# Ideal Gas Expansion Problem **Given:** - \( R = 287 \) J/kg·K (specific gas constant) - \( k = 1.4 \) (isentropic exponent) - Initial state: \( P_A = 5 \) bar, \( T_A = 450 \) K - Final pressure: \( P_B = 1 \) bar We analyze two paths from \( A \) to \( B \): - **Path I:** Isothermal (constant \( T \)) - **Path II:** Isentropic (adiabatic, reversible) from \( A \) to \( X \), then constant volume to \( B \). --- ## Step 1: Find \( T_B \) for Each Path ### **Path I: Isothermal Process (\( T_B = T_A \))** Since \( T \) is constant, \[ T_B = T_A = 450~\text{K} \] --- ### **Path II: Isentropic + Constant Volume** #### 1. **Isentropic (A → X):** Use the isentropic relation for an ideal gas: \[ \frac{T_X}{T_A} = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \] But \( P_X \) is unknown; we need to relate it to the final state. We know after isentropic expansion (A→X), the process continues at constant volume (X→B) until pressure reaches \( P_B \). For constant volume (X→B): \[ \frac{P_B}{T_B} = \frac{P_X}{T_X} \implies T_B = T_X \frac{P_B}{P_X} \] Let’s combine this with the isentropic step: 1. \( P_A, T_A \) known, \( P_B \) given. 2. Let \( v \) be specific volume. For isentropic, \( Pv^k = \text{const} \) or \( \frac{T_X}{T_A} = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \). But after isentropic, the volume is fixed for the second step, so \( v_X = v_B \). For an ideal gas: \[ P_X v_X = R T_X \quad \text{and} \quad P_B v_B = R T_B \] But at constant volume: \( v_X = v_B \). So: \[ \frac{P_B}{T_B} = \frac{P_X}{T_X} \implies P_X = P_B \frac{T_X}{T_B} \] Now, express everything in terms of \( T_B \): From isentropic: \[ \frac{T_X}{T_A} = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \] But above, \( P_X = P_B \frac{T_X}{T_B} \) so: \[ \frac{T_X}{T_A} = \left(\frac{P_B \frac{T_X}{T_B}}{P_A}\right)^{(k-1)/k} \] Take logarithms or solve directly: \[ \frac{T_X}{T_A} = \left( \frac{P_B}{P_A} \cdot \frac{T_X}{T_B} \right)^{(k-1)/k} \] Let \( x = \frac{T_X}{T_A} \), \( y = \frac{T_B}{T_A} \), so \( \frac{T_X}{T_B} = \frac{x}{y} \): \[ x = \left( \frac{P_B}{P_A} \cdot \frac{x}{y} \right)^{(k-1)/k} \] Take both sides to the power of \( k/(k-1) \): \[ x^{k/(k-1)} = \frac{P_B}{P_A} \cdot \frac{x}{y} \] \[ x^{k/(k-1) - 1} = \frac{P_B}{P_A} \cdot \frac{1}{y} \] But \( k/(k-1) - 1 = \frac{1}{k-1} \) So: \[ x^{1/(k-1)} = \frac{P_B}{P_A} \cdot \frac{1}{y} \] \[ x^{1/(k-1)} \cdot y = \frac{P_B}{P_A} \] But \( x = \frac{T_X}{T_A} \), \( y = \frac{T_B}{T_A} \): \[ \left(\frac{T_X}{T_A}\right)^{1/(k-1)} \cdot \frac{T_B}{T_A} = \frac{P_B}{P_A} \] Recall earlier that at constant volume, \[ T_B = T_X \frac{P_B}{P_X} \] But we've already incorporated this. Now, let's express everything in terms of \( y \): \[ x^{1/(k-1)} \cdot y = \frac{P_B}{P_A} \] But from above, \( x = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \), but that's circular. Let's instead solve numerically. Let \( k = 1.4 \), \( P_A = 5 \) bar, \( P_B = 1 \) bar. Let \( x = \frac{T_X}{T_A} \), \( y = \frac{T_B}{T_A} \): \[ x^{1/(k-1)} \cdot y = \frac{1}{5} \] But from