The notes are for the thermodynamics in chemistry and chemical engineering. They explain the adiabatic processes in detail and also they consists of worked examples and self test exercises
(2)
For an adiabatic reversible process (expansion or compression) the governing formula is:
𝑃𝑉 𝛾 = 𝐶
Rearranging in terms of P:
𝐶
𝑃=
𝑉𝛾
Followed by substitution in (2):
𝑉2
1
𝑤 = −𝐶 ∫ 𝑑𝑉
𝑉1 𝑉𝛾
(3)
Solve integral by applying power rule:
∫ n
x n+1
ax dx = a +C
n+1
𝑉2
𝑤 = −𝐶 ∫ 𝑉 −𝛾 𝑑𝑉
𝑉1
(4)
1 𝑉
𝑤 = −𝐶[ 𝑉 1−𝛾 ]𝑉21
1−𝛾
(5)
, 2
1 1
𝑤 = −𝐶[ 𝑉2 1−𝛾 − 𝑉 1−𝛾 ]
1−𝛾 1−𝛾 1
𝐶 1−𝛾
𝑤=− [𝑉2 − 𝑉1 1−𝛾 ]
1−𝛾
(6)
You may further want to get rid of the negative sign. Since 1-γ = -(γ-1), the equation then
changes to:
𝐶 1−𝛾
𝑤= [𝑉 − 𝑉1 1−𝛾 ]
𝛾−1 2
(7)
Work can also be evaluated in terms of the temperature change, since
𝑑𝑈 = 𝑑𝑤
(8)
𝑑𝑤 = 𝑛𝐶𝑉 𝑑𝑇
𝑇2
𝑤 = 𝑛𝐶𝑉 ∫ 𝑑𝑇
𝑇1
𝑤 = 𝑛𝐶𝑉 (𝑇2 − 𝑇1 )
(9)
It is assumed that CV is constant.
Example1: Work of adiabatic reversible expansion
Consider the adiabatic, reversible expansion of 0.020 mol Ar, initially at 25°C, from
0.50 dm3 to 1.00 dm3. Calculate the work of expansion.
Given: Cvm = 12.48 J K-1 mol-1
Method 1: Use of
𝐶 1−𝛾
𝑤= [𝑉2 − 𝑉1 1−𝛾 ]
𝛾−1
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