Problem statement

Find the minimum attainable pressure of an ideal gas in the process T = T₀ + αV², where T₀ and α are positive constants and V is the volume of one mole of the gas.

We assume sample values T₀ = 300 K and α = 0.2 K/L². This means the page can show real numbers. The sliders change them, and the formula holds for any positive values.

We assume pressure in atm and volume in litres, so R = 0.082057 L atm/(mol K). This means pressure comes out in atm. In SI, R = 8.314 J/(mol K).

We assume the gas is in equilibrium at every step. This means pV = RT holds at every point of the process.

Symbols: p is pressure, V is volume of one mole, T is temperature, R is the gas constant. pmin is the lowest pressure along the process.

The big idea in plain words

For one mole, p = RT/V. When the gas expands along this process, T rises, but V rises too. At small V the volume is tiny, so the pressure is huge. At large V the temperature grows like V², so the pressure RT/V grows again. In between, the pressure dips to a lowest point. On the p–V diagram this is where a horizontal line p = constant just touches the curve from below.

The solution in each frame

Lab frame

The gas sits in a cylinder at rest. The full solution is in the next section.

Other frames

No other frame is needed. Only the state of the gas matters.

The solution step by step

For one mole of an ideal gas pV = RT, so

Solve the process for V: V = √((T − T₀)/α). Put this into p:

The minimum is where dp/dT is zero. Differentiate T (T − T₀)^−1/2 with the product rule:

Multiply by 2 (T − T₀)^3/2: 2(T − T₀) − T = 0, so

At this temperature V = √(T₀/α). Then

For the sample values: T = 600 K, V = √(300/0.2) = 38.73 L, pmin = 2 × 0.082057 × √60 = 1.271 atm. For the state now on screen, pmin = atm.

Five methods

1. Force method

Not needed. The problem asks about the state of the gas, not about forces on the piston.

2. Energy method

Not needed. The minimum comes from the gas law and the shape of the process.

3. Graph method

Plot p against V. The lowest point of the curve is the answer. The pressure against temperature graph shows the same minimum at T = 2T₀.

4. Calculus method

Using V as the variable: p = R (T₀ + αV²)/V = R (T₀/V + αV). Then dp/dV = R (−T₀/V² + α) = 0 gives V = √(T₀/α), the same result.

5. Check method

This one is also quick by an inequality: T₀/V + αV is at least 2√(αT₀), with equality when T₀/V = αV. Units: atm L/(mol K) times √(K/L² × K) = K/L gives atm. Limits: pmin grows as the square root of T₀ and of α.

Fastest: the inequality in method 5. Most reliable: the numerical scan.

What each slider changes

Control	Changes the answer?	What it changes
Starting constant T0	Yes	The curve, V and T at the minimum, pmin and all graphs
Constant α	Yes	How fast T rises with V, so the minimum, pmin and all graphs
Volume of the gas now	No	The piston height, p and T now, and the cyan dot
Given-answer mode	Sets T0	The page finds the T0 that gives the chosen pmin

Parameters

Quantity	Symbol	Value used
Starting constant	T0	K
Process constant	α	K/L²
Volume at the minimum	V	L
Temperature at the minimum	T	K
Least pressure	pmin	atm

Exam traps

• Setting dT/dV = 0. That gives V = 0, which is not a pressure minimum.
• Forgetting that V is the volume of one mole. For n moles, pV = nRT.
• Mixing units. In pascals and cubic metres use R = 8.314. In atm and litres use 0.0821.
• Finding T = 2T₀ but not converting to p. The answer needs V as well.
• Thinking the minimum pressure is where the temperature is least. The temperature is least at V = 0, where the pressure is largest.