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.
Jitendra Harbola said:
1 decade ago
V^2 = 3RT/M = 3PV/M^2.
Put mass and volume equal to unity. Get the answer.
Put mass and volume equal to unity. Get the answer.
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)
Imran Ali said:
1 decade ago
To understand this relation please read kinetic theory of gases.
Jayant said:
9 years ago
I can't understand. Please help me to understand the solution.
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.
Dhinesh said:
1 decade ago
How the pressure is related K.E and please give the relation?
Sathish said:
8 years ago
How did you find KE=3/2 KT?
WHAT IS MEAN BY "KT"?
WHAT IS MEAN BY "KT"?
(1)
Ravi Uttamrao Chavan said:
1 decade ago
@Jitendra Harbolaji.
V^2 = 3RT/M, How this came?
V^2 = 3RT/M, How this came?
Sangram0015 said:
9 years ago
I can't understand how this relation comes out?
Mykal baron said:
7 years ago
How can I get the weight of a substance?
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