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

a)
`k_(avg)=6.22* 10^(-21)`

b)
`k_(avg)=8.61* 10^(-21)`

c)
`k_(mol)=3.74* 10^(3)J/mol`

d)
`k_(mol)=5.1* 10^(3)J/mol`

Step-by-step explanation:

Average translation kinetic energy (
`k_(avg)`) is given as

`k_(avg)=(3)/(2)* 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)=(3)/(2)* 1.38* 10^(-23)* 300.8`

`k_(avg)=6.22* 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)=(3)/(2)* 1.38* 10^(-23)* 416`

`k_(avg)=8.61* 10^(-21)J`

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

`k_(mol)=A_(v)* k_(avg)`

here
`A_(v)` = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

`k_(mol)=6.02* 10^(23)* 6.22* 10^(-21)`

`k_(mol)=3.74* 10^(3)J/mol`

d) now at T = 143° C

`k_(mol)=6.02* 10^(23)* 8.61* 10^(-21)`

`k_(mol)=5.1* 10^(3)J/mol`