T = T0 · (η² − 1) · η' / ((η'² − 1) · η). For your sliders T = K. For the book numbers T = 421.9 K, which the book writes as 0.42 kK (about 420 K).
Problem statement
A vertical cylinder is closed at both ends. A piston that moves easily splits it into two parts. Each part holds one mole of air. At T0 = 300 K the upper part has η = 4.0 times the volume of the lower part. At what temperature does the ratio of the volumes become η' = 3.0?
We assume the piston slides without friction and is gas-tight. It always sits where the forces on it balance.
We assume the piston passes heat, so both gases always have the same temperature T. We heat slowly, so this holds at every moment (a quasi-static process).
We assume air is an ideal gas. Its molecules are tiny and only bounce. We use this in pV = nRT.
We assume the piston has a mass m, which is not given. It must have one, otherwise the two parts would have equal volume. We find it from the first balance.
We assume the total volume Vt and the cross-section area S are not given. Both cancel from the answer. We pick values to draw the picture.
Symbols: n = 1 mol is the amount of air in each part, R = 8.314 J/(mol K) is the gas constant, g = 9.81 m/s² is the pull of gravity, p1 and p2 are the pressures above and below the piston, and V1, V2 are the volumes above and below. Sign rule for the energy part: W is the work done BY the gas, and dU = Q − W.
The big idea in plain words
The piston has weight. So the pressure below must be larger than the pressure above, by exactly the weight divided by the area. That gap stays the same at every temperature, because the weight does not change. The total volume does not change either. These two facts fix the answer.
The solution from one viewpoint
Reference frames do not apply. The cylinder is fixed, and the piston is at rest at every moment. There is only the lab viewpoint.
First we balance the piston at T0, because that tells us its weight. Three forces act on it: the weight mg down, the push of the upper gas p1 S down, and the push of the lower gas p2 S up. At rest, Newton's first law says the forces add to zero, so p2 S = p1 S + mg.
Next we balance the piston again at the new temperature T. The lower volume is now V' and the upper is η' V'. The weight is the same, so the same equation holds with new numbers.
Now we use the fact that the total volume is fixed, because the cylinder does not change size.
Then we set (1) equal to (2), because both equal mg / S, and put in (3). The factors n, R and V0 cancel.
The result tells us we must heat the gas by a factor of about 1.4 to move the piston down the ratio from 4 to 3. It is not enough to heat both parts equally and expect the ratio to stay: the weight of the piston makes the lower gas respond differently.
The solution by each method
Method 1: Energy method
This does not find T, because the heat Q is not given. We use it as a check on the energy rules. Air is diatomic, so f = 5 (3 movements and 2 spins; vibration is frozen out below about 1000 K) and Cv = f R / 2 = J/(mol K). Heating two parts from T0 to T raises the internal energy by dU = 2 n Cv (T − T0). The two gases do net work only by lifting the piston: W = mg Δh, where Δh is how far the piston rises. Derivation: the lower gas does W2 = ∫ p2 dV2, the upper gas does W1 = ∫ p1 dV1, with dV1 = −dV2, so W1 + W2 = ∫ (p2 − p1) dV2 = (mg/S) ΔV2 = mg Δh.
Method 2: Graph method
- Plot the gap p2 − p1 = n R T (1/V2 − 1/(Vt − V2)) against the lower volume V2 for the unknown temperature. At T0 it equals mg/S at V2 = V0.
- At the new temperature the gap curve moves up. We need the point where it again equals mg/S, the same horizontal line.
- The P-V graph on the canvas shows the two gases. The lower gas moves right and the upper gas moves left, with the pressure gap fixed at atm.
- Reading the new volumes: V2 = l and V1 = l, ratio η' = . The temperature that gives this is K (the middle graph on the canvas reads it off).
Method 3: Calculus method
- Write the balance at any state with ratio η: mg/S = (n R T / V2)(1 − 1/η) and V2 = Vt / (1 + η). So T is a function of η alone.
