Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 1 (Q.No. 14)
14.
The pressure exerted by an ideal gas is __________ of the kinetic energy of all the molecules contained in a unit volume of gas.
Discussion:
32 comments Page 1 of 4.
Sach said:
7 years ago
Assuming monoatomic ideal gas, we have the degree of freedom as 3.
So U = KE = 3/2RT.
and RT = PV. (Ideal gas equation)
So,
KE = 3/2PV.
V = 1. (Given)
KE = 3/2P.
P = 2/3KI.
So U = KE = 3/2RT.
and RT = PV. (Ideal gas equation)
So,
KE = 3/2PV.
V = 1. (Given)
KE = 3/2P.
P = 2/3KI.
(8)
Mandeep Vishnoi said:
7 years ago
K.E = 3/2KT.
Where K is Boltzmann's constant and T is temperature.
Where K is Boltzmann's constant and T is temperature.
(5)
Msdprasad said:
7 years ago
Kinetic theory of gases PV = 2/3KE.
(1)
Sathish said:
8 years ago
How did you find KE=3/2 KT?
WHAT IS MEAN BY "KT"?
WHAT IS MEAN BY "KT"?
(1)
Rahul sharma said:
9 years ago
KE = 3/2kT.
KT = RT.
and RT = PV.
So,
KE = 3/2PV.
V = 1.
KE = 3/2P.
P = 2/3KI.
KT = RT.
and RT = PV.
So,
KE = 3/2PV.
V = 1.
KE = 3/2P.
P = 2/3KI.
Mykal baron said:
7 years ago
How can I get the weight of a substance?
Mykal baron said:
7 years ago
How can I get the weight of a substance?
Mykal baron said:
7 years ago
How can I get the weight of a substance?
Mykal baron said:
7 years ago
How can I get the weight of a substance?
Harsh said:
7 years ago
Pressure excerpted by an ideal gas p equal to. 2/3 of kinetic energy per molecule.
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