ΔT = (p0 ΔS + m g) l / (n R). For your sliders ΔT = K. For the book numbers (ΔS = 10 cm², m = 5.0 kg, p0 = 1.0 atm, l = 5.0 cm, one mole) ΔT = 0.904 K, which the book rounds to 0.9 K.
Problem statement
A smooth vertical tube has two sections of different width. It is open at both ends. Two pistons slide in it, one in each section. A thread that cannot stretch ties them together. One mole of ideal gas is trapped between the pistons. The upper piston has an area ΔS = 10 cm² bigger than the lower one. The two pistons together have mass m = 5.0 kg. The outside air pressure is p0 = 1.0 atm. By how many kelvins must the gas be heated to move the pistons up by l = 5.0 cm?
We assume the pistons slide without friction and seal the gas perfectly. This means only gas, air, weight and thread push on them.
We assume the thread is light and does not stretch. This means the pistons always stay the same distance L apart and move together.
We assume we heat slowly. The pistons are then always in balance (a quasi-static process), so we can use the balance of forces at every moment.
We assume the gas is ideal. Its molecules are tiny and only bounce. We use it in pV = nRT.
We assume the pistons stay in their own sections while they move. The lower piston must not reach the step where the tube widens. The live warnings check this.
Symbols: S1 and S2 are the areas of the upper and lower piston, and ΔS = S1 − S2. The masses of the pistons are m1 and m2, and m = m1 + m2. Also p is the gas pressure, p0 is the outside air pressure, g = 9.81 m/s² is the pull of gravity, n = 1 mol is the amount of gas, R = 8.314 J/(mol K) is the gas constant, l is the shift of the pistons, and ΔT is the temperature rise. Sign rule: W is the work done BY the gas, and dU = Q − W.
The big idea in plain words
The pair of pistons is in balance. That fixes the gas pressure, because only the weight and the pressure gap act on the pair. The pressure stays the same while we heat. So this is a constant-pressure (isobaric) heating. Then the volume grows by ΔS times l, and the gas law gives ΔT.
The solution from one viewpoint
Reference frames do not apply. The tube is fixed and the pistons move slowly, so there is only the lab viewpoint.
First we find the gas pressure, because we need it for the gas law. We look at the two pistons and the thread as one body. Newton's first law says the forces on a body at rest add to zero. Pressure times area is the force of a gas on a surface.
Look at the result. The pressure has no temperature in it. So it stays the same while we heat, as long as the pistons are free.
Next we find the volume change, because the gas law links it to ΔT. When both pistons rise by l, the upper piston sweeps new gas volume S1 l. The lower piston moves up too, and removes the volume S2 l from the narrow part. The net gain is the difference.
Now we use the gas law, because the pressure is fixed. The law p V = n R T holds before and after. We subtract the two with the same p and n.
The result says the heating must raise the temperature by less than one kelvin. The answer needs no start temperature and no piston areas, only the difference ΔS.
The reference gives ΔT = (1/R)(p0 ΔS + m g) l = 0.9 K. We solved it from first principles and got K for the book numbers (0.904 K), so the two agree. In the reference, equation (2) is printed with p S1 on both sides. The right side should be p0 S1, since the air above pushes down with p0. The next line of the reference, (p − p0) ΔS = m g, shows the author used p0, so the result is not affected.
The solution by each method
Method 1: Energy method
- We want the work the gas does, because it equals the energy the gas gives up in pushing and lifting. The pistons move slowly, so their kinetic energy stays zero. The gas work goes into pushing the air back and lifting the pistons.
- Pushing the air: the air pressure p0 acts on the net extra area ΔS, so W_air = p0 ΔS l = N × m = J.
- Lifting the pistons by l: W_lift = m g l = N × m = J.
- Total work by the gas: W = W_air + W_lift = J. This is also p ΔV, which checks the pressure we found.
- For an ideal gas at constant p, W = p ΔV = n R ΔT, so ΔT = W / (n R) = K.
- The first law then gives the heat: dU = n Cv ΔT = J (f = , Cv = f R / 2), and Q = dU + W = J.
Method 2: Graph method
- On a P-V graph, the heating is a flat line at p = atm, from V0 to V0 + ΔV.
- The area under it is the work: W = p ΔV = J. The first graph on the canvas shades it.
- The same area equals n R ΔT, because the isotherms through the two end points differ by that amount of p V.
- So ΔT = area / (n R) = K. The middle graph shows T rising as a straight line with the shift s, with slope p ΔS / (n R) = K per cm.
Method 3: Calculus method
- Let the pistons move up a distance x. The volume changes by dV = ΔS dx.
