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JEE Main Physics Class 11 Kinetic Theory of Gases Part 1 Questions
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Question : 47
Total: 100
To raise the temperature of a certain mass of gas by 50°C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?
[Sep. 03, 2020 (II)]
5
6
3
7
Validate
Solution:
Let
C
p
and
C
v
be the specific heat capacity of the gas at constant pressure and volume.
At constant pressure, heat required
Δ
Q
1
=
n
C
p
Δ
T
⇒
160
=
n
C
p
.
50
...(i)
At constant volume, heat required
Δ
Q
2
=
n
C
v
Δ
T
⇒
240
=
n
C
v
.
100
...(ii)
Dividing (i) by (ii), we get
160
240
=
C
p
C
v
.
50
100
⇒
C
p
C
v
=
4
3
γ
=
C
p
C
v
=
4
3
=
1
+
2
f
(
Here
,
f
=
degree of freedom
)
⇒
f
=
6
.
© examsnet.com
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