Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.3

Charles's Law, Absolute Zero and Gay-Lussac's Pressure Law.

Why does a balloon left in the sun go tight, and why is a car tyre harder after a long drive? One answer covers both: when a gas gets hotter, something in it grows. This note finds out what grows, by how much, and why every calculation has to use kelvin instead of Celsius.

1. The experiment of Charles and Gay-Lussac

In 1787 the French scientist Charles, and in 1802 Gay-Lussac, each worked out separately how the volume of a gas changes with temperature. The set-up is simple (Figure 1.7): a fixed amount of air is trapped by a mercury plug in a glass tube closed at one end, and the tube is dipped in a water bath. The plug is free to slide and only the atmosphere presses on it, so the pressure of the gas stays constant.

Table 1.4 gives four readings for 0.04 mol of gas at 1 atm. Drag the handle through them one by one, and keep an eye on the last figure.

2. The graph, absolute zero and the kelvin

Take the line of Table 1.4 and pull it backwards, into the cold. Where does it stop?

The V against t°C graph is a straight line. Extend it backwards and it cuts the x-axis at −273°C, so in theory the volume is zero at that temperature. Lines for a different amount (0.02 mol) or a different pressure (4 atm) cut the axis at the very same point, −273°C. Change the gas or the pressure and that point stays put.

If the volume at 0°C is V₀, the volume at t°C is Vₜ = V₀ + V₀t/273 = V₀(1 + t/273). Put t = −273 and the volume is 0; put t = 273 and it is 2V₀. The ratio V₀/273 is called the volume expansion coefficient of the gas.

In 1854 William Thomson (Lord Kelvin) built the absolute temperature scale on this straight-line link between gas volume and temperature; it later took the name Kelvin scale. Its zero sits at −273.15°C. A degree is the same size on the Celsius and Kelvin scales, so ice melts at 273 K and water boils at 373 K.

The lowest temperature on the Kelvin scale, 0 K or −273.15°C, is absolute zero. Below it the volume of a gas would be negative, which cannot happen. The book's graphs and questions often write −273°C; both are fine, but keep the working rule: T = 273 + t.

3. Charles's law

Charles's law: at constant pressure the volume of a fixed mass of gas is proportional to its absolute (Kelvin) temperature. With n and P fixed, V ∝ T, that is V = KT or V/T = K. So for two states, V₁/T₁ = V₂/T₂. The nearly steady V/T in the table is this law showing through.

Why kelvin and not Celsius? Going from 200 K to 400 K doubles the volume exactly. Going from 200°C to 400°C raises it only (400 + 273)/(200 + 273) = 673/473 = 1.42 times. Doubling the temperature doubles the volume only on the kelvin scale, because its zero is a true zero.

Plot V against T (kelvin) and you get a straight line through the origin, V = KT + 0. Such graphs, drawn for various fixed pressures, are called isobars.

4. The order of working, and examples

Every Charles problem follows the same steps. Put them in order yourself first.

Problem 1.7: at what temperature does 906 mL of N₂ at 27°C shrink to 500 mL? T₁ = 300 K, so T₂ = V₂T₁/V₁ = 500 × 300/906 = 165.56 K = −107.44°C. Problem 1.8: the volume doubles from 17°C, so T₁ = 290 K and T₂ = 580 K = 307°C. The temperature rose by 307 − 17 = 290°C; the rise and the final temperature are different things.

Problem 1.9: 5.0 L of gas in a cylinder at 27°C is warmed to 30°C at constant pressure: V₂ = 5.0 × 303/300 = 5.05 L, so 0.05 L escapes. Problem 1.10: a balloon inflated to 950 cm³ at 17°C and 1 atm is 9/10 of its maximum volume. It bursts at T₂ = (950 × 10/9) × 290/950 = 322.2 K = 49.2°C, so it survives a 40°C room. Problem 1.11: to double 500 cm³ of air at 27°C, T₂ = 2 × 500 × 300/500 = 600 K = 327°C.

Cooling follows the same rule. At constant pressure, 300 mL of gas at 27°C taken to 7°C gives V₂ = 300 × 280/300 = 280 mL. And as a percentage: to raise 80 cm³ at 25°C by 20%, T₂ = 1.2 × 298 = 357.6 K = 84.6°C. Check them in the calculator.

5. Hotter gas is thinner

Density is d = W/V. At constant pressure a rise in temperature raises V, so d falls: d ∝ 1/T, that is d₁T₁ = d₂T₂. Problem 1.12: the density of CO₂ at STP is 22 kg m⁻³; at 11°C and constant pressure d₂ = 22 × 273/284 = 21.15 kg m⁻³. The temperature went from 273 K to 284 K and the density dipped a little.

6. Gay-Lussac's pressure law

Now lock the volume instead. Gay-Lussac's pressure law (also called Amontons's law): at constant volume the pressure of a fixed amount of gas is proportional to its Kelvin temperature. P ∝ T, so P₁/T₁ = P₂/T₂. The P against T (K) graph is a straight line through the origin, called an isochor. After a drive the tyre warms from friction; its volume cannot change, so its pressure climbs.

Problem 1.13: an oxygen cylinder withstands 250 atm and is filled at 125 atm and 27°C. It bursts at T₂ = 250 × 300/125 = 600 K = 327°C. Problem 1.14: an iron cylinder holds He at 250 kPa and 27°C and withstands 1000 kPa. T₂ = 1000 × 300/250 = 1200 K = 927°C, below the melting point of iron, 1500°C, so the cylinder bursts and does not melt. Problem 1.15: a methane cylinder is at 27°C and 1.3 atm, and its safety valve opens above 1.8×10³ torr. At 100°C P₂ = 988 × 373/300 = 1228.4 torr, under 1800 torr, so the valve stays shut.

Now tell the three laws' graphs apart: which is an isobar, which an isochor, which an isotherm.

7. Check yourself

Three questions in the exam's style. The first is about the law itself.

The second needs a calculating eye.

In the last one, remember what kelvin is for.

The whole note on seven cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.3.