The Adiabatic Expansion of an Ideal Gas — Question 4

Section 19-10 · Physics pages 3218–3219

Question 4 Endpoint measurements infer a constant adiabatic exponent Use pV = nRT, R = 8.314 J/(mol K), and ∆U = Q −W with W positive for work done by the gas. A gas follows a reversible adiabatic path with constant capacities. Its volume increases by a factor of 2.50 while pressure decreases to one-quarter of its initial value. ln V ln p slope −γ Axes and labels specify the model; the sketch is schematic. (a) Infer γ, CV /R and the temperature ratio. (b) If equipartition is used to define f = 2CV /R, calculate the effective count and discuss whether these endpoint data identify a unique pure molecular species.
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Question 4 - Solution Endpoint measurements infer a constant adiabatic exponent ln V ln p slope −γ (a) Method and calculation piV γ i = pfV γ f ⇒4 = (2.50)γ ⇒γ = ln 4 ln 2.50 ≈1.51294. CV /R = 1 γ −1 ≈1.94954, Tf/Ti = (pf/pi)(Vf/Vi) = .625. (b) Interpretation and further calculation f = 2 γ −1 ≈3.89908. Tf/Ti = (2.50)1−γ = .625. A noninteger effective count can reflect a mixture or internal-mode response. Identification also depends on the assumptions of reversibility and constant capacities; endpoint ratios alone do not establish them. Physical check The log–log slope identifies a path exponent only within the stated model.

Question 4

Endpoint measurements infer a constant adiabatic exponent

Use pV=nRTpV=nRT, R=8.314R=8.314 J/(mol K), and ΔU=Q−W\Delta U=Q-W with WW positive for work done by the gas. A gas follows a reversible adiabatic path with constant capacities. Its volume increases by a factor of 2.50 while pressure decreases to one-quarter of its initial value.

Diagram shown in the original SVG worksheet.

  1. Infer γ\gamma, CV/RC_V/R and the temperature ratio.

  2. If equipartition is used to define f=2CV/Rf=2C_V/R, calculate the effective count and discuss whether these endpoint data identify a unique pure molecular species.

Question 4 - Solution

Endpoint measurements infer a constant adiabatic exponent

Diagram shown in the original SVG worksheet.

(a) Method and calculation

piViγ=pfVfγ⇒4=(2.50)γ⇒γ=ln⁡4ln⁡2.50≈1.51294.p_iV_i^\gamma=p_fV_f^\gamma\ \Rightarrow\ 4=(2.50)^\gamma\ \Rightarrow\gamma=\frac{\ln4}{\ln2.50}\approx1.51294.CV/R=1γ−1≈1.94954,Tf/Ti=(pf/pi)(Vf/Vi)=.625.C_V/R=\frac1{\gamma-1}\approx1.94954,\quad T_f/T_i=(p_f/p_i)(V_f/V_i)=.625.

(b) Interpretation and further calculation

f=2γ−1≈3.89908.f=\frac2{\gamma-1}\approx3.89908.Tf/Ti=(2.50)1−γ=.625.T_f/T_i=(2.50)^{1-\gamma}=.625.A noninteger effective count can reflect a mixture or internal-mode response. Identification also depends on the assumptions of reversibility and constant capacities; endpoint ratios alone do not establish them.

Physical check

The log–log slope identifies a path exponent only within the stated model.