(a) For p = p0 − αV², Tmax = (2p0/3R) √(p0/3α). (b) For p = p0 e−βV, Tmax = p0/(eβR). With your numbers: (a) Tmax = K at V = L and p = atm. (b) Tmax = K at V = L and p = atm.
Problem statement
Find the maximum attainable temperature of an ideal gas in each of these processes: (a) p = p0 − αV², and (b) p = p0 e−βV. Here p0, α and β are positive constants, and V is the volume of one mole of the gas.
We assume sample values p0 = 3 atm, α = 0.002 atm/L² and β = 0.04 per litre. This means the page can show real numbers. The sliders change them, and the formulas hold for any positive values.
We assume pressure in atm and volume in litres, so R = 0.082057 L atm/(mol K). This means the temperature comes out in kelvin. 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. Tmax is the highest temperature the gas reaches along the process.
What α and β mean
In both processes the pressure starts at p0 when the volume is tiny, and then falls as the gas expands. The two constants tell us how fast.
α in process (a). The pressure lost is αV². Each litre-squared of V² takes α atm off the pressure. At small V the loss is tiny, so the pressure is nearly p0. As V grows the loss grows faster and faster, because V is squared. At V = √(p0/α) the whole of p0 is lost and the pressure reaches zero. So the curve ends there.
β in process (b). The pressure kept is p0 e−βV. Each extra litre multiplies the pressure by e−β. For β = per litre this factor is , so about the same share of the pressure is lost in every litre. The pressure keeps falling but never reaches zero.
On the page, the grey part of each pressure gauge shows the pressure lost, and the third graph plots p0 − p against V for both processes.
The big idea in plain words
For one mole, pV = RT. This one equation joins three quantities, p, V and T. If only the gas law is used, T can be anything, because we do not yet know how p and V go together. The process equation gives that missing link. It tells us what p is for every V. Then T = pV/R depends on one variable only, and we can ask where it is largest.
As the gas expands, p falls but V rises. At first V rises faster than p falls, so the product pV and the temperature grow. Later p falls faster, so the temperature drops. In between there is a peak. On the p–V diagram the thin curves are isotherms (pV = constant at a fixed temperature). The dot crosses from cooler isotherms to hotter ones. The hottest one it can reach is the one that just touches the curve, and the touching point is the answer.
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. Nothing moves except the piston, and only the gas state matters.
The solution step by step
For one mole of an ideal gas the state equation is pV = RT, so
(a) p = p0 − αV²
Solve the process for V: V = √((p0 − p)/α). Put this into T:
The maximum is where the slope dT/dp is zero. Differentiate p (p0 − p)1/2 with the product rule:
Multiply by 2 (p0 − p)1/2: 2(p0 − p) − p = 0, so
At this pressure V = √((p0 − 2p0/3)/α) = √(p0/(3α)). Then
For the sample values (3 atm, 0.002 atm/L²): p = 2 atm, V = 22.36 L, Tmax = 2 × 22.36 / 0.082057 = 545 K.
(b) p = p0 e−βV
Take the logarithm: βV = ln(p0/p), so V = ln(p0/p)/β. Then
Differentiate: d/dp [p ln(p0/p)] = ln(p0/p) − 1. Setting this to zero:
At this pressure V = 1/β, so
For the sample values (3 atm, 0.04 per litre): p = 1.104 atm, V = 25 L, Tmax = 1.104 × 25 / 0.082057 = 336 K.
A separate check does not use the derivative. For each process it computes T = pV/R at 4000 volumes, picks the largest, and refines it by a golden-section search. It finds (a) Tmax = K against the formula value K, and (b) Tmax = K against K. A second check compares the slope of each process with the slope of its isotherm at the peak. They agree.
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 maximum comes from the gas law and the shape of the process.
3. Graph method
On the p–V diagram, draw isotherms pV = RT. The highest one that still touches the process curve is the answer. The point of contact is the maximum. The temperature graph shows the same peak.
4. Calculus method
Write T as a function of one variable and set its derivative to zero, as in the steps above. Using V as the variable also works: in (a), d(p0V − αV³)/dV = 0 gives V = √(p0/3α). In (b), d(V e−βV)/dV = 0 gives V = 1/β.
5. Check method
Units: atm × L divided by atm L/(mol K) gives kelvin. Limits: in (a), Tmax grows as p0 to the power 1.5 and falls as α grows. In (b), Tmax is in proportion to p0 and falls as β grows. Both agree with the picture: a steeper fall of the pressure leaves less room for the gas to heat.
Fastest: differentiate pV with V as the variable. Most reliable: the numerical scan.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Pressure constant p0 | Yes, both | Both curves, V and p at both peaks, both Tmax values and all graphs |
| α (process a) | Yes, (a) | How much pressure each L² of V² removes, so curve (a), its peak and the graphs |
| β (process b) | Yes, (b) | The share of pressure kept per litre, so curve (b), its peak and the graphs |
| Volume in cylinder (a) or (b) | No | The piston height, the pressure and temperature there, and the dot on that curve |
| Given-answer mode | Sets α and β | The page finds the α and β that make both processes peak at the chosen temperature |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Pressure constant | p0 | atm |
| Constant of (a) | α | atm/L² |
| Constant of (b) | β | per L |
| (a) peak: V, p | L, atm | |
| (a) maximum temperature | Tmax | K |
| (b) peak: V, p | L, atm | |
| (b) maximum temperature | Tmax | K |
Exam traps
- Setting dp/dV = 0 instead of d(pV)/dV = 0. The maximum of the temperature is not the maximum of the pressure.
- Forgetting that V is the volume of one mole. For n moles, pV = nRT, and the answer is multiplied by 1/n.
- Mixing units. If p is in pascals and V in cubic metres, use R = 8.314 J/(mol K). If atm and litres, use 0.0821.
- In (a), using p = 2p0/3 but forgetting to find V from the process. Tmax needs both p and V.
- In (b), writing ln(p0/p) = 0 instead of 1. The factor of p in front gives the extra term.