Solved thermodynamics exercise

Textbook exercise: elementary transformations

Exercise 9 · Lesson 5Application to the Ideal Gas

  • textbook exercise
  • monatomic ideal gas
  • isochoric transformation
  • isothermal transformation
  • isobaric transformation
  • internal energy
  • work
  • heat
  • first law

Statement

Consider nn moles of a monatomic ideal gas, initially in state A(PA,VA,Ti)A(P_A,V_A,T_i). We study separately three quasi-static transformations, all starting from AA, with Tf>TiT_f>T_i:

  • ABA\to B: isochoric heating from TiT_i to TfT_f;
  • ACA\to C: isothermal compression at TiT_i, from VAV_A to VA/2V_A/2;
  • ADA\to D: isobaric heating from TiT_i to TfT_f.

We use the banker's convention: transfers received by the gas are positive.

  1. Plot the three transformations on the same Clapeyron diagram (P,V)(P,V) and formally determine states BB, CC, and DD.
  2. Formally calculate WW, ΔU\Delta U, and QQ for each of the three transformations.
  3. Comment on the signs of the energy transfers.
  4. Numerical application: n=1 moln=1\ \mathrm{mol}, Ti=300 KT_i=300\ \mathrm{K}, Tf=500 KT_f=500\ \mathrm{K}, and R=8,314 Jmol1K1R=8{,}314\ \mathrm{J\,mol^{-1}\,K^{-1}}.

Hint

Hint
For a monatomic ideal gas, ΔU=32nRΔT\Delta U=\frac32nR\Delta T. Use W=PdVW=-\int P\,\mathrm{d}V and the first law ΔU=Q+W\Delta U=Q+W.

Detailed solution

Solution
Question 1. The ideal-gas equation gives

StatePVTAPAVATiBPATfTiVATfC2PAVA2TiDPAVATfTiTf\begin{aligned} \begin{array}{c|c|c|c} \text{State} & P & V & T \\ \hline A & P_A & V_A & T_i \\ B & P_A\dfrac{T_f}{T_i} & V_A & T_f \\[4pt] C & 2P_A & \dfrac{V_A}{2} & T_i \\[4pt] D & P_A & V_A\dfrac{T_f}{T_i} & T_f \end{array} \end{aligned}

In the (P,V)(P,V) diagram, ABA\to B is vertical, ACA\to C follows the isotherm TiT_i to the left, and ADA\to D is horizontal to the right.

Question 2.

Isochoric heating ABA\to B. Since dV=0\mathrm{d}V=0,

WAB=0,ΔUAB=32nR(TfTi),QAB=32nR(TfTi).W_{AB}=0, \qquad \Delta U_{AB}=\frac32nR(T_f-T_i), \qquad Q_{AB}=\frac32nR(T_f-T_i).

Isothermal compression ACA\to C. The temperature is constant, so ΔUAC=0\Delta U_{AC}=0. Moreover,

WAC=VAVA/2nRTiVdV=nRTiln2,QAC=nRTiln2.W_{AC} =-\int_{V_A}^{V_A/2}\frac{nRT_i}{V}\,\mathrm{d}V =nRT_i\ln2, \qquad Q_{AC}=-nRT_i\ln2.

Isobaric heating ADA\to D. At constant pressure, PA(VDVA)=nR(TfTi)P_A(V_D-V_A)=nR(T_f-T_i), so

WAD=nR(TfTi),ΔUAD=32nR(TfTi),QAD=52nR(TfTi).W_{AD}=-nR(T_f-T_i), \qquad \Delta U_{AD}=\frac32nR(T_f-T_i), \qquad Q_{AD}=\frac52nR(T_f-T_i).

Question 3. Along the isochoric transformation, all the heat received increases the internal energy. During the isothermal compression, the gas receives work and releases exactly the same amount of energy as heat. Along the isobaric transformation, the gas receives heat, increases its internal energy, and does work as it expands.

Question 4. With TfTi=200 KT_f-T_i=200\ \mathrm{K}:

TransformationW (kJ)ΔU (kJ)Q (kJ)Isochoric AB0+2,49+2,49Isothermal AC+1,7301,73Isobaric AD1,66+2,49+4,16\begin{aligned} \begin{array}{c|r|r|r} \text{Transformation} & W\ (\mathrm{kJ}) & \Delta U\ (\mathrm{kJ}) & Q\ (\mathrm{kJ}) \\ \hline \text{Isochoric }A\to B & 0 & +2{,}49 & +2{,}49 \\ \text{Isothermal }A\to C & +1{,}73 & 0 & -1{,}73 \\ \text{Isobaric }A\to D & -1{,}66 & +2{,}49 & +4{,}16 \end{array} \end{aligned}