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Engineering › Thermodynamics › Ideal Gases ›

Variations in Entropy

Changes in Entropy that occur when an Ideal Gas is subjected to both reversible and irreversible operations

Contents

The Variations Of Entropy For A Perfect Gas,
Entropy For An Irreversible Operation
Sketching Operations On P.v. And T.$\phi$ Graphs
Page Comments

The Variations Of Entropy For A Perfect Gas,

$\delta \Phi = \frac{dq}{T}\;for\;a\;reversible\;operation$

13108/img_therm_14.jpg

$\delta \Phi \;T = \delta\,q$

At Constant Volume:-

$\delta \Phi = \frac{w\,C_V\,dt}{T}$

$\therefore\;\;\;\;\;\Phi _2 - \Phi _1 = \int_{T_1}^{T_2}\frac{w\;C_V\,dt}{T}$

$= wC_V\;Ln\frac{T_2}{T_1}\;\;\;\;\;or\;\;\;\;\;w\;C_V\;ln\frac{P_2}{P^1}$

At Constant Pressure

$\delta \Phi = \frac{w\;C_P\;dt}{T}$

$\therefore\;\;\;\;\;\Phi _2 - \Phi _1 = \int_{T_1}^{T_2}\;w\;C_P\;dt$

$= w\;C_P\;Ln\frac{T_2}{T_1}\;\;\;\;\;or\;\;\;\;\;w\;C_P\;Ln\frac{V_2}{V_1}$

For an Isothermal Operation

$Heat\;Supplied = Work\;Done + \delta\,U\,( Constant)$

$= P\;\delta \,V$

$\therefore\;\;\;\;\;\delta \Phi = \frac{p\delta V}{T} = w\;R\;\frac{dv}{V}$

$Thus\;\;\;\;\;\Phi _2 - \Phi_1 = w\,R\,Ln\frac{V_2}{V_1}\;\;\;\;\;or\;\;\;\;\;w\,R\,Ln\frac{P_1}{P_2}$

Entropy For An Irreversible Operation

Consider a gas changing from $\inline P_1V_1T_1\;to\;P_2V_2T_2$ . Now since the Function of State depends upon the values of P:V:&T at 1 & 2 and is independent of the process.

13108/img_therm_15.jpg

Stage 1

$\Phi _a - \Phi _1 = wC_V\;Ln\;\frac{T_a}{T_1}\;\;\;\;\;or\;\;\;\;\;wC_V\;Ln\;\frac{P_2}{P_1}$

$\Phi _2 - \Phi _a = wC_P\;Ln\;\frac{T_2}{T_a}\;\;\;\;\;or\;\;\;\;\;wC_V\;Ln\;\frac{V_2}{V_1}$

$\therefore\;\;\;\;\;\Phi _2 - \Phi _1 = w\left( C_P\;Ln\;\frac{P_2}{P_1} + C_V\;Ln\;\frac{V_2}{V_1} \right)$

Sketching Operations On P.v. And T.$\phi$ Graphs

NOTE

The Area under the curve on a P.V. graph is the Work Done

The Area under the curve on a T. $\inline \Phi$ graph is the heat Supplied

13108/img_therm16.jpg

Constant Volume. 1 - 2

$P\;V^n = Constant$

$V = Constant$

$n = infinity$

$\Phi _2 - \Phi _2 = w\;C_V\;Ln\;\frac{T_2}{T_1}$

$\therefore\;\;\;\;\;\Phi = w\;C_V\;Ln\;T$

Constant Pressure 1 - 3

$P\;V^n = Constant$

$P = Constant$

$n = 0$

$\Phi _2 - \Phi _1 = wC_P\;Ln\;\frac{T_2}{T_1}$

Isothermal 1 - 4

$P\;V^n = Constant$

$T = Constant$

$\therefore\;\;\;P\;V = Constant$

$n = 1$

Adiabatic 1 - 5

$n = \gamma$

$\Phi = Constant$

Last Modified: 25 Sep 10 @ 08:30 Page Rendered: 2022-03-14 15:42:49