Chapter 19:The Kinetic Theory of Gases
(Fundamental of Physics, 10th edition)
• Lesson -6: Adiabatic expansion of an ideal gas
and related problems.
Lesson 6
19-9 Adiabatic expansion of an ideal gas: p𝑉 𝛾 = a constant
Suppose that you remove some shots from the
piston, allowing the ideal gas to push the piston
and the remaining shots upward and thus to
increase the volume by a differential amount dV.
Since the volume change is tiny, we may
assume that the pressure p of the gas on the
piston is constant during the change. The work
dW done by the gas during the volume increase
is equal to W = p dV.
adiabatic process, Q = 0
1st law of thermodynamics, ΔEint = Q – W
nCV ΔT = 0 – p dV
p dV
nΔT = –
CV
[ΔEint = Q – W = nCV ΔT – pΔV = nCV ΔT– p(V – V) = nCV ΔT– p(0) = nCV ΔT]
Ideal gas equation, pV = nRT
𝑑
𝑑
(pV) = (nRT)
𝑑𝑇
𝑑𝑇
p
𝑑
[𝑑𝑥 (uv) = u
𝑑𝑢
𝑑𝑉
𝑑𝑝
𝑑𝑇
+V
= nR
𝑑𝑇
𝑑𝑇
𝑑𝑇
𝑑𝑥
]
𝑝 𝑑𝑉+𝑉 𝑑𝑝
= nR
𝑑𝑇
𝑝 𝑑𝑉+𝑉 𝑑𝑝
= n dT
𝑅
𝑝 𝑑𝑉+𝑉 𝑑𝑝
p dV
=–
𝑅
CV
[ n dT = –
R
𝑝 𝑑𝑉 + 𝑉 𝑑𝑝 = – ( ) p dV
CV
p dV
]
CV
[Cp – CV = R]
𝑑𝑣
𝑑𝑥
+v
C – CV
C
𝑝 𝑑𝑉 + 𝑉 𝑑𝑝 = – ( p
) p dV = – ( p – 1) p dV = – (𝛾 – 1) p dV = – 𝛾 p dV + p
CV
dV
CV
𝑉 𝑑𝑝 = – 𝛾 p dV
𝑑𝑝
=–𝛾
𝑝
∫
𝑑𝑉
𝑉
𝑑𝑝
= – ∫𝛾
𝑝
[divided by pV]
𝑑𝑉
= –𝛾∫
𝑉
𝑑𝑉
𝑉
ln p + C1= - 𝛾 ln V + C2
ln p + 𝛾 ln V = C
ln p + ln 𝑉 𝛾 = C
ln (p𝑉 𝛾 ) = C
γ)
ln
(
pV
𝑒
= 𝑒C
p𝑉 𝛾 = a constant
[adiabatic expansion or contraction]
pi𝑉𝑖 𝛾 = pf𝑉𝑓 𝛾
19-9 T𝑉 𝛾−1 = constant for an adiabatic process:
For an adiabatic process, p𝑉 𝛾 = constant
To write an equation for an adiabatic process in terms of T and V, we use the ideal
gas equation to eliminate p
Ideal gas equation, pV = nRT
p=
nRT
V
nRT
(
) V γ = constant
V
T (
Vγ
constant
)
=
V1
nR
[n and R are constants]
TV γ−1 = constant
When the gas goes from an initial state i to a final state f: TiViγ−1 = TfVf γ−1
𝑪𝒗 , 𝑪𝒑 and 𝜸:
Types of Gas
𝑓
𝐶𝑣 = 𝑅
2
𝐶𝑝 = 𝐶𝑣 + 𝑅
𝑪𝒑
𝛾=
𝑪𝒗
Monoatomic
3
𝑅
2
5
𝑅
2
5
= 1.67
3
Diatomic
5
𝑅
2
7
𝑅
2
7
= 1.4
5
Polyatomic
3𝑅
4𝑅
4
= 1.33
3
