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Byju's Answer
Standard XII
Physics
Equipartition Theorem
Work done dur...
Question
Work done during isothermal expansion of one mole of an ideal gas from
10
a
t
m
to
1
a
t
m
at
300
K
is
A
4938.8
J
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B
4138.8
J
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C
5744.1
J
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D
6257.2
J
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Solution
The correct option is
C
5744.1
J
Work done during isothermal expansion-
Given,
n
=
1
T
=
300
k
p
1
=
10
a
t
m
p
2
=
1
a
t
m
W
=
2.303
n
R
T
log
p
1
p
2
⇒
2.303
×
1
×
8.314
×
300
×
log
10
=
5744.1
J
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