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Gas Laws

Note: Pressure and temperature used in the gas laws must be absolute. Convert gage pressure to absolute pressure and temperature to kelvins or degrees Rankine before substitution.

Boyle's Law (pressure vs. volume, constant temperature):

$$P_1V_1=P_2V_2$$

Charles' Law (volume vs. temperature, constant pressure):

$$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$

Gay-Lussac's Law (pressure vs. temperature, constant volume):

$$\frac{P_1}{T_1}=\frac{P_2}{T_2}$$

Combined Gas Law:

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Boyle's law is the form most often used in hydraulics for trapped-air tubes, bubbles rising through water, and compressed gas pockets.

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.

Three-meter inverted closed tube submerged 1.2 meters in water
$$p_1V_1 = p_2V_2$$ $$98(A)(1.5) = [98 + 9.81(1.2 - x)]A(1.5 - x)$$ $$147 = (109.77 - 9.81x)(1.5 - x)$$ $$x \approx 0.14 \text{ m}$$

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.

Inverted Closed Tube Submerged 1.2 m

A vertical tube 3 m long, with one end closed, is inserted vertically with the open end down into a tank of water until the open end is submerged to a depth of 1.2 m. Neglecting vapor pressure, how far will the water level in the tube be below the level in the tank? Assume absolute atmospheric pressure is 101.356 kPa

See images:

Three-meter inverted closed tube submerged 1.2 meters in water
Three-meter inverted closed tube submerged 1.2 meters in water

Let x be the distance of the water level inside the tube below the tank water surface. Because the 3 m tube is submerged 1.2 m, its closed end is 1.8 m above the tank water surface. The final trapped-air length is therefore 1.8 + x.

Using absolute pressure and Boyle's law:

$$P_1V_1=P_2V_2$$ $$P_1=101.356\ \mathrm{kPa}$$ $$V_1=3A$$ $$P_2=101.356+9.81x$$ $$V_2=(1.8+x)A$$

Substitute the initial and final conditions. The constant tube area cancels.

$$101.356(3A)=(101.356+9.81x)(1.8+x)A$$ $$304.07=(101.356+9.81x)(1.8+x)$$ $$304.07=182.44+119.01x+9.81x^2$$ $$9.81x^2+119.01x-121.63=0$$ $$x=\frac{-119.01+\sqrt{(119.01)^2-4(9.81)(-121.63)}}{2(9.81)}$$ $$\boxed{x=0.948\ \mathrm{m}}$$

The negative quadratic root is rejected because it is not physically possible.

Balloon Expansion by Charles' Law

Gas in a balloon occupies 2.50 L at 300°K. At what temperature will the balloon expand to 7.50 L?

At constant pressure, apply Charles' law using absolute temperature.

$$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ $$T_2=T_1\frac{V_2}{V_1}=300\left(\frac{7.50}{2.50}\right)$$ $$\boxed{T_2=900\ \mathrm{K}}$$

New Volume of Gas in a Flexible Container

A gas at 1.2 atm and 30°C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised by 80°C and the pressure is increased to 4 atm, what is the new volume?

Convert both temperatures to kelvins. An increase of 80°C makes the final temperature 110°C.

$$T_1=30+273=303\ \mathrm{K},\qquad T_2=110+273=383\ \mathrm{K}$$ $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$ $$V_2=\frac{P_1V_1T_2}{P_2T_1}=\frac{(1.2)(2.00)(383)}{(4)(303)}$$ $$\boxed{V_2=0.758\ \mathrm{L}}$$

Compressibility and Bulk Modulus of Elasticity

Compressibility, β, is the extent to which a fluid particle changes volume when subjected to either a change in pressure or a change in temperature.

$$\beta=\frac{1}{K}$$

Bulk Modulus of Elasticity, K:

$$K=\frac{-\Delta P}{\Delta V/V_0}=\frac{-\Delta P}{\Delta v/v_0}$$

Percentage Volume Decrease of Pressurized Water

In a hydraulic press, water is subjected to a pressure of 10000 psia at 68°F. If the initial pressure is 15 psia, what is the percentage decrease in volume, considering that the average bulk modulus of elasticity of water is 365000 psi for this temperature range?

The pressure increase is the final absolute pressure minus the initial absolute pressure. From the definition of bulk modulus, compression gives a negative volume change.

$$\Delta P=10000-15=9985\ \mathrm{psi}$$ $$K=-\frac{\Delta P}{\Delta V/V_0}$$ $$\frac{\Delta V}{V_0}=-\frac{\Delta P}{K}=-\frac{9985}{365000}=-0.02736$$ $$\boxed{\frac{\Delta V}{V_0}(100)=-2.736\%}$$

The signed volume change is -2.736%; equivalently, the volume decreases by 2.736%.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t25

HGE - Hydraulics / Pressure and Boyle's Law / Civil Engineering Refresher

Find the headwater height H in meters such that an air bubble reaching the surface has a volume 3 times larger than it had at the bottom (Patm = 101.3 kPa).

  1. 20.65 m
  2. 10.33 m
  3. 30.98 m
  4. 15.20 m
Boyle's Law: $P_{bottom}V = P_{surface}(3V) \Rightarrow P_{bottom} = 3P_{surface}$.
$P_{surface} = 101.3$ kPa, so $P_{bottom} = 303.9$ kPa abs.
$P_{atm} + \gamma_w H = 303.9 \Rightarrow 101.3 + 9.81H = 303.9$
$H = \frac{202.6}{9.81}$
$\boxed{= 20.65 \text{ m}}$