Problem statement

Problem 2.22. One mole of nitrogen is contained in a vessel of volume V = 1.00 l. Find: (a) the temperature of the nitrogen at which the pressure can be calculated from an ideal gas law with an error η = 10% (as compared with the pressure calculated from the Van der Waals equation of state); (b) the gas pressure at this temperature.

Interpretation. Two formulas give a pressure for the same gas: the ideal gas law and the Van der Waals equation. The Van der Waals value is taken as the better one. Part (a) asks at what temperature the ideal value is higher than it by exactly 10%. Part (b) asks for the pressure of the gas at that temperature.

Interpretation, what happens as the temperature changes. A real gas differs from an ideal gas in two ways. The molecules have a size, so the free room is V − b, not V. This pushes the pressure up by RT/(V−b) − RT/V, which grows with T. The molecules also pull on each other. A molecule near the wall is pulled back by its neighbours, so it hits the wall more gently. This lowers the pressure by a/V², which does not depend on T. When the gas is cold the molecules are slow, the pull wins and the ideal value is too high. When the gas is hot the molecules are fast, the pull hardly matters, the size wins and the ideal value is a little too low. So the ideal law is worst at low temperature in a vessel this small, and it improves as T rises. The page shows this with two boxes of molecules, one real and one ideal, and with a graph of the two corrections against T.

We assume the Van der Waals constants of nitrogen from the usual tables: a = 1.35 l² atm/mol² and b = 0.039 l/mol, because the problem does not give them. We also assume R = 0.08206 l atm/(mol K). The constants are sliders.

Symbols: V is the volume of the vessel (one mole, so V is also the molar volume), T the temperature, p the Van der Waals pressure, η the allowed error, a and b the Van der Waals constants.

The big idea in plain words

The ideal gas law treats molecules as points that never touch and never attract. Real molecules take up room, so the gas has less free space, and this raises the pressure. They also pull on each other, and this lowers the pressure. The Van der Waals equation has one correction for each. At high temperature the molecules move so fast that the pull hardly matters, and the two laws agree. At low temperature the pull matters more, and the ideal law reads too high. We ask where the difference reaches exactly 10%.

What the molecules do as the temperature rises

Effect	What the molecules do	Change to the pressure	Depends on T?
Size	They take up room, so the free volume is V − b and they hit the walls more often	+ [RT/(V−b) − RT/V], about + RT b/V²	Yes, grows with T
Pull	Neighbours pull a molecule at the wall back into the gas, so it hits more gently	− a/V²	No, the same at every T

At low T the ideal pressure RT/V is small, so the fixed pull a/V² is a large part of it: the ideal law reads too high. At high T RT/V is large and the pull is a small part, while the size effect keeps its share b/V: the ideal law reads a little too low. Between them is a temperature where the two corrections cancel, T = a(V − b)/(R b V), and there the ideal and Van der Waals pressures are equal. The temperature T* asked for in the problem is on the cold side of it, where the pull makes the ideal value 10 percent too high.

The solution in each frame

The problem is about a gas at rest in a vessel, so only one frame is needed. The vessel is not moving or turning.

The solution step by step

Step 1, the two pressures. For one mole, the ideal law is p_id = RT/V. The Van der Waals equation (p + a/V²)(V − b) = RT gives p = RT/(V − b) − a/V².

Step 2, the condition (part a). The ideal value is larger than the Van der Waals value by the fraction η: p_id = (1 + η) p. So RT/V = (1 + η)[RT/(V − b) − a/V²].

Step 3, solve for T. Move the terms with T to one side: a(1 + η)/V² = RT[(1 + η)/(V − b) − 1/V]. Put the brackets over one denominator: (1 + η)V − (V − b) = ηV + b, so the bracket is (ηV + b)/(V(V − b)). Then T = a(1 + η)(V − b)/(R V (ηV + b)). With the numbers: T = 1.35 × 1.1 × 0.961/(0.08206 × 1 × 0.139) = K for the current sliders.

Step 4, the pressure (part b). Put RT/(V − b) = a(1 + η)/(V(ηV + b)) into p = RT/(V − b) − a/V². This gives p = (a/V²)[(1 + η)V/(ηV + b) − 1] = (a/V²)(V − b)/(ηV + b). Numbers: p = atm.

Step 5, the cases. The Van der Waals pressure is zero or negative below T₀ = a(V − b)/(R V²) = K, so a percentage error has no meaning there. The error is very large just above T₀ and falls as T rises, so one temperature gives the error η on the low side. Far above it the ideal law reads a little too low, by at most b/V. If η is smaller than b/V, a second temperature with an error of that size exists on the high side. The book answer is the low one.

Five methods

1. Force method

Pressure is the force of molecule hits on the wall. Less free room (V − b) means more hits, which raises it. Pull between molecules weakens the hits, which lowers it by a/V².

2. Energy method

The pull between molecules gives the gas a negative energy −a/V per mole. At low T this matters more than the thermal energy RT, which is why the two laws part ways there.

3. Graph method

The first graph shows both pressures against T. The orange line is straight. The blue line is steeper (smaller free volume) but starts lower. The second graph shows the error falling as T rises. It crosses the yellow line at T*.

4. Calculus method

The error is f(T) = RT/V divided by [RT/(V − b) − a/V²], minus 1. Its derivative is negative for T above T₀, so f falls steadily and crosses η once. This is why the low-side answer is unique.

5. Check method

Units: a/V² is atm, RT/V is atm. As a → 0 and b → 0, T* → 0, since the laws agree at every T. If η grows, T* gets lower, as the third graph shows.

Fastest: the formula in step 3. Most reliable: the formula with the numerical check.

What each slider changes

Control	Changes the answer?	What it changes
Allowed error η	Yes	T*, the pressure at T*, the target line and the dots on every graph
Volume V	Yes	Both pressures, T*, p*, the vessel size and every graph
Constant a	Yes	The Van der Waals pressure, T*, p* and the graphs
Constant b	Yes	The Van der Waals pressure, T*, p*, the molecule size and the graphs
Temperature shown T	No	The dots on the graphs, both gauges, the dots in the vessel and the live readings
Given-answer mode	Sets η	The page finds the error η that gives a chosen T*, and warns if η falls outside 2% to 30%
Play and time bar	No	Play raises T from 40 K, holds 3 s at T*, then runs on to 700 K and repeats. The bar sets T

Parameters

Quantity	Symbol	Value used
Allowed error	η	%
Volume	V	l
Constant a	a	l² atm/mol²
Constant b	b	l/mol
Temperature	T*	K
Pressure	p*	atm

Exam traps

• Using V instead of V − b in the first term, or forgetting that for one mole the molar volume is V.
• Defining the error against the ideal pressure. The problem measures it against the Van der Waals pressure, so p_id = (1 + η)p.
• Mixing units. With R = 0.08206 l atm/(mol K), V is in litres and p in atmospheres.
• Writing η = 10 instead of 0.10 in the formula.
• Expecting the answer to be near room temperature. Nitrogen is already a gas at 125 K, where it is well above its boiling point, and the two laws differ by 10% there.