m(H2) / m(He) = (M1 / M2) · (M2 − M) / (M − M1), where M = m R T / (p V) is the average molar mass. For your sliders the ratio is . For the book numbers it is about 0.50.
Problem statement
A vessel of volume V = 20 l holds a mixture of hydrogen and helium at t = 20 °C and p = 2.0 atm. The mass of the mixture is m = 5.0 g. Find the ratio of the mass of hydrogen to the mass of helium.
We assume both gases are ideal. Their molecules are tiny and only bounce. We use this in pV = νRT.
We assume the gases mix completely and share the same temperature. We use this to add their pressures (Dalton's law, below).
We assume hydrogen is the molecule H2 with molar mass M1 = g/mol and helium is the single atom He with M2 = g/mol. The book uses 2 and 4. The toggle on the left uses the exact values.
Symbols: ν is the number of moles, R = 8.314 J/(mol K) is the gas constant, T = t + 273.15 is the temperature in kelvin, and 1 atm = 1.013 × 10⁵ Pa. A mole is a fixed big number of molecules (6.022 × 10²³). Molar mass is the mass of one mole, in grams.
Sign rule for the energy part: W is the work done BY the gas, and dU = Q − W. Heating at fixed volume means W = 0.
The big idea in plain words
We do not know how many moles of each gas there are. But we know two things: the total number of moles (from the gas law) and the total mass. Two facts, two unknowns. Hydrogen is light and helium is heavier, so the average molar mass of the mixture tells us the blend.
The solution from one viewpoint
Reference frames do not apply. The vessel sits still, so there is only the lab viewpoint.
First we need the total moles, because the mass alone does not tell us. Each gas obeys pV = νRT. Dalton's law says the total pressure is the sum of the partial pressures, p = p1 + p2, where each partial pressure is what that gas would give alone in the vessel. So pV = (ν1 + ν2) R T, where ν1 and ν2 are the moles of H2 and He.
Next we find the average molar mass, because it is the mass per mole of the mixture.
Now we write the two facts for the unknowns. Mass: ν1 M1 + ν2 M2 = m. Moles: ν1 + ν2 = ν. We multiply the second by M2 and subtract the first, so ν2 disappears.
Last, the masses. Each is moles times molar mass, so the ratio follows.
The result tells us there is about half as much hydrogen mass as helium mass. Since hydrogen molecules are half as heavy, there are about as many hydrogen molecules as helium atoms. The book rounds M to 3.0 g/mol before the last step and gets exactly 0.50.
The solution by each method
Method 1: Energy method
This does not suit the question. Energy does not tell us the blend. As a check on the energy rules, suppose we heat the mixture at constant volume to tf = °C. Each gas has Cv = f R / 2, where f counts the ways a molecule stores energy (3 for He, 5 for H2). The mixture Cv is the mole-fraction weighted average: Cv = x1 (5R/2) + x2 (3R/2), where x1 = ν1 / ν and x2 = ν2 / ν. Vibration of H2 is frozen out below about 1000 K, so f = 5 holds.
Method 2: Graph method
- The average molar mass is a mole-fraction average: M = x1 M1 + x2 M2 with x1 + x2 = 1. Plot M against x1: a straight line from M2 (x1 = 0) to M1 (x1 = 1).
- Find M = g/mol on the line and read x1 = (M2 − M) / (M2 − M1) = .
- The ratio is the mass fractions: m1 / m2 = x1 M1 / (x2 M2) = .
- The middle graph on the canvas plots this ratio against M. The shaded sides are impossible mixtures.
Method 3: Calculus method
No integral is needed for the answer. We use calculus only to see how sensitive it is. With r(M) = (M1 / M2)(M2 − M) / (M − M1), differentiate:
So a small error in M changes the ratio a lot when M is close to M1. Fix: measure M carefully when the gas is nearly pure hydrogen.
Method 4: State-variable method
| State | p | V | T | moles |
|---|---|---|---|---|
| Start (given) | atm | l | K | ν = |
| After heating (optional) | atm | l | K | ν = |
- Equation of state: pV = νRT gives ν = mol.
- Two mass-and-mole equations give ν1 = and ν2 = .
- Answer: m1 / m2 = .
- Process check: heating at fixed volume is isochoric (constant volume), with law p / T = constant. So pf = p Tf / T = atm. The blend does not change on heating, so the ratio stays put.
Method 5: Check method
- Units: ν in mol, M in g/mol, masses in g, so the ratio has no unit.
- Add the masses: m1 + m2 = + = g. Good.
- Add the moles: ν1 + ν2 = mol. Good.
- Limit M → M1 (pure hydrogen): the ratio grows without bound, as it should. Limit M → M2 (pure helium): the ratio goes to 0.
- If M is below M1 or above M2, no mixture can have that molar mass. The simulation says "impossible" there.
- Quick shortcut: for the book numbers M is exactly halfway between M1 and M2. So the moles are equal, and the mass ratio is M1 / M2 = 0.50.
Fastest: the halfway shortcut, when it applies. Most reliable in an exam: the state-variable method followed by the check method.
What each slider changes
| Slider | Outputs it changes | How | Why |
|---|---|---|---|
| V | ratio, M, ν, dots, vessel size | M inverse to V | More room at the same p holds more moles, so the average mass per mole falls. |
| t | ratio, M, dot speed | M proportional to T | Hotter gas at the same p has fewer moles. |
| p | ratio, M, ν, dots | M inverse to p | More pressure means more moles. |
| m | ratio, M, mass bars | M proportional to m | The same moles weigh more. |
| tf (final t) | Q, dU, pf, dot speed | Q proportional to tf − t. Ratio none. | Heating does not change the blend. |
| Heating time | heating rate, warning | none for the ratio | It only checks the slow-heating assumption. |
| Target ratio | p (given-answer mode) | solved backwards | The formula is inverted. |
Parameter table
| Symbol | Meaning | Value now |
|---|---|---|
| V, t, p, m | volume, temperature, pressure, mixture mass | l, °C, atm, g |
| M1, M2 | molar mass of H2, He | , g/mol |
| ν, ν1, ν2 | total, H2 and He moles | , , mol |
| p1, p2 | partial pressures of H2, He | , atm |
| 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 |
| Cv, γ | mixture heat capacity, Cp / Cv | J/(mol K), |
Common exam traps
- Using the molar mass of atomic hydrogen (1) instead of H2 (2).
- Using °C in the gas law. Convert to kelvin.
- Mixing atm with pascals, or litres with cubic metres, inside ν = pV / (RT).
- Giving the ratio as moles instead of masses. Here the moles are nearly equal but the masses are not.
- Putting the ratio upside down. We want hydrogen over helium.
- Forgetting that M must lie between M1 and M2.