p = p0 · e^(−C t / V). For your sliders at t = s the pressure is atm, which is times lower than p0. The book numbers (V = 10 l, C = 2 l/s, p0 = 1 atm, t = 5 s) give p = 0.3679 atm.
Problem statement
Find the pressure of air in a vessel that is being pumped out, as a function of the time t. The vessel has volume V. The start pressure is p0. The process is isothermal. The pumping rate C does not depend on pressure. The rate C is the gas volume removed per second, measured at the pressure the gas has at that moment.
We assume the temperature stays the same. Heat flows in as needed, so the gas never cools. We use it in the gas law with T fixed.
We assume the gas is ideal. Its molecules are tiny and only bounce. We use it in pV = nRT.
We assume the gas in the vessel is mixed evenly, so one pressure p describes all of it. This means the pump takes gas at the same density as the vessel average.
We assume C is constant. This is the problem's own condition, and it fails at very low pressure, as the live warnings show.
Symbols: V is the vessel volume, p0 is the start pressure, p is the pressure at time t, C is the evacuation rate (a volume per second), m is the mass of gas in the vessel, M is the molar mass, n is the number of moles, 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
The pump removes the same volume each second, but that volume holds less gas as the pressure falls. So the pump takes a fixed share of the gas every second, not a fixed amount. A fixed share per second gives an exponential decay.
The solution from one viewpoint
Reference frames do not apply. The vessel and pump are fixed, so there is only the lab viewpoint.
First we find how fast the mass in the vessel falls, because pressure follows the mass. In a short time dt the pump removes the volume C dt. That gas has the density the vessel gas has right now, which is m/V (mass divided by volume). So the mass removed is density times volume.
The book gets the same line by writing the removed volume as v = (V/m_N)(m_(N−1) − m_N) for a small step and letting the step shrink. The density way above is shorter.
Next we turn mass into pressure, because the question asks for pressure. The ideal gas law with fixed T and V says p = (m/M)(RT/V). The factor RT/(MV) is a constant, so p is proportional to m.
Now we solve equation (2), because it tells us p changes at a rate proportional to p. We separate p and t, then add up (integrate) from the start (t = 0, p = p0) to now (time t, pressure p).
The result says the pressure falls by the same factor in equal times. The time V/C = s is the time constant: after it, the pressure is p0/e = atm. The pressure never reaches exactly zero.
The solution by each method
Method 1: Energy method
This is not suited to finding p(t). The gas is ideal and isothermal, so its internal energy per mole does not change (dU = 0), and energy gives no equation for p. We use it to find the work and heat.
- The gas pushes out the volume C dt at the pressure p. The work is dW = p C dt.
- Put in p = p0 e^(−C t / V) and integrate: W = ∫ p0 e^(−C t / V) C dt from 0 to t = p0 V (1 − e^(−C t / V)) = (p0 − p) V.
- Numbers: W = ( − ) atm × 1.013 × 10⁵ Pa/atm × × 10⁻³ m³ = 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. Equation (2) says it is a straight line through the origin.
- The slope is −C/V = − per second.
- Read the line at t = s: ln(p/p0) = −, so p = atm. The middle graph on the canvas shows this.
- On the p against t graph, the tangent at any point reaches p = 0 exactly one time constant later. This gives a quick way to read V/C = s.
Method 3: Calculus method
- Set up the differential equation from the rate: dp/dt = −(C/V) p.
- Separate the variables: dp/p = −(C/V) dt.
- Integrate both sides: ln p = −(C/V) t + constant. At t = 0, p = p0, so the constant is ln p0.
- Result: p = p0 e^(−C t / V) = atm.
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 |
| After two | s | atm | n0 / e² | p0 V / e² |
| Now | s | atm | p V |
- Equation of state for the gas left in the vessel: 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, which is p C dt / (R T) = (C/V) n dt. So dn/dt = −(C/V) n.
- This has the solution n = n0 e^(−C t / V), so p = n R T / V = p0 e^(−C t / V) = atm.
- Process check: V and T are fixed while n falls. The p V column equals n R T at each state, so the isothermal law holds.
Method 5: Check method
- Units: C/V is volume per second divided by volume, so it is 1 per second. Then C t / V has no units, as an exponent must.
- Limit t = 0: p = p0. Correct.
- Limit t very large: p tends to 0 but never reaches it. A pump that removes a share of the gas cannot take the last molecule.
- Limit small C t / V: p ≈ p0 (1 − C t / V). The pressure falls at the steady rate p0 C / V at the start, which matches dp/dt at t = 0 from equation (2).
- Link to question 7: with a stroke volume ΔV every time ts, C = ΔV / ts. The stroke result p = p0 (1 + ΔV/V)^(−t/ts) tends to p0 e^(−C t / V) as ΔV becomes small.
- Direction check: a bigger V or a smaller C makes the exponent smaller, so the pressure falls more slowly. The sliders agree.
Fastest: the formula p = p0 e^(−C t / V). Most reliable in an exam: write dm = −(m/V) C dt first, then integrate, then run the check method.
What each slider changes
| Slider | Outputs it changes | How | Why |
|---|---|---|---|
| V (vessel) | time constant, p(t), vessel size, molecules, W | time constant proportional to V | A bigger vessel holds more gas, so the same C removes a smaller share per second. |
| C (rate) | time constant, p(t), pump spin, flow, W | time constant inverse to C | A faster pump removes a bigger share per second. |
| p0 | p(t), W, dots, mean free path | p proportional to p0 | The ratio p/p0 depends only on C t / V, so p scales with the start pressure. |
| T | dot speed, molecule count, mean free path | none for p/p0 or time constant | T cancels, because p and the pumped share both use the same RT. |
| t (time now) | p(t), reduction, W, markers | p falls exponentially with t | Each equal step in time multiplies p by the same factor. |
| η (target, given-answer mode) | time t | t proportional to ln η | The formula is solved backwards: t = (V/C) ln η. |
Parameter table
| Symbol | Meaning | Value now |
|---|---|---|
| V | vessel volume | l |
| C | evacuation rate (volume per second at the current pressure) | l/s |
| V/C | time constant | s |
| p0, T | start pressure, temperature | atm, K |
| t, p(t) | time now, pressure now | s, atm |
| η = p0/p | reduction reached | |
| 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
- Thinking C removes the same mass each second. It removes the same volume, which holds less mass as p falls.
- Writing e^(+C t / V), or forgetting the minus sign.
- Mixing units: C in l/s with V in m³. Use the same volume unit for both.
- Saying the pressure reaches zero at a finite time. It does not.
- Using the stroke formula of question 7 without a count of strokes, or the reverse.
- Thinking a higher start pressure p0 changes the time constant. It does not.