INGENIA

THM-08

Isentropic ideal-gas relation

T2/T1 = (P2/P1)^{(γ−1)/γ}.

Reading speed
Ideal gasLaplace / Poisson

Governing equation

T2T1=(P2P1)(γ−1)/γ\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}

P2∈[10,8000]P_2 \in \left[10, 8000\right]

where

T_1
Inlet T (K)
P_1
Inlet P (kPa)
P_2
Exit P (kPa)
\gamma
Heat-capacity ratio (—)
T_2
Exit temperature (K)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

THM-08 · phase
00:0 / 00:08
①

Governing law — red

T2T1=(P2P1)(γ−1)/γ\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}

P2∈[10,8000]P_2 \in \left[10, 8000\right]

P2=10P_2 = 10T2=155.4T_2 = 155.4

②

Domain of the variable — blue

P_2_{\min} = 10,\quad P_2_{\max} = 8000

P2∈[10,8000]P_2 \in \left[10, 8000\right]

P2=10P_2 = 10P2=10P_2 = 10

③

Response curve — green

T2T1=(P2P1)(γ−1)/γ\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}

P2∈[10,8000]P_2 \in \left[10, 8000\right]

P2=10P_2 = 10T2=155.4T_2 = 155.4

④

Rate of change — amber

dT2dP2\frac{\mathrm{d}T_2}{\mathrm{d}P_2}

P2∈[10,8000]P_2 \in \left[10, 8000\right]

P2=60.25P_2 = 60.25T2′=1.525T_2^{\prime} = 1.525

Lecture brief

Historical brief

Carnot (1824), Clausius entropy, the first law and later van der Waals and Gibbs potentials turned heat into a state science. The sheets compute work, efficiency and vapour pressure. This sheet (THM-08 — Isentropic ideal-gas relation) is the form associated with Laplace / Poisson. Working symbols: T1T_1, P1P_1, P2P_2, γ\gamma →\rightarrow T2T_2. For a reversible adiabatic of an ideal gas with constant γ, Poisson's relations link T, P and V.

Purpose

Purpose: compute T2T_2 from T1T_1, P1P_1, P2P_2, γ\gamma in Thermodynamics via T2T1=(P2P1)(γ−1)/γ\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma} T2/T1 = (P2/P1)^{(γ−1)/γ}. Use it when a real thermodynamics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given T1=300.000 KT_1 = 300.000\,\mathrm{K}, P1=100.000 kPaP_1 = 100.000\,\mathrm{kPa}, P2=400.000 kPaP_2 = 400.000\,\mathrm{kPa}, γ=1.400 —\gamma = 1.400\,\mathrm{—}, the governing relation T2T1=(P2P1)(γ−1)/γ\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma} yields T2=445.798 KT_2 = 445.798\,\mathrm{K}. Calorically perfect gas, reversible adiabatic. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Exit temperature T_2445.798 K
Reading speed

Watch on YouTube

Free library

Full library

Free PDF / open book

YouTube channels

THM-08 · phase
00:0 / 00:08

Narration of this film

Calorically perfect gas, reversible adiabatic.

For a reversible adiabatic of an ideal gas with constant γ, Poisson's relations link T, P and V.

Reading speed

Watch on YouTube