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Boyle's Law Core Concept

For a given mass of gas at constant temperature, pressure varies inversely with volume and directly with density. In hydraulics, this appears in trapped-air tubes, air bubbles rising through water, and compressed gas pockets.

$$p_1V_1 = p_2V_2$$ $$\frac{p_1}{p_2} = \frac{V_2}{V_1}$$ $$\frac{p_1}{p_2} = \frac{\rho_1}{\rho_2}$$
Use absolute pressure in Boyle's Law. Convert gage pressure to absolute pressure before applying $p_1V_1 = p_2V_2$.

Boyle's Law with Hydrostatic Pressure

When gas is located below a liquid surface, its absolute pressure is atmospheric pressure plus the hydrostatic pressure of the liquid above it.

$$p_{gas} = p_{atm} + \gamma h$$ $$p_{gas} = p_{atm} + SG\,\gamma_w h$$

If the gas is trapped in a tube with constant cross-sectional area, volume may be written as $V = AL$.

$$p_1(AL_1) = p_2(AL_2)$$ $$p_1L_1 = p_2L_2$$

Problem: Air Bubble Depth

Assuming normal barometric pressure, how deep is the ocean at a point where an air bubble has six times its bottom volume when it reaches the surface? Use salt water with $SG = 1.03$.

$$p_1V_1 = p_2V_2$$ $$p_{surface} = 101.356 \text{ kPa}$$ $$p_{bottom} = 101.356 + 9.81(1.03)h$$ $$V_{surface} = 6V_{bottom}$$ $$(101.356)(6V) = (101.356 + 10.104h)V$$ $$h = 50.16 \text{ m}$$

Answer: The ocean depth is approximately 50.16 m.

Problem: Closed Tube Water Rise

A glass tube 1.5 m long has one end closed and is inserted vertically, open end down, into water until the open end is submerged 1.20 m. If $p_{atm} = 98 \text{ kPa}$, find the water rise $x$ in the tube.

$$p_1V_1 = p_2V_2$$ $$98(A)(1.5) = [98 + 9.81(1.2 - x)]A(1.5 - x)$$ $$147 = (109.77 - 9.81x)(1.5 - x)$$ $$x \approx 0.14 \text{ m}$$

Answer: Water rises approximately 140 mm in the tube.

Problem: Air Compressed by Pumping Water

An airtight pressure vessel initially has a total volume of 0.050 m³ filled entirely with air at atmospheric pressure of 101.3 kPa. Water is pumped into the vessel until the air occupies only 0.020 m³. Assuming isothermal compression, find the absolute pressure of the trapped air and the gage pressure in kPa.

$$p_1 V_1 = p_2 V_2$$ $$p_2 = \frac{p_1 V_1}{V_2} = \frac{101.3(0.050)}{0.020} = 253.25 \text{ kPa (abs)}$$ $$p_{gage} = 253.25 - 101.3 = 151.95 \text{ kPa}$$

Answer: Absolute pressure is 253.25 kPa. Gage pressure is 151.95 kPa. The air acts like a hydraulic accumulator — it stores energy as it is compressed.

Problem: Trapped Air Pocket at a Pipe Summit

A pipeline has a high point (summit). Initially, an air pocket 3.0 m long occupies a section of constant cross-section pipe at the summit, where the absolute air pressure is 60 kPa. Water is then forced through the pipe, compressing the air pocket to a length of 1.5 m. Determine the new absolute air pressure and the resulting gage water pressure at the base of the air pocket. Atmospheric pressure is 101.3 kPa.

Since cross-section is constant, volume is proportional to length:

$$p_1 L_1 = p_2 L_2$$ $$60(3.0) = p_2(1.5)$$ $$p_2 = 120 \text{ kPa (abs)}$$ $$p_{gage} = 120 - 101.3 = 18.7 \text{ kPa}$$

Answer: The compressed air pocket has absolute pressure 120 kPa (gage: 18.7 kPa). Air pockets at pipe summits reduce flow capacity and must be vented in real pipeline design.

Problem: Bubble Volume as a Diver Ascends

A diver releases an air bubble at a depth of 20 m in a fresh-water lake. The bubble has a volume of 6 cm³ at that depth. Atmospheric pressure is 101.3 kPa. Assuming isothermal expansion as the bubble rises, determine the volume of the bubble when it reaches the surface. Express the answer in cm³.

$$p_1 = p_{atm} + \gamma_w h = 101.3 + 9.81(20) = 297.5 \text{ kPa (abs)}$$ $$p_2 = p_{atm} = 101.3 \text{ kPa (abs)}$$ $$p_1 V_1 = p_2 V_2$$ $$V_2 = \frac{297.5 \times 6}{101.3} = 17.63 \text{ cm}^3$$

Answer: The bubble expands from 6 cm³ to 17.63 cm³ — nearly three times its original size. This rapid expansion near the surface is why divers must exhale continuously when ascending.

Problem: Two Chambers Equalizing Through a Valve

Two rigid airtight chambers are initially isolated from each other by a closed valve. Chamber 1 has a volume of 0.030 m³ at an absolute pressure of 250 kPa. Chamber 2 has a volume of 0.050 m³ at an absolute pressure of 80 kPa. When the valve is opened and pressures equalize at constant temperature, determine the final equilibrium pressure in both chambers (in kPa absolute).

Conservation of mass of gas (isothermal): total amount of gas (pV) is conserved.

$$p_1 V_1 + p_2 V_2 = p_f (V_1 + V_2)$$ $$250(0.030) + 80(0.050) = p_f(0.030 + 0.050)$$ $$7.5 + 4.0 = p_f(0.080)$$ $$p_f = \frac{11.5}{0.080} = 143.75 \text{ kPa (abs)}$$ $$p_{gage} = 143.75 - 101.3 = 42.45 \text{ kPa}$$

Answer: The equilibrium pressure is 143.75 kPa absolute (42.45 kPa gage), between the original pressures of the two chambers.

Problem: Density Ratio and Volume of Deep-Sea Sample

A rigid container traps 0.10 m³ of air at the sea surface (atmospheric pressure 101.3 kPa). The container is sealed and lowered to a depth of 500 m in salt water (SG = 1.025). Assuming the container is not rigid but perfectly flexible (like a balloon), find the new volume of air at depth. Also find the ratio of air density at depth to that at the surface.

$$p_2 = p_1 + \gamma_{sw} h = 101.3 + 9.81(1.025)(500) = 101.3 + 5027.6 = 5128.9 \text{ kPa}$$ $$V_2 = \frac{p_1 V_1}{p_2} = \frac{101.3(0.10)}{5128.9} = 0.001975 \text{ m}^3$$ $$\frac{\rho_2}{\rho_1} = \frac{p_2}{p_1} = \frac{5128.9}{101.3} = 50.6$$

Answer: At 500 m depth, the balloon volume shrinks from 0.10 m³ to only about 0.00198 m³. The air density increases by a factor of 50.6. This illustrates why deep-sea organisms have rigid bodies rather than air-filled cavities.