How do I find the gas pressure in a tilted tube?

Only the vertical height of the mercury counts. If the open end makes angle θ with vertical-up and the column is h long, p = p₀ + h·cosθ in cmHg: open up gives p₀ + h, horizontal gives p₀, open down gives p₀ − h.

How do I tell whether mercury spills when the tube is turned?

Assume it does not, find L₂ from p₁L₁ = p₂L₂, then check whether L₂ + h exceeds the tube length. If it does, call the remaining column h′, so the gas length is the tube length minus h′, and solve p₁L₁ = (p₀ + h′cosθ)(tube length − h′) for the root between 0 and h.

Why does the process become constant-volume once the piston reaches the stops?

The stops block the piston, so the volume can no longer change. Further heating raises the pressure with p/T constant, and the stops push down on the piston. Find the temperature at which the piston reaches the stops first, then use the pressure law for the rest.

What sign does W take in ΔU = W + Q?

Here W is the work done on the gas: positive when the gas is compressed, negative when it expands and does work. Q is positive when the gas absorbs heat. An ideal gas’s internal energy depends only on temperature, so ΔU = 0 for an isothermal process.

Why do isochores on a p–T graph pass through the origin?

At constant volume p/T is constant (T in kelvin), so p is proportional to T and the line passes through the origin; a larger volume gives a smaller slope. Likewise isobars on a V–T graph pass through the origin, and isotherms on a p–V graph are hyperbolas, farther out for higher temperatures.

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🧪Trapped Gas Columns & Pistons - mercury tube, U-tube, cylinder, p–V

Built for the gas-law problems in high-school physics. A tube closed at one end turns from open-up to horizontal, tilted or open-down: the pressure follows p = p₀ ± h·cosθ, Boyle’s law gives the new column length, and spilling mercury is detected automatically. You can also add or remove mercury in a U-tube, heat a gas under a piston until it hits the stops, add springs or weights, or heat one of two chambers split by a piston. Every process is animated with p–V, p–T and V–T graphs and a step-by-step solution.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Trapped Gas Columns & Pistons

1Choose the tube, U-tube, cylinder or two-chamber model, then a preset or your own p₀, column lengths and temperatures
2Press Run or drag the slider to watch the mercury or piston move and the red dot travel along the three graphs
3Read the results and the worked solution (cmHg and cm for tubes, Pa for cylinders)
4Try random problems in Practice with instant marking

FAQ

How do I find the gas pressure in a tilted tube?: Only the vertical height of the mercury counts. If the open end makes angle θ with vertical-up and the column is h long, p = p₀ + h·cosθ in cmHg: open up gives p₀ + h, horizontal gives p₀, open down gives p₀ − h.
How do I tell whether mercury spills when the tube is turned?: Assume it does not, find L₂ from p₁L₁ = p₂L₂, then check whether L₂ + h exceeds the tube length. If it does, call the remaining column h′, so the gas length is the tube length minus h′, and solve p₁L₁ = (p₀ + h′cosθ)(tube length − h′) for the root between 0 and h.
Why does the process become constant-volume once the piston reaches the stops?: The stops block the piston, so the volume can no longer change. Further heating raises the pressure with p/T constant, and the stops push down on the piston. Find the temperature at which the piston reaches the stops first, then use the pressure law for the rest.
What sign does W take in ΔU = W + Q?: Here W is the work done on the gas: positive when the gas is compressed, negative when it expands and does work. Q is positive when the gas absorbs heat. An ideal gas’s internal energy depends only on temperature, so ΔU = 0 for an isothermal process.
Why do isochores on a p–T graph pass through the origin?: At constant volume p/T is constant (T in kelvin), so p is proportional to T and the line passes through the origin; a larger volume gives a smaller slope. Likewise isobars on a V–T graph pass through the origin, and isotherms on a p–V graph are hyperbolas, farther out for higher temperatures.

Features

Mercury column turned to any angle or heated, with automatic spill handling
U-tube: add or remove mercury and heat the trapped gas
Cylinder: isobaric heating, stops then constant volume, spring, isothermal loading, locked piston
Synchronised p–V, p–T and V–T graphs with isotherms, isochores and isobars
First law ΔU = W + Q with work done and heat absorbed or released
Random problems covering the common exam patterns, each with a full solution

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