Cyclic Process Energy Balance Tool

Physics Thermodynamics • First Law of Thermodynamics

Written by STEM Calculators Team Published January 2, 2026 Updated February 24, 2026

6. Cyclic Process Energy Balance Tool

A closed thermodynamic cycle returns to its starting state, so $\Delta U_{\text{cycle}}=0$ and therefore $W_{\text{net}}=Q_{\text{net}}$ . On a PV diagram, $W_{\text{net}}=\oint P\,dV$ equals the signed area enclosed by the loop (sign depends on direction).

Inputs support: pi, e, sqrt(), sin, cos, exp, log. Use * for multiplication.

Cycle builder (states + process types)

Number of states: Preset: Volume unit (inputs): Pressure unit (inputs): Adiabatic index $\gamma$: University mode (efficiency): $Q_{in}$ (J, total heat input):

Define corner states $(V_i,P_i)$. For each state $i$, choose the process from state $i\to i+1$ (the last connects back to state 1). The calculator computes each segment work and totals $W_{\text{net}}$. Then $\Delta U_{\text{cycle}}=0\Rightarrow Q_{\text{net}}=W_{\text{net}}$.

State	Volume $V$	Pressure $P$	Process to next

Preset loaded. Edit values if needed.

PV graph + loop progress

Progress $\tau\in[0,1]$: Play speed: Shade enclosed loop: Direction arrows:

PV graph supports drag-to-pan, wheel-to-zoom, double-click or “Reset view” to restore.

Ready

Steps

Define a cycle and click Solve.

PV graph (cycle loop + area)

Closed PV loop: shaded area = net work (sign by direction)

Hover: (V,P)=… Marker: $\tau=1$

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Theory: Cyclic Process Energy Balance

A thermodynamic cycle is a sequence of processes that returns the system to its initial state. Because internal energy $U$ is a state function, the net change over one complete cycle is:

\[ \Delta U_{\text{cycle}} = 0. \]

1) First law over a cycle

Using the convention $W>0$ for work done by the system:

\[ \Delta U = Q - W. \]

Over a cycle, $\Delta U_{\text{cycle}}=0$, so:

\[ 0 = Q_{\text{net}} - W_{\text{net}} \quad\Rightarrow\quad Q_{\text{net}} = W_{\text{net}}. \]

2) Net work equals PV-loop area

For quasi-static boundary work:

\[ W_{\text{net}} = \oint P\,dV. \]

On a PV diagram, $\oint P\,dV$ equals the signed area enclosed by the loop: clockwise cycles yield $W_{\text{net}}>0$ (engine-like), while counterclockwise cycles yield $W_{\text{net}}<0$ (work input).

3) Common segment work formulas

\[ \begin{aligned} \text{Isochoric }(V=\text{const})&:\quad W=\int P\,dV = 0\\ \text{Isobaric }(P=\text{const})&:\quad W=P(V_2-V_1)\\ \text{Linear }P(V)&:\quad W=\frac{P_1+P_2}{2}(V_2-V_1)\\ \text{Isothermal (ideal gas)}&:\quad W=K\ln\!\left(\frac{V_2}{V_1}\right),\ K=P_1V_1\\ \text{Adiabatic }(PV^\gamma=\text{const})&:\quad W=\frac{K}{1-\gamma}\left(V_2^{1-\gamma}-V_1^{1-\gamma}\right),\ K=P_1V_1^\gamma \end{aligned} \]

4) Efficiency (university mode)

If the cycle is a heat engine and the total heat input is $Q_{in}$, a common definition is:

\[ \eta = \frac{W_{\text{net}}}{Q_{in}}. \]

Computing $Q_{in}$ exactly generally requires heat transfer along each segment; here you can supply $Q_{in}$ to estimate $\eta$.

5) Units of PV area

The loop area has units of pressure × volume. In SI, $\text{Pa}\cdot\text{m}^3=\text{J}$. Useful equivalences:

PV unit	equals
$\text{Pa}\cdot\text{m}^3$	$1\ \text{J}$
$\text{kPa}\cdot\text{L}$	$1\ \text{J}$
$\text{bar}\cdot\text{L}$	$100\ \text{J}$
$\text{atm}\cdot\text{L}$	$101.325\ \text{J}$

Website tip: connect this PV-loop interpretation to engine cycles (Carnot, Otto, Diesel, Brayton).

PV unit	equals
\(\text{Pa}\cdot\text{m}^3\)	\(1\ \text{J}\)
\(\text{kPa}\cdot\text{L}\)	\(1\ \text{J}\)
\(\text{bar}\cdot\text{L}\)	\(100\ \text{J}\)
\(\text{atm}\cdot\text{L}\)	\(101.325\ \text{J}\)

Steps

PV graph (cycle loop + area)

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