Zoezi la thermodynamiki lililofanyiwa kazi

Zoezi la somo: sheria za Laplace

Zoezi 10 · Somo 5 — Matumizi kwa Gesi Bora

  • zoezi la somo
  • sheria ya Laplace
  • kielezo cha adiabati
  • gamma
  • upanuzi wa adiabati
  • kanuni ya kwanza katika umbo la diferenshali

Swali

Tunazingatia moli nn za gesi kamilifu yenye kielezo cha adiabati γ=cP/cV\gamma = c_P/c_V kinachodhaniwa kuwa thabiti. Mfumo unadhaniwa kuwa umefungwa. Tunakubali uhusiano wa Mayer cP−cV=Rc_P - c_V = R (tazama zoezi kuhusu ugunduzi wa cPc_P).

  1. Onyesha kuwa cV=Rγ−1,cP=γRγ−1.c_V = \frac{R}{\gamma - 1}, \qquad c_P = \frac{\gamma R}{\gamma - 1}.
  2. Gesi hupitia badiliko la adiabati la nusutuli ambapo tunadhani Pext=PP_{\text{ext}} = P. Kwa kuanzia kanuni ya kwanza katika umbo la diferenshali, onyesha kuwa dTT+(γ−1)dVV=0,\frac{\mathrm{d}T}{T} + (\gamma - 1)\frac{\mathrm{d}V}{V} = 0, kisha dedua kuwa TVγ−1=thabitiTV^{\gamma-1} = \text{thabiti} katika badiliko lote.
  3. Dedua maumbo mengine mawili ya sheria ya Laplace: PVγ=thabitiPV^\gamma = \text{thabiti} na TγP1−γ=thabitiT^\gamma P^{1-\gamma} = \text{thabiti}.
  4. Fikiria upanuzi kama huo unaoongeza mara mbili ujazo wa gesi ya atomi moja (γ=5/3\gamma = 5/3), yenye n=1n = 1 moli, kwenye halijoto ya mwanzo Ti=300T_i = 300 K. Kokotoa TfT_f, kisha ΔT\Delta T na WW, na utoe maoni kuhusu ishara za ΔT\Delta T na WW.
  5. Kwenye mchoro wa Clapeyron (P,V)(P,V), linganisha katika nukta ileile mteremko wa adiabati kama hiyo na wa isothemali.

Kidokezo
Kwa swali la 2: dU=ncV dT\mathrm{d}U = n c_V\,\mathrm{d}T na δW=−P dV\delta W = -P\,\mathrm{d}V huku P=nRT/VP = nRT/V, kisha tenga vigezo. Kwa swali la 5, fanya diferenshali ya PVγ=thabitiPV^\gamma = \text{thabiti} na PV=thabitiPV = \text{thabiti}.

Jibu la kina

Suluhisho
Swali la 1. Kutoka cP=γcVc_P = \gamma c_V na cP−cV=Rc_P - c_V = R, tunapata cV(γ−1)=Rc_V(\gamma - 1) = R, na hivyo fomula zote mbili.

Swali la 2. Adiabati: δQ=0\delta Q = 0, kwa hiyo dU=δW\mathrm{d}U = \delta W, yaani ncV dT=−P dV=−nRTVdVn c_V\,\mathrm{d}T = -P\,\mathrm{d}V = -\dfrac{nRT}{V}\mathrm{d}V. Kwa cV=R/(γ−1)c_V = R/(\gamma-1):

dTT=−(γ−1)dVV.\frac{\mathrm{d}T}{T} = -(\gamma-1)\frac{\mathrm{d}V}{V}.

Kwa kuintegrali: ln⁡T+(γ−1)ln⁡V=thabiti\ln T + (\gamma-1)\ln V = \text{thabiti}, yaani TVγ−1=thabitiTV^{\gamma-1} = \text{thabiti}.

Swali la 3. Kwa kubadilisha T=PV/(nR)T = PV/(nR) tunapata PVγ=thabitiPV^\gamma = \text{thabiti} kwa kuwa nn ni thabiti. Kwa kubadilisha V=nRT/PV = nRT/P, tunapata T(TP)γ−1=thabitiT \left(\frac{T}{P}\right)^{\gamma-1} = \text{thabiti}, na hivyo TγP1−γ=thabitiT^\gamma P^{1-\gamma} = \text{thabiti}.

Swali la 4. Njia rahisi zaidi ni kutumia sheria ya Laplace TVγ−1=thabitiT V^{\gamma - 1} = \text{thabiti}, ambayo inatoa: TiViγ−1=TfVfγ−1T_i V_i^{\gamma - 1} = T_f V_f^{\gamma - 1}, yaani Tf=Ti 2−(γ−1)=300×2−2/3≈189T_f = T_i\,2^{-(\gamma-1)} = 300 \times 2^{-2/3} \approx 189 K. Kumbuka kwamba ni lazima kutumia halijoto katika kelvini kwenye hesabu hii ili kuepuka kosa! Tunapata ΔT=Tf−Ti≈−111\Delta T = T_f-T_i \approx -111 K. Kwa kazi, kwa kuwa Q=0Q = 0:

W=ΔU=32RΔT≈32×8,314×(−111)≈−1,4 kJ.W = \Delta U = \tfrac{3}{2}R\Delta T \approx \tfrac{3}{2}\times 8{,}314 \times (-111) \approx -1{,}4\ \text{kJ}.

W<0W < 0: gesi hufanya kazi kwa mazingira ya nje wakati wa upanuzi. Kwa kuwa haiwezi kuchota nishati hiyo kutoka chanzo chochote cha joto (Q=0Q=0), huitoa katika nishati yake ya ndani, na hivyo kupoa kwa ΔT<0\Delta T < 0.

Swali la 5. Kwa kufanya diferenshali ya PV=thabitiPV = \text{thabiti}: (dPdV)isoT=−PV\left(\frac{\mathrm{d}P}{\mathrm{d}V}\right)_{\text{isoT}} = -\dfrac{P}{V}. Kwa kufanya diferenshali ya PVγ=thabitiPV^\gamma = \text{thabiti}: (dPdV)adiab=−γPV\left(\frac{\mathrm{d}P}{\mathrm{d}V}\right)_{\text{adiab}} = -\gamma\dfrac{P}{V}. Adiabati ina mteremko mkali mara γ\gamma kuliko isothemali: kubana bila kuondoa joto huongeza shinikizo kwa kasi zaidi, kwa sababu halijoto pia huongezeka.