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calculate the kinetic energy of translation of molecules of nitrogen gas at STP moving with the most probable speed ​

Sagot :

PreetKumarSangdot

Answer:

1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at 27

∘

C is given as

E

T

​

=

2

3

​

kT=

2

3

​

×1.38×10

−23

×300

=6.21×10

−21

J/mol

2.  An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given as

E

T

​

=

2

5

​

kT=

2

5

​

×1.38×10

−23

×300

=10.35×10

−21

J/mol

3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given as

U=

2

f

​

μRT=

2

5

​

×1×8.314×300=6235.5J/mol1. An oxygen molecule has three translational degrees of freedom, thus the average translational kinetic energy of an oxygen molecule at 27

∘

C is given as

E

T

​

=

2

3

​

kT=

2

3

​

×1.38×10

−23

×300

=6.21×10

−21

J/mol

2.  An oxygen molecule has total five degrees of freedom, hence its total kinetic energy is given as

E

T

​

=

2

5

​

kT=

2

5

​

×1.38×10

−23

×300

=10.35×10

−21

J/mol

3. Total kinetic energy of one mole of oxygen is its internal energy, which can be given as

U=

2

f

​

μRT=

2

5

​

×1×8.314×300=6235.5J/mol

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