constant volume heating, \[ y = x \frac{P_B}{P_X} \] Or, \[ P_X = P_B \frac{x}{y} \] Now, the isentropic relation: \[ \frac{T_X}{T_A} = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \] So, \[ x = \left(\frac{P_X}{P_A}\right)^{(k-1)/k} \] But \( P_X = P_B \frac{x}{y} \) So, \[ x = \left(\frac{P_B}{P_A} \frac{x}{y}\right)^{(k-1)/k} \] Let \( a = \frac{P_B}{P_A} = .2 \), \( k = 1.4 \), so \( (k-1)/k = .4/1.4 \approx .2857 \). Let’s write: \[ x = (a \frac{x}{y})^{.2857} \] Take logarithms: \[ \ln x = .2857 \ln (a \frac{x}{y}) = .2857 (\ln a + \ln x - \ln y) \] \[ \ln x - .2857 \ln x = .2857 \ln a - .2857 \ln y \] \[ (1 - .2857) \ln x = .2857 \ln a - .2857 \ln y \] \[ .7143 \ln x = .2857 (\ln a - \ln y) \] \[ \ln x = \frac{.2857}{.7143} (\ln a - \ln y) \] \[ \ln x = .4 (\ln a - \ln y) \] \[ \ln x = .4 \ln a - .4 \ln y \] \[ \ln x + .4 \ln y = .4 \ln a \] But from earlier, \[ x^{1/(k-1)} \cdot y = a \] \( 1/(k-1) = 1/.4 = 2.5 \) \[ x^{2.5} y = a \] Now, substitute \( \ln x \) from above: \[ \ln x + .4 \ln y = .4 \ln a \] Let’s write \( x = e^u \), \( y = e^v \): \[ u + .4 v = .4 \ln a \] So \( u = .4 \ln a - .4 v \) Now, from the other equation: \[ x^{2.5} y = a \implies e^{2.5u} e^v = a \implies 2.5u + v = \ln a \] Substitute \( u \): \[ 2.5 (.4 \ln a - .4 v) + v = \ln a \] \[ 1. \ln a - 1. v + v = \ln a \] \[ 1. \ln a = \ln a \] This is always satisfied, implying as long as the earlier relation is satisfied. So, from above, \[ \ln x + .4 \ln y = .4 \ln a \] Let’s pick \( y \), solve for \( x \): \[ \ln x = .4 \ln a - .4 \ln y \] \[ x = a^{.4} y^{-.4} \] Now, from the earlier equation: \[ x^{2.5} y = a \] \[ (a^{.4} y^{-.4})^{2.5} y = a \] \[ a^{1.} y^{-1.} y = a \] \[ a^{1.} = a \] So everything is consistent. Thus, the solution is: \[ x = a^{.4} y^{-.4} \] But \( x = \frac{T_X}{T_A} \), \( y = \frac{T_B}{T_A} \). From the other relation, \[ x^{2.5} y = a \] So, substituting \( x \): \[ (a^{.4} y^{-.4})^{2.5} y = a \] \[ a^{1.} y^{-1.} y = a \] \[ a = a \] Thus, for any \( y \), the system is satisfied. But \( T_B = T_X \frac{P_B}{P_X} \), with \( P_X = P_B \frac{x}{y} \). But, let's take a more direct approach: 1. Isentropic (A→X): \( \frac{T_X}{T_A} = \left(\frac{P_X}{P_A}\right)^{.2857} \) 2. Constant volume (X→B): \( \frac{T_B}{T_X} = \frac{P_B}{P_X} \implies \frac{T_B}{T_A} = \frac{T_X}{T_A} \cdot \frac{P_B}{P_X} \) Let’s denote \( r = \frac{P_X}{P_A} \): - \( \frac{T_X}{T_A} = r^{.2857} \) - \( \frac{T_B}{T_A} = r^{.2857} \cdot \frac{P_B}{P_X} \) But \( P_X = r P_A \), so \( \frac{P_B}{P_X} = \frac{P_B}{r P_A} \): \[ \frac{T_B}{T_A} = r^{.2857} \cdot \frac{P_B}{r P_A} = r^{.2857 - 1} \cdot \frac{P_B}{P_A} \] \[ \frac{T_B}{T_A} = r^{-.7143} \cdot \frac{P_B}{P_A} \] But at the end of the process, we want the state at \( P_B \). To find the value of \( r \), recall that after isentropic step, a constant volume process brings \( P_X \) to \( P_B \) as \( T_X \) is heated to \( T_B \). Let’s assume the entire process is A→X (isentropic), then X→B (constant volume). But wait, for the given two pressures, we can make a table for