- The pressure is the constant p = p0 + m g/ΔS, so the work is W = ∫ p dV = ∫ from 0 to l of p ΔS dx = p ΔS l.
- For the gas, d(p V) = n R dT. With p constant, p dV = n R dT, so ∫ n R dT = ∫ p ΔS dx.
- n R ΔT = (p0 ΔS + m g) l, so ΔT = K.
Method 4: State-variable method
| State | p | V | T |
|---|---|---|---|
| Start | atm | l | K |
| End | atm | l | K |
- Equation of state: p V = n R T, with n = mol.
- The process is isobaric, so its law is V / T = constant (Charles's law). That gives ΔT = T0 ΔV / V0.
- With V0 = n R T0 / p this becomes ΔT = p ΔV / (n R) = K. The start temperature cancels.
- Process check: p is equal in both rows, so the heating is isobaric. The ratio V1/V0 = equals T1/T0 = .
Method 5: Check method
- Units: p ΔS l is pascal × m² × m = newton × metre = joule. Joule divided by (mol × J/(mol K)) is kelvin.
- Limit l = 0: ΔT = 0. Correct.
- Limit m = 0: the pistons are light, and ΔT = p0 ΔS l / (n R) = K. Only the air has to be pushed back.
- Limit ΔS → 0: the volume change ΔS l goes to zero, but the pressure p = p0 + m g/ΔS grows without limit. Their product stays finite, so ΔT stays at m g l / (n R) = K.
- Direction check: the upper piston is the larger one. Heating makes the gas push harder, so the pistons rise, as asked. If the lower piston were larger, they would fall.
- Link to the reference: it gives 0.9 K, and the live book value is 0.904 K.
Fastest: ΔT = (p0 ΔS + m g) l / (n R). Most reliable in an exam: balance the forces on both pistons first, then use p ΔV = n R ΔT, and run the check method.
What each slider changes
| Slider | Outputs it changes | How | Why |
|---|---|---|---|
| ΔS (area difference) | ΔT, pressure, tube step, forces | ΔT grows with ΔS, and tends to m g l/(nR) as ΔS → 0 | A bigger step gives more volume gain per metre, but less pressure. |
| m (piston mass) | ΔT, pressure, weight arrow, W | ΔT grows in proportion to m g for the weight term | A heavier pair needs a higher gas pressure to hold it. |
| p0 (air pressure) | ΔT, pressure, W against air | ΔT grows in proportion to p0 for the air term | Pushing back more air takes more work. |
| l (shift) | ΔT, piston positions, W, Q | proportional | The volume gain is ΔS l. |
| n (moles) | ΔT, start temperature, dU, Q, dots | ΔT inverse to n | More gas shares the same work, so each mole warms less. |
| S2 (lower area) | start temperature, tube width, volume | none for ΔT | Only the difference S1 − S2 matters, and S2 cancels. |
| L (thread length) | start temperature, tube height | none for ΔT | L fixes the gas volume, but ΔT does not use it. |
| f0 (start position) | start temperature, picture | none for ΔT | The start state cancels in p ΔV = n R ΔT. |
| f (gas type) | dU, Q, γ | none for ΔT | f sets the heat needed, not the temperature rise for a given volume change. |
| Heating time | heating rate, warning | none for ΔT | It only checks that the heating is slow. |
| Target ΔT (given-answer mode) | shift l | l proportional to ΔT | The formula is solved backwards. |
Parameter table
| Symbol | Meaning | Value now |
|---|---|---|
| ΔS, S2, S1 | area difference, lower area, upper area | cm², cm², cm² |
| m | combined piston mass | kg |
| p0, p | air pressure, gas pressure | atm, atm |
| l, L | shift of the pistons, thread length | cm, cm |
| n | moles of gas | mol |
| T0, T1 | start and end temperature (set by the geometry and p) | K, K |
| 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 |
| f, Cv, γ | degrees of freedom (3 monatomic, 5 diatomic, 6 polyatomic), Cv = fR/2, γ = (f + 2)/f | , J/(mol K), (vibration frozen out below about 600 K) |
Common exam traps
- Using p0 as the gas pressure. The gas pressure is p0 + m g/ΔS, which is higher.
- Writing the volume change as S l instead of ΔS l. One piston gains volume and the other loses it.
- Forgetting the weight term m g, or forgetting the air term p0 ΔS.
- Mixing units: cm² and cm with pascals. Convert to m² and m first.
- Using a constant-volume or isothermal law. The pressure is fixed, so the process is isobaric.
- Thinking the answer depends on the start temperature or on the single areas. It depends on ΔS only.