Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Thermodynamic standard conditions of temperature and pressure are
Q. Thermodynamic standard conditions of temperature and pressure are
2628
178
Thermodynamics
Report Error
A
0
∘
C
and
101.3
k
P
a
6%
B
298
K
and
1
a
t
m
56%
C
273
K
and
101.3
k
P
a
22%
D
0
∘
C
and
1
a
t
m
16%
Solution:
Thermodynamic standard conditions of temperature and pressure are
T
=
2
5
∘
C
=
25
+
273
=
298
K
,
P
=
1
atm