Degrees of Freedom and Molar Specific Heats — Question 8

Section 19-8 · Physics pages 3186–3187

Question 8 A gas restricted to a plane has two translational energy terms Consider a classical monatomic ideal gas confined strictly to a two-dimensional plane of area A. There is no out-of-plane motion or internal excitation. Use the two-dimensional equation of state ΠA = NkBT = nRT, where Π is force per unit length. The work is δW = Π dA. strictly planar motion Schematic; labels, not drawing lengths, specify the data. (a) Derive U, molar CA, and the mean square speed in the plane. (b) Derive the molar heat capacity CΠ at fixed two-dimensional pressure and its ratio to CA. Explain why Π is not a pressure measured in pascals.
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Question 8 - Solution A gas restricted to a plane has two translational energy terms strictly planar motion (a) Method and calculation f = 2, U = NkBT = nRT, CA = R. 1 2m⟨v2 x + v2 y⟩= kBT ⇒⟨v2⟩= 2kBT/m. (b) Interpretation and further calculation Π dA = nR dT (Π fixed), δQ = dU + Π dA = 2nR dT. CΠ = 2R, CΠ/CA = 2, [Π] = J/m2 = N/m. The geometry changes both the mode count and the mechanical pressure variable. Physical check The two-dimensional model must not be combined with the three-dimensional factor 3/2.

Question 8

A gas restricted to a plane has two translational energy terms

Consider a classical monatomic ideal gas confined strictly to a two-dimensional plane of area AA. There is no out-of-plane motion or internal excitation. Use the two-dimensional equation of state ΠA=NkBT=nRT\Pi A=Nk_BT=nRT, where Π\Pi is force per unit length. The work is δW=ΠdA\delta W=\Pi\,dA.

Diagram shown in the original SVG worksheet.

  1. Derive UU, molar CAC_A, and the mean square speed in the plane.

  2. Derive the molar heat capacity CΠC_\Pi at fixed two-dimensional pressure and its ratio to CAC_A. Explain why Π\Pi is not a pressure measured in pascals.

Question 8 - Solution

A gas restricted to a plane has two translational energy terms

Diagram shown in the original SVG worksheet.

(a) Method and calculation

f=2,U=NkBT=nRT,CA=R.f=2,\quad U=Nk_BT=nRT,\quad C_A=R.12m⟨vx2+vy2⟩=kBT⇒⟨v2⟩=2kBT/m.\tfrac12m\langle v_x^2+v_y^2\rangle=k_BT\ \Rightarrow\langle v^2\rangle=2k_BT/m.

(b) Interpretation and further calculation

ΠdA=nRdT(Π fixed),δQ=dU+ΠdA=2nRdT.\Pi\,dA=nR\,dT\quad(\Pi\text{ fixed}),\quad\delta Q=dU+\Pi\,dA=2nR\,dT.CΠ=2R,CΠ/CA=2,[Π]=J/m2=N/m.C_\Pi=2R,\quad C_\Pi/C_A=2,\quad[\Pi]=\mathrm{J/m^2}=\mathrm{N/m}.The geometry changes both the mode count and the mechanical pressure variable.

Physical check

The two-dimensional model must not be combined with the three-dimensional factor 3/23/2.