clarity. #### Let’s Choose a Value for \( r \): But since \( P_X \) is arbitrary, the only way to satisfy the final state \( P_B \) is when the final equation holds. Let’s try \( r = 1 \) (i.e., no isentropic process) -- then \( T_B = T_A \cdot (1)^{-.7143} \cdot .2 = .2 T_A \), which is not physical. Alternatively, the isentropic process could be all the way to \( P_B \), then no constant volume heating. Try \( r = .2 \): - \( r = .2 \) - \( r^{.2857} = (.2)^{.2857} \approx e^{.2857 \ln .2} \approx e^{-.4605} \approx .631 \)

# Ideal Gas Expansion Summary ## Final Temperature \( T_B \) ### **Path I: Isothermal Process** - \( T_B = T_A = 450~\text{K} \) ### **Path II: Isentropic Process followed by Constant Volume Heating** - Using the derived relation: \[ T_B = T_A \cdot \left( \frac{P_B}{P_A} \right)^{\frac{k-1}{k}} = 450~\text{K} \cdot \left( \frac{1~\text{bar}}{5~\text{bar}} \right)^{\frac{0.4}{1.4}} \approx 450~\text{K} \cdot (0.2)^{0.2857} \approx 450~\text{K} \cdot 0.631 \approx 284~\text{K} \] --- ## P-v Diagram - **Path I**: Isothermal from \( A \) to \( B \) (horizontal line). - **Path II**: Isentropic drop from \( A \) to \( X \) (curved line) followed by constant volume to \( B \) (vertical line). *(P-v diagram showing the isothermal and isentropic paths)* *Alt text: P-v diagram illustrating isothermal and isentropic expansion processes from state A to state B.* --- ## Entropy Change \( \Delta s \) ### **Path I: Isothermal Process** \[ \Delta s = R \ln\left(\frac{T_B}{T_A}\right) + R \ln\left(\frac{P_B}{P_A}\right) = R \ln\left(\frac{1~\text{bar}}{5~\text{bar}}\right) = 287 \ln(0.2) \approx -287 \cdot 1.609 = -461.7~\text{J/kg·K} \] ### **Path II: Isentropic Process** - For isentropic, \( \Delta s = 0 \). - Constant volume heating leads to: \[ \Delta s = R \ln\left(\frac{T_B}{T_X}\right) = R \ln\left(\frac{T_B}{T_A \cdot (0.2)^{0.2857}}\right) \approx 0 \text{ for heat transfer} \] --- ## Specific Work Produced ### **Path I: Isothermal Process** \[ W = R(T_B - T_A) \ln\left(\frac{V_B}{V_A}\right) = R \cdot T_A \ln\left(\frac{P_A}{P_B}\right) \approx 287 \cdot 450 \cdot \ln(5) \approx 287 \cdot 450 \cdot 1.609 \approx 260,251~\text{J/kg} \] ### **Path II: Isentropic Process** - Work done during isentropic process: \[ W_{isentropic} = \frac{T_A}{T_B} R (P_B - P_A) \approx \text{(using the pressures and temperatures derived)} \] --- ## Specific Heat Exchanged ### **Path I: Isothermal Process** \[ Q = mC_v(T_B - T_A) \text{ (no heat transfer in isothermal)} \] ### **Path II: Isentropic followed by Constant Volume** - During constant volume heating: \[ Q = mC_v(T_B - T_X) \] - Total heat exchanged requires knowing specific volumes and state changes. --- ### Final Answer Summary - **Temperatures**: \( T_B \) is 450 K for Path I and approximately 284 K for Path II. - **Entropy Change**: \( \Delta s \) is negative for Path I and zero for Path II. - **Specific Work**: Significant work produced in Path I, lesser in Path II. - **Heat Exchange**: Specific heat quantities depend on the specific volumes and require further calculations.