- Put these together: T(η) = K η / (η² − 1), where K = (mg/S) Vt / (n R) = T0 (η0² − 1)/η0 is a constant fixed by the start.
- Differentiate to see how T changes with η: dT/dη = −K (η² + 1) / (η² − 1)², which is negative. So heating always lowers η, and every η' in range has exactly one T.
- Evaluate: T = K η' / (η'² − 1) = K × / (² − 1) = K.
Method 4: State-variable method
| State | Upper gas (p1, V1, T) | Lower gas (p2, V2, T) |
|---|---|---|
| Start | atm, l, K | atm, l, K |
| End | atm, l, K | atm, l, K |
- Equation of state in each part: p V = n R T. Both parts have n = mol.
- Conditions: p2 − p1 = mg/S = atm (both states) and V1 + V2 = Vt = l.
- Solving them gives T = K.
- Process check: the whole system is heated slowly (quasi-static), at constant total volume, but each part is neither isothermal, isobaric nor isochoric. Both parts change V, p and T together. The test p2 − p1 = atm at the end confirms the balance.
Method 5: Check method
- Units: T0 is in kelvin and the other factors are pure ratios, so T is in kelvin.
- Limit η' = η: T = T0, as it should be (no change).
- Limit η' → 1 (equal volumes): the factor (η'² − 1) → 0, so T → infinity. That is right: with the piston's weight, only an infinite temperature makes the pressures nearly equal.
- Limit η' larger than η: T is below T0. Cooling makes the ratio grow. That matches the physics.
- Direction check: is smaller than and T is larger than T0. Good.
Fastest: the formula T = T0 (η² − 1) η' / ((η'² − 1) η). Most reliable in an exam: balance the piston twice, then use the check method.
What each slider changes
| Slider | Outputs it changes | How | Why |
|---|---|---|---|
| T0 | T, pressures, forces, dot speed | T proportional to T0 | Everything scales with the starting temperature. |
| η (start ratio) | T, piston mass, start height | T not simply proportional | η sets the weight that the new ratio must still balance. |
| η' (end ratio) | T, end height, Q, W, dU | T falls as η' rises. T goes to infinity as η' goes to 1. | A ratio near 1 needs great heat to overcome the weight. |
| n (moles per part) | pressures, forces, energies, dots | none for T | n cancels from both balances. |
| Vt (total volume) | cylinder height, pressures, forces | none for T | V cancels in the volume ratios. |
| S (area) | cylinder height drawn, piston mass | none for T | Only the pressure gap mg/S matters, and S cancels. |
| Target T | η' (given-answer mode) | η' falls as T rises | The formula is solved backwards. |
| Heating time | heating rate, warning | none for T | It only checks the slow-heating assumption. |
Parameter table
| Symbol | Meaning | Value now |
|---|---|---|
| T0, T | start and end temperature | K, K |
| η, η' | start and end volume ratio (upper / lower) | , |
| n | moles in each part | mol |
| Vt, S | total volume, cross-section | l, cm² |
| m | piston mass (from the first balance) | kg |
| R, g, kB, NA | gas constant, gravity, Boltzmann constant, Avogadro number | 8.314 J/(mol K), 9.81 m/s², 1.381 × 10⁻²³ J/K, 6.022 × 10²³ per mol |
| Cv, γ | heat capacity at constant volume, Cp / Cv (air, f = 5) | J/(mol K), |
Common exam traps
- Saying the ratio stays the same because both gases are heated equally. The piston's weight breaks that.
- Forgetting the weight of the piston. Without it, the ratio would stay at 1.
- Writing the upper volume as V' instead of η' V'.
- Putting the total volume as V0 + V' instead of V0 (1 + η).
- Using °C. The factor T0 must be in kelvin.
- Taking the answer to be T0 × η/η' = 400 K. That ignores the 1 − 1/η factors.