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बनाया गया: 15 जनवरी 2026

का उपयोग करके उत्तर दिया गया GPT-5.2 Thinking द्वारा Chat01

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Calculating work for thermodynamic cycle

In Russian, "Найдите разность теплоты в процессах 1-2 и 2-3" likely refers to Q_12 - Q_23, asking for the difference in heat. I'll compute the work for the cycle, starting with the segments—Segment 1-2 has no work, Segment 2-3 has 6p0V0, and Segment 3-1 is linear for pressure-volume changes.

Calculating work for thermodynamic cycle

I derived the expression for pressure along segment 3-1: p = p0*(V/V0). Then, I computed the work during compression from 3V0 to V0, which gives W31 = -4 p0 V0. The net work for the cycle is 2 p0 V0. Next, I’ll find the heats Q12 and Q23 using the first law for a monoatomic ideal gas.

Calculating heat difference and final result

For process 1-2, the heat Q12 is 3 p0 V0, and for process 2-3, it’s 15 p0 V0. The difference Q12 - Q23 is -12 p0 V0, and when we take the absolute value, the result is 600 kJ. I’ll make sure to show all steps clearly and concisely in Russian.

Thought for 1m 2s

उत्तर

По графику:
1: (V0,p0)(V_0,p_0), 2: (V0,3p0)(V_0,3p_0), 3: (3V0,3p0)(3V_0,3p_0). Газ одноатомный, поэтому

U=32 pV,Q=ΔU+A.U=\frac32\,pV,\qquad Q=\Delta U + A.

1) Найдём p0V0p_0V_0 из работы цикла

Работа за цикл равна площади фигуры на pVpV-диаграмме.

  • 1→21\to2: V=const⇒A12=0V=\text{const}\Rightarrow A_{12}=0
  • 2→32\to3: p=3p0⇒A23=3p0(3V0−V0)=3p0⋅2V0=6p0V0p=3p_0\Rightarrow A_{23}=3p_0(3V_0-V_0)=3p_0\cdot2V_0=6p_0V_0
  • 3→13\to1: прямая от (3V0,3p0)(3V_0,3p_0) к (V0,p0)(V_0,p_0). Её уравнение p=p0V0Vp=\frac{p_0}{V_0}V.
A31=∫3V0V0p0V0V dV=p0V0⋅V22∣3V0V0=−4p0V0A_{31}=\int_{3V_0}^{V_0}\frac{p_0}{V_0}V\,dV=\frac{p_0}{V_0}\cdot\frac{V^2}{2}\Big|_{3V_0}^{V_0}=-4p_0V_0

Итого:

Aцикла=0+6p0V0−4p0V0=2p0V0=100 кДж⇒p0V0=50 кДж.A_{\text{цикла}}=0+6p_0V_0-4p_0V_0=2p_0V_0=100\text{ кДж} \Rightarrow p_0V_0=50\text{ кДж}.

2) Теплоты на участках 1–2 и 2–3

1→2 (изохора):

ΔU12=32(3p0V0−p0V0)=32⋅2p0V0=3p0V0\Delta U_{12}=\frac32(3p_0V_0-p_0V_0)=\frac32\cdot2p_0V_0=3p_0V_0 Q12=ΔU12+A12=3p0V0Q_{12}=\Delta U_{12}+A_{12}=3p_0V_0

2→3 (изобара p=3p0p=3p_0):

ΔU23=32(9p0V0−3p0V0)=32⋅6p0V0=9p0V0\Delta U_{23}=\frac32(9p_0V_0-3p_0V_0)=\frac32\cdot6p_0V_0=9p_0V_0 Q23=ΔU23+A23=9p0V0+6p0V0=15p0V0Q_{23}=\Delta U_{23}+A_{23}=9p_0V_0+6p_0V_0=15p_0V_0

Разность:

∣Q12−Q23∣=∣3−15∣p0V0=12p0V0=12⋅50=600 кДж.|Q_{12}-Q_{23}|=|3-15|p_0V_0=12p_0V_0=12\cdot50=600\text{ кДж}.

Ответ: 600600 кДж.

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