Question

The gas law for a fixed mass $ m $ of an ideal gas at absolute temperature $ T $, pressure $ P $, and volume $ V $ is $ PV = mRT $, where $ R $ is the gas constant. Show that $$ \dfrac{\partial P}{\partial V} \dfrac{\partial V}{\partial T} \dfrac{\partial T}{\partial P} = -1 $$

Name: Problem 88, Chapter 14: Calculus: Early Transcendentals
Uploaded: 2021-04-19T22:00:11-08:00
Duration: 4 min 20 s
Channel: Chris Trentman
Description: The gas law for a fixed mass m of an ideal gas at absolute temperature T , pressure P , and volume V is PV = mRT , where R is the gas constant. Show that ∂P∂V ∂V∂T ∂T∂P = -1

   The gas law for a fixed mass $ m $ of an ideal gas at absolute temperature $ T $, pressure $ P $, and volume $ V $ is $ PV = mRT $, where  $ R $ is the gas constant. Show that $$ \dfrac{\partial P}{\partial V} \dfrac{\partial V}{\partial T} \dfrac{\partial T}{\partial P} = -1 $$

Calculus: Early Transcendentals

James Stewart 8th Edition

Chapter 14, Problem 88

Instant Answer

Solved by Expert Chris Trentman

04/19/2021

Step 1

We use the product rule for differentiation, which gives us: \[ \frac{\partial P}{\partial V} V + P = 0 \] Solving for $ \frac{\partial P}{\partial V} $, we get: \[ \frac{\partial P}{\partial V} = -\frac{P}{V} \] Show more…

Show all steps

Thanks for your feedback!

The gas law for a fixed mass $ m $ of an ideal gas at absolute temperature $ T $, pressure $ P $, and volume $ V $ is $ PV = mRT $, where $ R $ is the gas constant. Show that $$ \dfrac{\partial P}{\partial V} \dfrac{\partial V}{\partial T} \dfrac{\partial T}{\partial P} = -1 $$