Question

The average translational kinetic energy
of 1.5 moles of oxygen at 21°C is(in joule)​

Answer 1

Given:

No of moles of oxygen $=1.5 moles$

Temperature $=21^{o}C$

To find:

The average Translational Kinetic energy.

Solution:

The Translational Kinetic energy, in general terms, is the energy of a body by virtue of the velocity it possesses.

 $KE=\frac{1}{2}mv^{2}$

For an ideal gas with mass $m$ in surrounding temperature $T$, the velocity is given by,

$v=\sqrt{\frac{3KT}{m} }$

where, $K$ is the Boltzmann Constant ($1.38$ × $10^{-23}m^{2} kg s^{-2} K^{-1}$).

Hence, the Translational Kinetic energy becomes

$KE=\frac{1}{2}m(\sqrt{\frac{3KT}{m} } )^{2}$

$KE=\frac{1}{2}m(\frac{3KT}{m} )$

$KE=\frac{3}{2}KT$

According to the question, we have

$T=21^{o}C=294K$

$n=1.5$

Substituting the required values, we get

$KE=\frac{3}{2}(1.38*10^{-23}) (294)$

$KE=\frac{3}{2}(405.72) 10^{-23}$

$KE=608.58$ × $10^{-23} J$

Final answer :

Hence, the average translational kinetic energy is 6.0858 × 10⁻²¹ J.

Answer 2

Answer:

Given:

No of moles of oxygen =1.5 moles=1.5moles

Temperature =21^{o}C=21

o

C

To find:

The average Translational Kinetic energy.

Solution:

The Translational Kinetic energy, in general terms, is the energy of a body by virtue of the velocity it possesses.

KE=\frac{1}{2}mv^{2}KE=

2

1

mv

2

For an ideal gas with mass mm in surrounding temperature TT , the velocity is given by,

v=\sqrt{\frac{3KT}{m} }v=

m

3KT

where, KK is the Boltzmann Constant (1.381.38 × 10^{-23}m^{2} kg s^{-2} K^{-1}10

−23

m

2

kgs

−2

K

−1

).

Hence, the Translational Kinetic energy becomes

KE=\frac{1}{2}m(\sqrt{\frac{3KT}{m} } )^{2}KE=

2

1

m(

m

3KT

)

2

KE=\frac{1}{2}m(\frac{3KT}{m} )KE=

2

1

m(

m

3KT

)

KE=\frac{3}{2}KTKE=

2

3

KT

According to the question, we have

T=21^{o}C=294KT=21

o

C=294K

n=1.5n=1.5

Substituting the required values, we get

KE=\frac{3}{2}(1.38*10^{-23}) (294)KE=

2

3

(1.38∗10

−23

)(294)

KE=\frac{3}{2}(405.72) 10^{-23}KE=

2

3

(405.72)10

−23

KE=608.58KE=608.58 × 10^{-23} J10

−23

J

Final answer :

Hence, the average translational kinetic energy is 6.0858 × 10⁻²¹ J.