Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
One mole of a non-ideal gas undergoes a change of state (.2.0 atm,3.0 L,95 K.) arrow (.4 atm, 5 L, 245 K.) with a change in internal energy, Δ E=30.0 L atm. The change in enthalpy, Δ H , of the process in L atm is
Q. One mole of a non-ideal gas undergoes a change of state
(
2.0
a
t
m
,
3.0
L
,
95
K
)
→
(
4
a
t
m
,
5
L
,
245
K
)
with a change in internal energy,
Δ
E
=
30.0
L
a
t
m
.
The change in enthalpy,
Δ
H
, of the process in L atm is
5043
239
NTA Abhyas
NTA Abhyas 2022
Report Error
A
40.0
B
42.3
C
44.0
D
not defined, because pressure is not constant
Solution:
Δ
H
=
Δ
E
+
Δ
(
P
V
)
Δ
H
=
30
+
(
4
×
5
−
2
×
3
)
=
44
L
a
t
m