Ideal Gas Thermometer

Introduction

            If we heat an ideal gas it will want to expand or increase its pressure.  If we want to have a measure of its new temperature we could either employ a pressure gauge or we could have the expansion raise a movable piston.  The simplest implementation of the gas thermometer is movable piston with the working gas trapped below a movable piston which has a vacuum space above it.  The temperature of the gas is changed by adding heat to it.  This results in an increase in internal kinetic energy.  But when the piston raises to expand the gas back to its old pressure, the gas gives up some of this heat because the atoms that hit the rising piston have their vertical speed reduced.  The following calculations will document how much the energy is reduced during expansion.

 

Calculation of Final Energy and Pressure

To start we will assume a 2 dimensional gas which has a simple ideal gas equation.

Before adding internal energy δU we have the ideal gas law:

P 0 V 0 = U 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfadaWgaa WcbaGaaGimaaqabaGccaWGwbWaaSbaaSqaaiaaicdaaeqaaOGaeyyp a0JaamyvamaaBaaaleaacaaIWaaabeaaaaa@3C41@  

(1.1)

When we add internal energy δU to the gas this expression changes to:

P 1 V 0 = U 0 +δU U 1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfadaWgaa WcbaGaaGymaaqabaGccaWGwbWaaSbaaSqaaiaaicdaaeqaaOGaeyyp a0JaamyvamaaBaaaleaacaaIWaaabeaakiabgUcaRiabes7aKjaadw facqGHHjIUcaWGvbWaaSbaaSqaaiaaigdaaeqaaaaa@4337@  

(1.2)

After expanding the volume to V2 so that the pressure returns to P0 we have the following equation :

P 0 V 2 = U 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfadaWgaa WcbaGaaGimaaqabaGccaWGwbWaaSbaaSqaaiaaikdaaeqaaOGaeyyp a0JaamyvamaaBaaaleaacaaIYaaabeaaaaa@3C45@  

(1.3)

where

U 2 = U 0 +δU V 0 V 2 P(V)dV MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwfadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpcaWGvbWaaSbaaSqaaiaaicdaaeqa aOGaey4kaSIaeqiTdqMaamyvaiabgkHiTmaapehabaGaamiuaiaacI cacaWGwbGaaiykaiaadsgacaWGwbaaleaacaWGwbWaaSbaaWqaaiaa icdaaeqaaaWcbaGaamOvamaaBaaameaacaaIYaaabeaaa0Gaey4kIi paaaa@499C@  

(1.4)

 

Problem is that we don't know P(v) in the integral. 

Let's assume that

U= U 1 V 0 V MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwfacqGH9a qpcaWGvbWaaSbaaSqaaiaaigdaaeqaaOWaaSaaaeaacaWGwbWaaSba aSqaaiaaicdaaeqaaaGcbaGaamOvaaaaaaa@3C4C@  

(1.5)

where U1 =U0+δU.

Then using the expression for U2:

P 0 V 2 =( U 0 +δU) V 0 V 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfadaWgaa WcbaGaaGimaaqabaGccaWGwbWaaSbaaSqaaiaaikdaaeqaaOGaeyyp a0JaaiikaiaadwfadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcqaH0o azcaWGvbGaaiykamaalaaabaGaamOvamaaBaaaleaacaaIWaaabeaa aOqaaiaadAfadaWgaaWcbaGaaGOmaaqabaaaaaaa@44A5@  

(1.6)

Then we may solve for V2 and obtain:

V 2 = U 1 V 0 P 0 = ( P 0 V 0 +δU) V 0 P 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAfadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpdaGcaaqaamaalaaabaGaamyvamaa BaaaleaacaaIXaaabeaakiaacAfadaWgaaWcbaGaaGimaaqabaaake aacaWGqbWaaSbaaSqaaiaaicdaaeqaaaaaaeqaaOGaeyypa0ZaaOaa aeaadaWcaaqaaiaacIcacaWGqbWaaSbaaSqaaiaaicdaaeqaaOGaam OvamaaBaaaleaacaaIWaaabeaakiabgUcaRiabes7aKjaadwfacaGG PaGaaiOvamaaBaaaleaacaaIWaaabeaaaOqaaiaadcfadaWgaaWcba GaaGimaaqabaaaaaqabaaaaa@4B2D@  

(1.7)

Then U2 is:

U 2 = U 1 V 0 V 2 = U 1 V 0 U 1 V 0 P 0 = U 1 P 0 V 0 = U 1 U 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwfadaWgaa WcbaGaaGOmaaqabaGccqGH9aqpdaWcaaqaaiaadwfadaWgaaWcbaGa aGymaaqabaGccaWGwbWaaSbaaSqaaiaaicdaaeqaaaGcbaGaamOvam aaBaaaleaacaaIYaaabeaaaaGccqGH9aqpdaWcaaqaaiaadwfadaWg aaWcbaGaaGymaaqabaGccaWGwbWaaSbaaSqaaiaaicdaaeqaaaGcba WaaOaaaeaadaWcaaqaaiaadwfadaWgaaWcbaGaaGymaaqabaGccaWG wbWaaSbaaSqaaiaaicdaaeqaaaGcbaGaamiuamaaBaaaleaacaaIWa aabeaaaaaabeaaaaGccqGH9aqpdaGcaaqaaiaadwfadaWgaaWcbaGa aGymaaqabaGccaWGqbWaaSbaaSqaaiaaicdaaeqaaOGaamOvamaaBa aaleaacaaIWaaabeaaaeqaaOGaeyypa0ZaaOaaaeaacaWGvbWaaSba aSqaaiaaigdaaeqaaOGaamyvamaaBaaaleaacaaIWaaabeaaaeqaaa aa@5369@  

(1.8)

The general expression for P during the expansion is:

P= U V = U 1 V 0 V 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadcfacqGH9a qpdaWcaaqaaiaadwfaaeaacaWGwbaaaiabg2da9maalaaabaGaamyv amaaBaaaleaacaaIXaaabeaakiaadAfadaWgaaWcbaGaaGimaaqaba aakeaacaWGwbWaaWbaaSqabeaacaaIYaaaaaaaaaa@3FFB@  

(1.9)

 

Conclusion

It is important to note that the expressions for V2, U2 and general expressions for U and P above agree with the plotted results for the ballistic gas animation.