Question

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. What is the translational kinetic energy per mole of an ideal gas at (c) 27.8°C and (d) 143°C?

Answer 1

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c)  $k_{mol}=3.74\times 10^{3}J/mol$

d)   $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ($k_{avg}$) is given as

$k_{avg}=\frac{3}{2}\times kT$    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$  

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$  

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

       $k_{mol}=A_{v}\times k_{avg}$

here   $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        $k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

          $k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

        $k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

          $k_{mol}=5.1\times 10^{3}J/mol$