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.
Suman said:
8 years ago
What is rt and kt?
Abishek said:
8 years ago
RT=KT AND RT=PV.
SO KE=3/2KT,
THEN KE = 3/2 PV.
Where v is unit volume =1.
So KE = 3/2 P(1).
SO KE=3/2KT,
THEN KE = 3/2 PV.
Where v is unit volume =1.
So KE = 3/2 P(1).
Ejazurrahman said:
8 years ago
How you find KE=3/2P?
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.
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.
Sangram0015 said:
9 years ago
I can't understand how this relation comes out?
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.
Ayobami said:
9 years ago
PV = 2/3 K.E.
Manish kumar said:
10 years ago
What is the mass of ideal gas?
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