t = (V/C) ln η. For your sliders t = s ( min). For the book numbers (V = 87 l, C = 10 l/s, η = 1000) t = 8.7 s × 6.908 = 60.1 s, about 1.0 min.
Problem statement
A chamber of volume V = 87 l is pumped out by a pump with evacuation rate C = 10 l/s. The rate is the gas volume removed per second, measured at the pressure the gas has at that moment. How soon does the pressure fall by η = 1000 times? (The book prints the unit of C as "1/s". It means litres per second, l/s.)
We assume the temperature stays the same. Heat flows in as needed, so the gas never cools. We use it in the link between pressure and density.
We assume the gas is ideal. Its molecules are tiny and only bounce. At fixed T the pressure is then proportional to the density.
We assume the gas in the chamber is mixed evenly, so one density describes all of it.
We assume C is constant, as the problem says. This fails at very low pressure, as the live warnings show.
Symbols: V is the chamber volume, C is the evacuation rate, ρ (rho) is the gas density (mass per volume), p is the pressure, p0 is the start pressure, η (eta) is the factor by which p falls, and t is the time. Also R = 8.314 J/(mol K) is the gas constant, and T is the temperature. Sign rule: W is the work done BY the gas, and dU = Q − W.
The big idea in plain words
In each short moment the pump removes a fixed volume, C dt. That volume holds the gas at the current density. So the pump takes a fixed share of the gas every second, C/V of it. A fixed share per second makes the pressure fall exponentially. We turn that into the time it takes to fall η times.
The solution from one viewpoint
Reference frames do not apply. The chamber and pump are fixed, so there is only the lab viewpoint.
First we find how the density changes in a short time dt, because the pressure follows the density. The mass in the chamber at this moment is V ρ. During dt the pump removes the volume C dt. We can say the gas that stays has the same mass, but it now spreads over V + C dt, with the density changed to ρ + dρ. Mass stays the same, so:
Next we turn density into pressure, because the question is about pressure. For an ideal gas at fixed T the pressure is proportional to the density (p = ρ R T / M, with M the molar mass). So the same relation holds for p.
Now we add up the changes from the start (pressure p1 = p0, time 0) to the end (pressure p2 = p0/η, time t), because we want the time for that fall.
The book writes ln(1/η) with the limits turned round. It gets the same positive time. Our version keeps p1/p2 = η, which is bigger than 1, so ln η is positive and t is positive.
The result says the time grows with the logarithm of η. Each extra factor of 10 in η adds the same time, (V/C) ln 10 = s.
The solution by each method
Method 1: Energy method
This is not suited to finding t. The gas is ideal and isothermal, so its internal energy does not change (dU = 0), and energy gives no equation for the time. We use it to find the work and heat.
- The gas pushes out the volume C dt at pressure p, so dW = p C dt. With p = p0 e^(−C t / V) this gives W = ∫ p0 e^(−C t / V) C dt = p0 V (1 − e^(−C t / V)) = (p0 − p) V.
- At the end p = p0/η, so W = p0 V (1 − 1/η) = atm × 1.013 × 10⁵ Pa/atm × × 10⁻³ m³ × (1 − 1/) = J.
- Since dU = 0, Q = W + dU = J. The gas takes in this heat from the surroundings.
Method 2: Graph method
- Plot ln(p/p0) against t. The equation dp/p = −(C/V) dt says it is a straight line through the origin with slope −C/V = − per second.
- Draw the horizontal line at ln(1/η) = −.
- It meets the straight line at t = / = s. The middle graph on the canvas shows this.
Method 3: Calculus method
- Set up the equation from the rate: dp/dt = −(C/V) p.
- Separate the variables: dp/p = −(C/V) dt.
- Integrate from p0 at time 0 to p0/η at time t: ln(1/η) = −C t / V.
- Solve for t: t = (V/C) ln η = s.
Method 4: State-variable method
| State | t | p | n (moles) | p V = n R T |
|---|---|---|---|---|
| Start | 0 | atm | p0 V | |
| After one time constant | s | atm | n0 / e | p0 V / e |
| End | s | atm | p0 V / η |
- Equation of state for the gas left in the chamber: p V = n R T. V and T are fixed, so p is proportional to n.
- The pump removes the moles carried by C dt: p C dt / (R T) = (C/V) n dt. So dn/dt = −(C/V) n.
- The solution is n = n0 e^(−C t / V). The end state needs n = n0/η, so C t / V = ln η, and t = s.
- Process check: V and T stay fixed while n falls. The p V column equals n R T in every state.
Method 5: Check method
- Units: V/C is litres divided by litres per second, so it is in seconds. The log is a pure number. So t is in seconds.
- Limit η = 1: ln 1 = 0, so t = 0. No reduction needs no time.
- Limit η very large: t grows only as ln η. A reduction of 10⁹ takes only 3 times as long as 10³, but it is far beyond what a real pump does.
- Direction check: a bigger chamber V or a slower pump C gives a longer time. The sliders agree.
- Number check: C t / V = × / = , and e^() = .
- Link to question 7: with a piston stroke ΔV every time ts, C = ΔV/ts, and the number of strokes n = t/ts. The stroke formula n = ln η / ln(1 + ΔV/V) gives the same time when ΔV is small.
Fastest: the formula t = (V/C) ln η. Most reliable in an exam: write V ρ = (V + C dt)(ρ + dρ) first, integrate, then run the check method.
What each slider changes
| Slider | Outputs it changes | How | Why |
|---|---|---|---|
| V (chamber) | answer t, time constant, chamber size, molecules, W | t proportional to V | A bigger chamber holds more gas, so the same C removes a smaller share per second. |
| C (rate) | answer t, time constant, pump spin, flow, W | t inverse to C | A faster pump removes a bigger share per second. |
| η (reduction) | answer t, final pressure, W | t proportional to ln η | Each extra factor of 10 adds the same time. |
| t (given-answer mode) | reduction η reached | η = e^(C t / V), grows exponentially | The formula is solved backwards. |
| p0 | pressures, dots, W, mean free path | none for t | Only the ratio p0/p matters, so p0 cancels. |
| T | dot speed, molecule count, mean free path | none for t | T cancels, because pressure and density both carry RT. |
Parameter table
| Symbol | Meaning | Value now |
|---|---|---|
| V | chamber volume | l |
| C | evacuation rate (volume per second at the current pressure) | l/s |
| V/C | time constant | s |
| η | factor the pressure falls by | |
| p0, T | start pressure, temperature | atm, K |
| t | answer: time needed | s |
| R, kB, NA | gas constant, Boltzmann constant, Avogadro number | 8.314 J/(mol K), 1.381 × 10⁻²³ J/K, 6.022 × 10²³ per mol |
| f, Cv, γ | degrees of freedom (air, vibration frozen), Cv = fR/2, γ = (f + 2)/f | 5, J/(mol K), (not needed, T is fixed) |
Common exam traps
- Using log base 10 instead of the natural log ln. With log10 the time comes out 2.3 times too small.
- Reading "10 1/s" as a rate of one tenth. It means 10 litres per second.
- Thinking C removes the same mass each second. It removes the same volume, which holds less mass as p falls.
- Mixing units: C in l/s with V in m³.
- Dividing the pressure ratio the wrong way, which gives a negative time.
- Thinking the answer depends on p0 or T. Neither does.