Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
An ideal gas at a pressure of 1 atmosphere and temperature of 27°C is compressed adiabatically until its pressure becomes 8 times the initial pressure. Then the final temperature is (γ=(3/2))
Q. An ideal gas at a pressure of 1 atmosphere and temperature of
2
7
°
C
is compressed adiabatically until its pressure becomes 8 times the initial pressure. Then the final temperature is
(
γ
=
2
3
)
5745
188
Thermodynamics
Report Error
A
62
7
°
C
14%
B
52
7
°
C
22%
C
42
7
°
C
19%
D
32
7
°
C
45%
Solution:
Here,
P
1
=
1
atm,
T
1
=
2
7
°
C
=
27
+
273
=
300
K
P
2
=
8
P
1
,
T
2
=
?
,
γ
=
2
3
As changes are adiabatic,
∴
P
1
γ
−
1
T
1
−
γ
=
P
2
γ
−
1
T
2
−
γ
(
T
1
T
2
)
−
γ
=
(
P
2
P
1
)
γ
−
1
T
2
=
T
1
(
P
1
P
2
)
γ
−
1/
γ
=
300
(
8
)
(
1.5
−
1
)
/1.5
=
300
(
8
)
1/3
T
2
=
600
K
=
(
600
−
273
)
°
C
=
32
7
°
C