Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is (α2/4) R J / mol K; then the value of α will be . (Assume that the given diatomic gas
Q. One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is
4
α
2
R
J
/
m
o
l
K
; then the value of
α
will be ____. (Assume that the given diatomic gas
64
0
JEE Main
JEE Main 2022
Kinetic Theory
Report Error
Answer:
3
Solution:
C
v
/
mi
x
=
n
1
+
n
2
n
1
C
v
1
+
n
2
C
v
2
=
1
+
3
1
⋅
2
3
R
+
3
⋅
2
5
R
=
4
9
R
=
4
α
2
R
α
=
3