Find the entropy increment of $
u$ moles of an ideal gas whose adiabatic exponent is $\gamma$ if, a

Question

Find the entropy increment of $
u$ moles of an ideal gas whose adiabatic exponent is $\gamma$ if, as a result of a certain process, the gas volume increased $\alpha$ times while the pressure dropped $\beta$ times.

Professor PC Thomas Classes · Answer

Deriving the entropy change equation in terms of pressure and volume 
Here is the step-by-step derivation for an ideal gas, starting from the very basics.
Step 1: The First Law of Thermodynamics and Entropy
We start with the differential form of the First Law of Thermodynamics:
$dQ = dU + dW$
For $
u$ moles of an ideal gas, the change in internal energy is $dU = 
u C_v dT$, and the work done is $dW = P dV$. Substituting these in:\
$dQ = 
u C_v dT + P dV$
The change in entropy $dS$ for a reversible process is defined as:\
$dS = \frac{dQ}{T}$
Dividing our $dQ$ equation by $T$, we get:\
$dS = \frac{
u C_v dT}{T} + \frac{P dV}{T}$
Step 2: Substituting the Ideal Gas Equation
To make the second term integrable, we use the ideal gas equation $PV = 
u RT$, which rearranges to $\frac{P}{T} = \frac{
u R}{V}$.
Substitute this into the $dS$ equation:\
$dS = 
u C_v \frac{dT}{T} + 
u R \frac{dV}{V}$
Now, integrate both sides from the initial state $(T_1, V_1)$ to the final state $(T_2, V_2)$:\
$\int_{S_1}^{S_2} dS = 
u C_v \int_{T_1}^{T_2} \frac{dT}{T} + 
u R \int_{V_1}^{V_2} \frac{dV}{V}$\
$\Delta S = 
u C_v \ln\left(\frac{T_2}{T_1}ight) + 
u R \ln\left(\frac{V_2}{V_1}ight)$
(This is the first standard form of the entropy equation, written in terms of $T$ and $V$.)

Step 3: Converting to Pressure and Volume ($P, V$) Form
Question  gave us data in terms of pressure and volume changes, so we need to eliminate the temperature term.

From the ideal gas law, $T = \frac{PV}{
u R}$. Therefore, the ratio of temperatures can be written as:\
$\frac{T_2}{T_1} = \frac{P_2 V_2}{P_1 V_1}$\Substitute this ratio back into our entropy equation:\
$\Delta S = 
u C_v \ln\left(\frac{P_2 V_2}{P_1 V_1}ight) + 
u R \ln\left(\frac{V_2}{V_1}ight)$
Using the logarithmic property $\ln(ab) = \ln(a) + \ln(b)$, expand the first term:\
$\$\Delta S = 
u C_v \ln\left(\frac{P_2}{P_1}ight) + 
u C_v \ln\left(\frac{V_2}{V_1}ight) + 
u R \ln\left(\frac{V_2}{V_1}ight)$\
Now, group the volume terms together:\
$\Delta S = 
u C_v \ln\left(\frac{P_2}{P_1}ight) + 
u (C_v + R) \ln\left(\frac{V_2}{V_1}ight)$

Step 4: Applying Mayer's Relation
Using Mayer's relation for ideal gases, $C_p = C_v + R$, substitute $C_p$ into the equation:\
$\Delta S = 
u C_v \ln\left(\frac{P_2}{P_1}ight) + 
u C_p \ln\left(\frac{V_2}{V_1}ight)$

$\Delta S = 
u C_v \ln(P_2/P_1) + 
u C_p \ln(V_2/V_1) = 
u \frac{R}{\gamma-1} \ln(1/\beta) + 
u \frac{\gamma R}{\gamma-1} \ln(\alpha) = \frac{
u R}{\gamma-1}[\gamma \ln \alpha - \ln \beta] = \frac{
u R}{\gamma-1} \ln\left(\frac{\alpha^\gamma}{\beta}ight)$.