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 2 of 4.
Anil shukla said:
1 decade ago
Pressure energy is related to macro level of system of gas. It is related to entire gas system. Whereas kinetic energy is energy at micro level energy, means at individual molecule level. And individual molecules having different K.E. , so how to compare them.
Krishna said:
1 decade ago
Theory of Kinetic energy gives us: PV = 1/3 NC^2.
Where NC^2 is root mean velocity.
Now if you multiply and divide by 2.
It gives => PV = 2/3*1/2NC^2 (K.E).
Where NC^2 is root mean velocity.
Now if you multiply and divide by 2.
It gives => PV = 2/3*1/2NC^2 (K.E).
Manish kumar said:
10 years ago
What is the mass of ideal gas?
Ayobami said:
9 years ago
PV = 2/3 K.E.
Dhiraj said:
9 years ago
I think the answer is b.
Because relation is P = N/2 mv^2/V.
Because relation is P = N/2 mv^2/V.
Jayant said:
9 years ago
I can't understand. Please help me to understand the solution.
Sangram0015 said:
9 years ago
I can't understand how this relation comes out?
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.
NIKHIL DESHMUKH said:
8 years ago
KE=3/2KT.
BUT PV=MRT FOR unit mass.
pv = rt.
So,
K.E = 3/2P.
Therefore,
p = 2/3 K.E.
BUT PV=MRT FOR unit mass.
pv = rt.
So,
K.E = 3/2P.
Therefore,
p = 2/3 K.E.
Ejazurrahman said:
8 years ago
How you find KE=3/2P?
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