Summary: four-stage cycle

Problem

One mole of monatomic ideal gas undergoes a cycle: A→B isobaric at $p_1 = 1\ \text{atm}$ from $V = 10\ \text{L}$ to $V = 20\ \text{L}$; B→C isochoric at $V = 20\ \text{L}$ down to $p_2 = 0{,}5\ \text{atm}$; C→D isobaric at $p_2$ down to $V = 10\ \text{L}$; D→A isochoric back to $p_1$. Calculate the net work and check it against the area of the rectangle.

Solution

Key idea. In a cycle, the net work is the area enclosed in the $p$-$V$ plane. Here the cycle is a rectangle with sides $\Delta p$ and $\Delta V$:

$L_\text{netto}=(p_1-p_2)(V_2-V_1)$

Conversion to SI.

$\begin{aligned}\Delta p &= (1-0{,}5)\cdot 1{,}013\cdot 10^5 = 50650\ \text{Pa}\\ \Delta V &= (20-10)\cdot 10^{-3} = 0{,}01\ \text{m}^3\end{aligned}$

Net work.

$L_\text{netto}=50650\cdot 0{,}01\approx 506{,}5\ \text{J}$

The result is positive, so the cycle runs clockwise: it is a heat engine that produces work.

$\ev{L_\text{netto}\approx 506\ \text{J}}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Skills: Solving a thermodynamic cycle Methods: P-V and ln P - ln V diagram Objects: Ideal gas