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Question

The average translational kinetic energy and the root mean square speed of the molecules in a sample of oxygen gas at 300 K are 6.21×1021 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

A
2.42×1020 J and 968 m/s
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B
8.78×1020 J and 684 m/s
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C
6.21×1020 J and 968 m/s
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D
1.24×1020 J and 684 m/s
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Solution

The correct option is D 1.24×1020 J and 684 m/s
Given:
Initial temperature, T1=300 K
Initial average translational kinetic energy, KE1=6.21×1021 J
Initial root mean square speed, v1=484 m/s
Final temperature, T2=600 K
To find:
Final average translational kinetic energy, KE2=?
Final root mean square speed, v2=?

We know that, average translational kinetic energy is given by,
KE=nRT2
where,
n number of moles
R universal gas constant
T temperature

So, KE1KE2=T1T2
[ n remain same because gas is same.]
KE2=KE1×T2T1
KE2=6.21×1021×600300
KE2=1.24×1020 J

Now,
Root mean square speed is given by,
v=3RTM
So, v1v2=T1T2
[M remains same]
v2=v1×T2T1
v2=484×600300
v2684 m/s

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