Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
For silver, Cp(JK-1 mol-1) = 23 + 0.01T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of Δ H will be close to
Q. For silver,
C
p
(
J
K
−
1
m
o
l
−
1
)
=
23
+
0.01
T
. If the temperature (T) of
3
moles of silver is raised from
300
K
to
1000
K
at
1
a
t
m
pressure, the value of
Δ
H
will be close to
3298
236
JEE Main
JEE Main 2019
Thermodynamics
Report Error
A
21 kJ
21%
B
16 kJ
7%
C
13 kJ
11%
D
62 kJ
62%
Solution:
Δ
H
=
n
T
1
∫
T
2
C
p
,
m
d
T
=
3
×
∫
300
1000
(
23
+
0.01
T
)
d
T
=
3
[
23
(
1000
−
300
)
+
2
0.01
(
100
0
2
−
30
0
2
)
]
=
61950
J
≈
2
k
J