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©Selfstu
Relation
Let us cpressuregas rises
Given:
P = cons
α = Area
ΔT= rise
dx = The
dv = α d
Cp & Cv respecti
Q be the
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Step 1: T
Let Q1 b
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Let Q2 =
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Page 1
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Relation Between Cp and Cv
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 2
Since Q1 and Q2 are the part of the total heat supplied to the gas Q hence
Q = Q1 + Q2 (3)
But if we look at the change as a whole we find that Q heat rises the temperature of
1 kg mole of the gas by dT at constant pressure.
Q = 1. Cp.dT (4)
Putting equation (1), (2) and (4) in (3)
CpdT = CvdT + Pdv
Cp – Cv = P.dv/dT (5)
For perfect gas the equation is PV = nRT
For 1 kg Mole n = 1 PV = RT
Differentiating with P constant P dv/dT = R (6)
Putting equation (6) in (5) we get
Cp – Cv = R (7)
If M be the molecular weight of the gas and cp & cv are the ordinary specific heat of the gas at constant pressure and constant volume respectively then
Mcp – Mcv = R
cp – cv =R/M = γ
©Selfstu
Transfor
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(ii) Volum
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Let ΔQ = Am
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Page 3
nstant
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Relation Between Cp and Cv
©Selfstudy.in IEMS ‐ High School Tutorial Class Notes General Physics Page 4
Q1 = 1.Cv.dT Joules (1)
Step II: Temperature remains constant at T + dT and the volume of the gas increases by dv
Let Q2 = Amount of heat energy absorbed by gas in doing the mechanical work, in pushing the piston up by a distance dx against a force (Pxα)
Since both the parts of heat absorbed are obtained from the supplied heat
( ) ( ) (2) Joules PdvdxαPdxαPQ2 →=××=×=∴
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