19-9 W𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑓𝑜𝑟 𝑎𝑛 𝑖𝑑𝑒𝑎𝑙 𝑔𝑎𝑠 𝑖𝑛 𝑎𝑛 𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑝𝑟𝑜𝑐𝑒𝑠𝑠: 𝑊 =
𝑉
𝑊 = ∫𝑉 𝑓 𝑝𝑑𝑉 =
𝑖
𝑉 𝑎
= ∫𝑉 𝑓 𝛾 𝑑𝑉 = 𝑎
𝑖 𝑉
𝑉𝑓
𝑎
−𝛾+1
W=
[𝑉
]𝑉
−𝛾+1
𝑖
−𝛾+1
−𝛾+1
𝑎𝑉𝑓
− 𝑎𝑉𝑖
−𝛾+1
−𝛾+1
𝑎𝑉𝑓
W=
−
−𝛾+1
𝑎𝑉𝑖
𝑣𝑖
=
−𝛾+1
𝑃𝑓 𝑉𝑓𝛾 𝑉𝑓
−
−𝛾+1
𝑉𝑓
𝑉
−𝛾
𝑉 𝑑𝑉 = 𝑎[
]
−𝛾+1 𝑉𝑖
𝑎
−𝛾+1
=
(𝑉𝑓
−𝛾+1
−𝛾+1
𝛾−𝛾+1
𝑃𝑓 𝑉𝑓
− 𝑃𝑖𝑉𝑖𝛾−𝛾+1
𝑃𝑓 𝑉𝑓 −𝑃𝑖 𝑉𝑖
− (𝑃𝑖 𝑉𝑖 −𝑃𝑓 𝑉 )
𝑓
W = −𝛾+1
=
−𝛾+1
−(𝛾−1)
𝑃𝑖 𝑉𝑖 − 𝑃𝑓 𝑉𝑓
W=
𝛾−1
𝑣𝑓
−𝛾+1
𝑉𝑖
) =
−𝛾+1
𝑃𝑖 𝑉𝑖 𝑉𝑖
−𝛾+1
𝑃𝑖 𝑉𝑖 −𝑃𝑓 𝑉𝑓
𝛾−1
Adiabatic process
of
an ideal gas:
𝑎
𝛾
𝑝𝑉 = 𝑎𝑝 = 𝑉 𝛾
𝑃𝑖 𝑉𝑖 𝛾 = 𝑃𝑓 𝑉𝑓 𝛾 = a
=
55. A certain gas occupies a volume of 4.3 L at a pressure of 1.2 atm and a
temperature of 310 K. It is compressed adiabatically to a volume of 0.76 L.
Determine (a) the final pressure and (b) the final temperature, assuming the gas
to be an ideal gas for which γ = 1.4.
Solution:
Here, Vi = 4.3 L
(b) TV γ−1 = constant
TiViγ−1 = TfVf γ−1
pi = 1.2 atm = 1.2x105 Pa
Ti = 310 K
Vf = 0.76 L
𝛾 = 1.4
Tf =
TiViγ−1
Vfγ−1
=
𝑉𝑖 𝛾−1
𝑇𝑖 ( )
𝑉𝑓
4.3 L 1.4−1
)
0.76 L
= 310(
= 310(2.00) = 6
(a) p𝑉 𝛾 = constant
pi𝑉𝑖 𝛾 = pf𝑉𝑓 𝛾
pf =
pi𝑉𝑖 𝛾
𝑉𝑓𝛾
=pi(
1.36x106 Pa
𝑉𝑖 𝛾
)
𝑉𝑓𝛾
𝑉
𝑉𝑓
4.3 L 1.4
)
= 1.2x105 (11.3166) =
0.76 L
= 𝑝𝑖 ( 𝑖 )𝛾 = 1.2x105 (
Home work
54. We know that for an adiabatic process pVγ = a constant. Evaluate “a constant” for an
adiabatic process involving exactly 2.0 mol of an ideal gas passing through the state having
exactly p = 1.0 atm and T = 300 K. Assume a diatomic gas whose molecules rotate but do not
oscillate.
Solution:
Here, n = 2 mol
T = 300 K
P = 1.0 atm = 1.0x105 Pa
p𝑉 𝛾 = constant
𝑐𝑃
𝛾=
𝑐𝑣
Diatomic gas whose molecules rotate
but do not oscillate, f = 3+2 = 5
CV = (𝑓2)R = (52)R
Cp – CV = R
Cp = CV + R = (52)R + R = (72)R
𝑐𝑃
𝛾=
𝑐𝑣
=
7
𝑅
2
5
𝑅
2
=
7
5
= 1.4
p𝑉 𝛾 = constant
a = p𝑉 𝛾
nRT γ
a = p(
)
P
[Ideal gas law, pV = nRT]
2(8.31)(300) 1.4
5
= 1.0x10 {
}
1.0×105
𝑛𝑅𝑇
[V =
]
𝑃
= 1.0x105{0.04986}1.4
a =1.5x103 Nm2.2
N
𝑚
Unit of a = p𝑉 𝛾 = ( 2)(𝑚3 )γ
= 𝑁𝑚3𝛾−2 = 𝑁𝑚3 1.4 −2
= 𝑁𝑚2.2