Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.2

Boyle's Law.

Ever blocked the nozzle of a bicycle pump with your thumb and pushed the handle? The more you push, the smaller the trapped air gets, and the harder the push becomes. A neat piece of arithmetic hides inside that squeeze, and in 1660 one chemist caught it.

1. Boyle and his law

After Torricelli invented the barometer, the English chemist Robert Boyle kept the temperature fixed and studied how pressure affects the volume of different gases. He published his results in 1660.

Boyle's law: at constant temperature, the volume of a fixed mass of a gas is inversely proportional to the pressure applied on it.

With n and T constant, V ∝ 1/P, that is V = K·(1/P). Tidied up, PV = K, a constant. For one gas in two states, then,

P₁V₁ = P₂V₂

2. Boyle's experiment

Pour mercury into a J-shaped glass tube and some gas is trapped (n and T constant). The total pressure on the trapped gas = P_atm + Δh, where P_atm is atmospheric pressure and Δh is the difference in mercury levels (mm) between the two arms.

In Table 1.3, for V = 30, 25, 20, 15 and 10 mL the total pressure is P = 760, 912, 1140, 1520 and 2280 mm(Hg), with Δh = 0, 152, 380, 760 and 1520. Drag the handle through the five stops. Every time, the product is PV = 2.28×10⁴.

The table gives the law in one sentence: at constant temperature, doubling the pressure on a gas halves its volume. At 20 mL the pressure is 1140; at 10 mL it is exactly double, 2280 mm(Hg).

3. Three graphs, three shapes

Block the nozzle and push the plunger. Each push of Table 1.3 drops a point on the graph.

Boyle's law has three graphs. V against P is a hyperbola. V against 1/P is a straight line through the origin. PV against P is a line parallel to the x-axis. Now drop the chips into the boxes by shape.

These graphs are called isotherms. Greek iso means same, therm means temperature. A process at constant temperature is called isothermal. An isotherm for a higher temperature lies farther from the origin.

Why does the V against 1/P line pass through the origin? It matches y = mx + c with y = V, x = 1/P, slope m = K and c = 0. A zero c means the line goes straight through the origin.

4. When the law holds

Boyle's law holds especially well at high temperature and low pressure. At low temperature and high pressure, gases deviate from it.

5. The book's calculations

One trick serves every problem: T and n are fixed, so P₁V₁ = P₂V₂ gives V₂ = P₁V₁/P₂. Problem 1.1: an engine cylinder holds 410 mL of an air and gasoline mixture. At 9.5 atm instead of 1 atm, V₂ = (1 atm × 410 mL)/9.5 atm = 43.16 mL. Problem 1.2: 3.15 L of oxygen at 1 atm is raised to 2.50 atm at constant temperature: V₂ = (1 atm × 3.15 L)/2.50 atm = 1.26 L. Both are preloaded below; change the numbers and see.

Problem 1.3: 50 L of hydrogen at 10 atm fills 2 L balloons, each at 2 atm. The n balloons together hold 2n L, so 10 × 50 = 2 × 2n, giving n = 125 balloons.

Problem 1.4: gas with marbles occupies 200 mL at 1 atm and 105 mL at double the pressure. If the marbles take volume V, then (200 − V) × 1 = (105 − V) × 2, so V = 10 mL. The marbles are solid and their volume does not change with pressure. Only the gas part shrinks.

Problem 1.5: 100 mL of H₂ at 1 atm must be brought down to 76 mL. P₂ = 760 × 100/76 = 1000 mm(Hg). The pressure must rise by 240 mm(Hg), that is 240/760 = 6/19 atm.

Problem 1.6: a balloon filled with 2.0 m³ of H₂ at 1.8 atm rises and bursts when it reaches 2.22 m³. P₂ = 1.8 × 2/2.22 = 1.62 atm. The pressure has dropped by 0.18 atm = 136.8 mm(Hg). At 0.044 mm(Hg) per metre, the height is 136.8/0.044 = 3109.1 m.

Pressure is proportional to 1/V, so values of 1/V give pressure too. If 1/V = 0.05 at 1 atm, then at 1/V = 0.042, P₂ = 1 × 0.042/0.05 = 0.84 atm.

To cut the volume at 1 atm by one third, from V to 2V/3, P₂ = 760 × 3/2 = 1140 mm(Hg). That is a rise of 380 mm(Hg).

6. Pressure and density

Boyle's law has a corollary. Density d = W/V, and for a fixed mass W is constant. So at constant temperature P ∝ d, that is P₁/d₁ = P₂/d₂. More pressure means less volume and more density.

Example: if the density is 2.5 g/L at 25°C and 1.01 atm, then at 780 mm(Hg) and the same temperature d₂ = 2.5 × (780/760)/1.01 = 2.54 g/L.

One more: a 50 L cylinder holds air at 15 atm and 25°C. Opened to 1 atm, all that air would fill 15 × 50 = 750 L. The cylinder keeps 50 L, so 700 L comes out.

7. Check yourself

Three questions in the exam's style. The first.

The second.

The last, take your time.

The whole note on seven cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.2.