Give expression for mono/ diatomic gases
Answers
Answer:
Explanation:
SPECIFIC HEAT CAPACITY OF MONATOMIC GAS - FORMULA
The molecules of a monatomic gas have 3 degrees of freedom.
The average energy of a molecule at temperature T is
2
3K
B
T
.
The total internal energy of a mole is:
2
3K
B
T
×N
A
.
The molar specific heat at constant volume C
v
is
For an ideal gas,
C
v
(monatomicgas)=
dT
dU
=
2
3RT
FOr an ideal gas, C
p
−C
v
=R
where C
p
is molar specific heat at constant pressure.
Thus, C
p
=
2
5R
SPECIFIC HEAT CAPACITY OF DIATOMIC GAS - FORMULA
The molecules of a monatomic gas have 5 degrees of freedom, 3 translational and 2 rotational.
The average energy of a molecule at temperature T is
2
5K
B
T
.
The total internal energy of a mole is:
2
5K
B
T
×N
A
.
The molar specific heat at constant volume C
v
is
For an ideal gas,
C
v
(monatomicgas)=
dT
dU
=
2
5RT
FOr an ideal gas, C
p
−C
v
=R
where C
p
is molar specific heat at constant pressure.
Thus, C
p
=
2
7R
Answer:
SPECIFIC HEAT CAPACITY OF MONATOMIC GAS - FORMULA
The molecules of a monatomic gas have 3 degrees of freedom.
The average energy of a molecule at temperature T is
2
3K
B
T
.
The total internal energy of a mole is:
2
3K
B
T
×
N
A
.
The molar specific heat at constant volume
C
v
is
For an ideal gas,
C
v
(monatomicgas)=
d T
d U
=
2
3RT
FOr an ideal gas,
C
p
C
v
=
R
where
C
p
is molar specific heat at constant pressure.
Thus,
C
p
2
5R
SPECIFIC HEAT CAPACITY OF DIATOMIC GAS - FORMULA
The molecules of a monatomic gas have 5 degrees of freedom, 3 translational and 2 rotational.
The average energy of a molecule at temperature T is
2
5K
B
T
.
The total internal energy of a mole is:
2
5K
B
T
×
N
A
.
The molar specific heat at constant volume
C
v
is
For an ideal gas,
C
v
(monatomicgas)=
d T
d U
=
2
5RT
FOr an ideal gas,
C
p
C
v
=
R
where
C
p
is molar specific heat at constant pressure.
Thus,
C
p
2
7R
SPECIFIC HEAT CAPACITY OF POLYATOMIC GAS - FORMULA
In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f) of vibrational modes.
According to the law of equipartition of energy, it is easily seen that one mole of such a gas has
U=
(
2
3K
B
T
+
2
3K
B
T
+
fK
B
T)N
A
or
C
v
=
(3+
f)R
,
C
v
=
(4+
f)R
Explanation:
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