Virtual laboratory/Physics/Thermodynamics/Gas processes — what stays constant

Gas processes — what stays constant

Pick a type of gas process and vary a parameter. Watch the p-V diagram and check what stays constant: temperature (isotherm), pressure (isobar) or volume (isochore). All of them satisfy the equation of state pV/T=const.

Gas processes — what stays constant
Physics · Thermodynamics · Gas processes (FIZ.TERM.PP.02)
Process type
Temperature TT (p = 150 kPa fixed)300 K
In an isothermal process the constant quantity is:
pressure p
150 kPa
volume V
2.00 L
temperature T
300 K
pV/T
1.00
🎯 Challenge: hit the pressure
streak: 0

I will give you a target pressure — choose the process and parameter (slider) to reach it. pV/T = nR.

What just happened?

Every ideal-gas process satisfies the same equation of state: the ratio pV/TpV/T does not change. You fix one quantity as constant, and the other two change together so that the ratio is preserved.

Ideal-gas equation of state:

pVT=const=nR\dfrac{pV}{T} = \text{const} = nR

— p is pressure (Pa), V is volume (m³), T is absolute temperature (K), n is the number of moles, R is the gas constant. For two states of the same gas: p1V1T1=p2V2T2\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}.

Three special processes:

pV=const⏟isotherm. T=constVT=const⏟isobar. p=constpT=const⏟isochor. V=const\underbrace{pV=\text{const}}_{\text{isotherm. }T=\text{const}} \qquad \underbrace{\tfrac{V}{T}=\text{const}}_{\text{isobar. }p=\text{const}} \qquad \underbrace{\tfrac{p}{T}=\text{const}}_{\text{isochor. }V=\text{const}}

On the p–V diagram the isotherm is a hyperbola, the isobar is a horizontal line, and the isochore is a vertical line. Always write down first what is constant — that quantity dictates the shape of the curve.

Why must the temperature be in kelvins?

The ratio pV/TpV/T works only for absolute temperature. If you plugged in degrees Celsius, at 0 °C you would divide by zero — that is why always T[K]=t[∘C]+273.15T[\text{K}] = t[^\circ\text{C}] + 273.15.

Why you need this for the exam:

The equation p1V1T1=p2V2T2\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2} solves almost any gas problem. Trick: mark which quantity is constant (cross it out on both sides), and a simple proportion between the other two remains.

Remember: pV/T = const for every ideal-gas process. Isothermal → T constant, isobaric → p constant, isochoric → V constant. Temperature always in kelvins.

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Theory & formulas

Equation of state of a gas

Every ideal-gas process satisfies: pVT=const\dfrac{p V}{T} = \text{const}

Three special processes

  • isothermal (T=constT=\text{const}): pV=constpV = \text{const} — Boyle-Mariotte law; the pp-VV diagram is a hyperbola,
  • isobaric (p=constp=\text{const}): VT=const\dfrac{V}{T} = \text{const}, i.e. V∝TV \propto T,
  • isochoric (V=constV=\text{const}): pT=const\dfrac{p}{T} = \text{const}, i.e. p∝Tp \propto T.

Common origin

Each of these processes is a special case of the ideal-gas equation of state (Clapeyron): pV=nRTpV = nRT.

Notation: pp — pressure, VV — volume, TT — absolute temperature (in kelvins), nn — number of moles, RR — gas constant.

Frequently asked questions

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What stays constant in an isothermal process?

Temperature; then pV=const (Boyle-Mariotte law), and the p-V diagram is a hyperbola.

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How do an isobar and an isochore differ?

In an isobar the pressure is constant (V∝T), in an isochore — the volume (p∝T).

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Where does the equation pV/T=const come from?

It is the ideal-gas equation of state (Clapeyron) — it unifies all three processes as special cases.

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