Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Isentropic process
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== Derivation of the isentropic relations === For a closed system, the total change in energy of a system is the sum of the work done and the heat added: : <math>dU = \delta W + \delta Q.</math> The reversible work done on a system by changing the volume is :<math>\delta W = -p \,dV,</math> where <math>p</math> is the [[pressure]], and <math>V</math> is the [[Volume (thermodynamics)|volume]]. The change in [[enthalpy]] (<math>H = U + pV</math>) is given by :<math>dH = dU + p \,dV + V \,dp.</math> Then for a process that is both reversible and adiabatic (i.e. no heat transfer occurs), <math> \delta Q_\text{rev} = 0</math>, and so <math>dS = \delta Q_\text{rev}/T = 0 </math> All reversible adiabatic processes are isentropic. This leads to two important observations: : <math>dU = \delta W + \delta Q = -p \,dV + 0,</math> :<math>dH = \delta W + \delta Q + p \,dV + V \,dp = -p \,dV + 0 + p \,dV + V \,dp = V \,dp.</math> Next, a great deal can be computed for isentropic processes of an ideal gas. For any transformation of an ideal gas, it is always true that :<math>dU = n C_v \,dT</math>, and <math>dH = n C_p \,dT.</math> Using the general results derived above for <math>dU</math> and <math>dH</math>, then : <math>dU = n C_v \,dT = -p \,dV,</math> : <math>dH = n C_p \,dT = V \,dp.</math> So for an ideal gas, the [[heat capacity ratio]] can be written as :<math>\gamma = \frac{C_p}{C_V} = -\frac{dp/p}{dV/V}.</math> For a calorically perfect gas <math>\gamma</math> is constant. Hence on integrating the above equation, assuming a calorically perfect gas, we get : <math> pV^\gamma = \text{constant},</math> that is, : <math>\frac{p_2}{p_1} = \left(\frac{V_1}{V_2}\right)^\gamma.</math> Using the [[Equation of state#Classical ideal gas law|equation of state]] for an ideal gas, <math>p V = n R T</math>, : <math> TV^{\gamma-1} = \text{constant}.</math> (Proof: <math>PV^\gamma = \text{constant} \Rightarrow PV\,V^{\gamma-1} = \text{constant} \Rightarrow nRT\,V^{\gamma-1} = \text{constant}.</math> But ''nR'' = constant itself, so <math>TV^{\gamma-1} = \text{constant}</math>.) : <math> \frac{p^{\gamma-1}}{T^\gamma} = \text{constant} </math> also, for constant <math>C_p = C_v + R</math> (per mole), : <math> \frac{V}{T} = \frac{nR}{p}</math> and <math>p = \frac{nRT}{V}</math> : <math> S_2-S_1 = nC_p \ln\left(\frac{T_2}{T_1}\right) - nR\ln\left(\frac{p_2}{p_1}\right)</math> : <math> \frac{S_2-S_1}{n} = C_p \ln\left(\frac{T_2}{T_1}\right) - R\ln\left(\frac{T_2 V_1}{T_1 V_2}\right ) = C_v\ln\left(\frac{T_2}{T_1}\right)+ R \ln\left(\frac{V_2}{V_1}\right)</math> Thus for isentropic processes with an ideal gas, : <math> T_2 = T_1\left(\frac{V_1}{V_2}\right)^{(R/C_v)}</math> or <math> V_2 = V_1\left(\frac{T_1}{T_2}\right)^{(C_v/R)}</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)