One mole of a certain gas is contained in a vessel of volume V = 0.250 l. At a temperature T1 = 300 K the gas pressure is p1 = 90 atm, and at a temperature T2 = 350 K the pressure is p2 = 110 atm. Find the Van der Waals parameters for this gas.
Two pressure readings at one volume fix both constants: b = V − R(T2 − T1)/(p2 − p1) and a = V²(T1p2 − p1T2)/(T2 − T1). With the numbers on screen (V = l, atm at K, atm at K): a = l² atm/mol² and b = l/mol.
Problem statement
Problem 2.23. One mole of a certain gas is contained in a vessel of volume V = 0.250 l. At a temperature T1 = 300 K the gas pressure is p1 = 90 atm, and at a temperature T2 = 350 K the pressure is p2 = 110 atm. Find the Van der Waals parameters for this gas.
Interpretation. The gas is shut in a vessel of fixed volume and heated from 300 K to 350 K. The pressure rises from 90 atm to 110 atm. The task is to work backwards: which two numbers a and b in the Van der Waals equation make both readings true?
Interpretation, why the pressure rises with temperature. The pressure comes from the molecules hitting the walls. When the gas is heated the molecules move faster. Each hit then pushes the wall harder, and the hits come more often, so the pressure grows with T. For an ideal gas this gives p = RT/V, a line through zero. A real gas differs in two ways. The molecules have a size, so the free room is V − b. Fewer places to go means more hits, so the line is steeper, with slope R/(V−b). The molecules also pull on each other, so a molecule at the wall is pulled back and hits more gently. This lowers the pressure by a/V², the same at every T, so the line is shifted down. With only two readings at one volume we can read the slope, which gives b, and the offset, which gives a.
We assume R = 0.08206 l atm/(mol K). The result for a does not depend on R. The result for b does, so a rounded R = 0.082 gives b = 0.045 l/mol, while 0.08206 gives 0.0448. All five given numbers are sliders.
Symbols: V is the volume (one mole, so also the molar volume), T1 and T2 the two temperatures, p1 and p2 the two pressures, a and b the Van der Waals constants.
The big idea in plain words
The Van der Waals equation has two unknowns, a and b. One reading gives one equation, which is not enough. Two readings give two equations, and two equations fix two unknowns. The trick is that the volume is the same in both readings. When the two equations are subtracted, the attraction term a/V² is identical in both and cancels. What is left holds only b. Then a follows from either reading.
What the molecules do when the gas is heated
| Change | What the molecules do | What it does to the pressure |
|---|---|---|
| Higher T | They move faster, speed grows as the square root of T | Each hit pushes harder (in proportion to the speed) and hits come more often (also in proportion to the speed), so p grows in proportion to T |
| Size (b) | The free room is V − b, so a molecule meets a wall sooner | More hits: the slope of p against T is R/(V−b), larger than the ideal R/V |
| Pull (a) | Neighbours pull a molecule at the wall inward | Weaker hits: p is lowered by a/V², the same at every T |
Two readings at the same volume give two points on the straight line p = [R/(V−b)] T − a/V². The slope between them is (p2 − p1)/(T2 − T1) = R/(V−b), so b follows. The pressure at T1 then fixes the offset a/V², so a follows.
The solution in each frame
The gas is at rest in a closed vessel, so only one frame is needed.
The solution step by step
Step 1, the two equations. For one mole at volume V: p = RT/(V − b) − a/V². Written for the two readings: p1 = RT1/(V − b) − a/V² and p2 = RT2/(V − b) − a/V².
Step 2, the constant b. Subtract: p2 − p1 = R(T2 − T1)/(V − b). So V − b = R(T2 − T1)/(p2 − p1) = l, and b = V − R(T2 − T1)/(p2 − p1) = l/mol.
Step 3, the constant a. The same subtraction gives RT1/(V − b) = T1(p2 − p1)/(T2 − T1). Put this into the first equation: a/V² = T1(p2 − p1)/(T2 − T1) − p1 = (T1p2 − p1T2)/(T2 − T1) = atm. Then a = V² times this = l² atm/mol².
Step 4, the cases. The problem has one situation. The result is a real gas only if 0 < b < V and a > 0. The second condition means p2/p1 > T2/T1: the pressure must rise by a larger factor than the temperature does. For a = 0 the pressure would rise in step with T, if b were also zero.
A separate calculation finds b by repeated halving of an interval, so that R(T2 − T1)/(V − b) equals p2 − p1. It then gets a from the first reading, and puts both constants back into both equations. The largest relative difference from the formulas and from the two readings is .
Five methods
1. Force method
Pressure is the push of molecule hits on the wall. Heating makes the hits harder, and the extra push per kelvin is R/(V − b). Measuring the extra push gives V − b directly.
2. Energy method
The attraction lowers the energy by a/V per mole. It does not change with T, which is why it drops out of the difference of two readings.
3. Graph method
At fixed V the Van der Waals pressure is a straight line in T. Its slope is R/(V − b) and its value at T = 0 is −a/V². The two readings are two points. The line through them gives both constants. This is the first graph.
4. Calculus method
The slope (∂p/∂T) at constant V is R/(V − b), so b comes from the slope alone. The intercept then gives a.
5. Check method
Units: RΔT/Δp is (l atm/mol K) × K / atm = l/mol, the unit of b. The first graph shows the straight line through both points. For the book numbers, b is about 18% of V.
Fastest: the two formulas. Most reliable: the straight line through both points, together with the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Volume V | Yes | b, a, the free volume bar, the vessel width and the graphs |
| Temperatures T1, T2 | Yes | b, a, the points on the line and the graphs |
| Pressures p1, p2 | Yes | b, a, the points on the line and both constant graphs |
| Temperature shown T | No | The gauge, the dots in the vessel, the dot on the line and the live readings |
| Given-answer mode | Sets p2 | The page finds the pressure p2 that gives a chosen b, and warns if it needs more than 200 atm |
| Play and time bar | No | Play raises T from 200 K, holds 2.5 s at T1 and again at T2, then runs to 500 K and repeats. The bar sets T |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Volume | V | l |
| First reading | T1, p1 | K, atm |
| Second reading | T2, p2 | K, atm |
| Constant a | a | l² atm/mol² |
| Constant b | b | l/mol |
Exam traps
- Using the ideal law to get b. The slope R/(V − b) contains V − b, not V.
- Forgetting that V is the molar volume. There is one mole, so V = 0.250 l is also the volume per mole.
- Mixing up p1p2 in the formula for a. The order is T1p2 − p1T2.
- Using Celsius. The temperatures must be in kelvin.
- Rounding R early. The value of b changes in the third digit.