What is the total translational kinetic energy of 1.0 mol of molecules at 25°C?

The pressure of the nitrogen gas at STP is \(P = 1\;{\rm{atm}}. = 1.013 \times {10^{ - 5}}\;{\rm{Pa}}\).

The temperature of the nitrogen gas at STP is \(T = 0^\circ {\rm{C}} = 273\;{\rm{K}}\).

The number of moles of nitrogen gas is n = 1.0 mol.

The temperature is \(T' = 25^\circ {\rm{C}} = \left( {25 + 273} \right)\;{\rm{K}} = 29{\rm{8}}\;{\rm{K}}\).

According to the Kinetic theory of gases,the average translational kinetic energy of the molecules in random motion in an ideal gas is directly proportional to the absolute temperature of the gas.

The average translational kinetic energy is given as

\(\overline {KE} = \frac{3}{2}kT\).

Here, k is the Boltzmann constant, whose value is \(1.38 \times {10^{ - 23}}\;{\rm{J/K}}\).

The average translational kinetic energy of a nitrogen molecule at STP can be calculated in the following manner:

\(\begin{aligned}{c}\overline {KE} &= \frac{3}{2}kT\\ &= \frac{3}{2}\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {273\;{\rm{K}}} \right)\\ &= 5.65 \times {10^{ - 21}}\;{\rm{J}}\end{aligned}\)

Thus, the average translational kinetic energy of a nitrogen molecule at STP is \(5.65 \times {10^{ - 21}}\;{\rm{J}}\).

The average translational kinetic energy of a molecule at \(25^\circ {\rm{C}}\) can be calculated in the following manner:

\(\begin{aligned}{c}\overline {KE} &= \frac{3}{2}kT'\\ &= \frac{3}{2}\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {298\;{\rm{K}}} \right)\\ &= 616.9 \times {10^{ - 23}}\;{\rm{J}}\end{aligned}\)

One mole of an ideal gas contains an Avogadro number of molecules, that is,\({N_{\rm{A}}} = 6.02 \times {10^{23}}\;{\rm{molecules}}\).

The total translational kinetic energy of one mole of molecules at \(25^\circ {\rm{C}}\) can be calculated in the following manner:

\(\begin{aligned}{c}K{E_{{\rm{Total}}}} &= {{\rm{N}}_{\rm{A}}}\left( {\overline {KE} } \right)\\ &= \left( {6.02 \times {{10}^{23}}} \right)\left( {616.9 \times {{10}^{ - 23}}\;{\rm{J}}} \right)\\ &= 3713.7\;{\rm{J}}\\ &= 3.{\rm{7}} \times {10^3}\;{\rm{J}}\end{aligned}\)

Thus, the total translational kinetic energy of 1.0 mole of molecules at \(25^\circ {\rm{C}}\) is \(3.{\rm{7}} \times {10^3}\;{\rm{J}}\).