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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 35
Total: 54
List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.
.
List-I
List-II
(I)
10
−
3
‌
kg
of water at
100
∘
C
is converted to steam at the same temperature, at a pressure of
10
5
‌
Pa
. The volume of the system changes from
10
−
6
m
3
to
10
−
3
m
3
in the process. Latent heat of water
=
2250
‌
kJ
∕
kg
.
(P)
2KJ
(II)
0.2
moles of a rigid diatomic ideal gas with volume
V
at temperature
500
K
undergoes an isobaric expansion to volume
3
V
. Assume
R
=
8.0
Jmol
−
1
K
−
1
.
(Q)
7KJ
(III)
On mole of a monatomic ideal gas is compressed adiabatically from volume
V
=
‌
1
3
m
3
and pressure
2
‌
kPa
to volume
‌
v
8
(R)
4KJ
(IV)
Three moles of a diatomic ideal gas whose molecules can vibrate, is given
9
‌
kJ
of heat and undergoes isobaric expansion
(S)
5KJ
(T)
3KJ
Which one of the following options is correct ?
[JEE Adv 2022 P1]
I
→
T
,
II
→
R
,
III
→
S
,
IV
→
Q
I
→
S
,
II
→
P
, III
→
T
,
IV
→
P
I
→
P
,
II
→
R
,
III
→
T
,
IV
→
Q
I
→
Q
,
II
→
R
,
III
→
S
,
IV
→
T
Validate
Solution:
(I)
∆
U
=
∆
Q
−
∆
W
=
{
(
10
−
3
×
2250
)
−
‌
10
5
(
10
−
3
−
10
−
6
)
10
3
}
‌
kJ
=
(
2.25
−
0.0999
)
‌
kJ
=
(
2.1501
)
‌
kJ
(II)
∆
U
=
nC
V
∆
T
=
‌
5
2
‌
nR
∆
T
=
‌
5
2
â‹…
(
0.2
)
(
8
)
(
1500
−
500
)
J
=
4
‌
kJ
(III)
P
1
V
2
γ
=
P
2
V
2
γ
⇒
2
(
‌
1
3
)
5
∕
3
=
P
2
(
‌
1
24
)
5
∕
3
⇒
P
2
=
64
‌
kPa
∆
U
=
nC
V
∆
T
=
‌
3
2
â‹…
(
P
2
V
2
−
P
1
V
1
)
=
‌
3
2
(
64
×
‌
1
24
−
2
×
‌
1
3
)
‌
kJ
=
3
‌
kJ
(IV)
∆
U
=
nC
v
∆
T
=
n
â‹…
‌
7
2
R
∆
T
=
‌
7
9
∆
Q
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