Pneumatics and Gas Laws

15 questions

Question 1Question →

A sealed pneumatic cylinder initially contains air at a pressure of 40 psi40 \text{ psi}. A mechanical linkage forces the cylinder's internal volume to be reduced to exactly half (1/21/2) of its original volume. Simultaneously, an external heat source causes the absolute temperature of the confined air to double. If this newly pressurized air is then directed into a separate pneumatic actuator with a piston surface area of 5 in25 \text{ in}^2, what is the total theoretical force exerted by this actuator?

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Answer: 800 lbs800 \text{ lbs}

Answer

The correct force is 800 lbs800 \text{ lbs}.
The correct answer accurately combines two gas law principles before applying fluid power mathematics. First, halving the volume of a gas doubles its pressure (Boyle's Law). Second, doubling the absolute temperature doubles that pressure again (Gay-Lussac's Law). This results in a final system pressure of 160 psi160 \text{ psi}. When applied to a 5 in25 \text{ in}^2 piston, the total force is 160 psi×5=800 lbs160 \text{ psi} \times 5 = 800 \text{ lbs}.

Step-by-Step Solution

1
Determine the effect of the volume change on the gas pressure using Boyle's Law (P1×V1=P2×V2P_1 \times V_1 = P_2 \times V_2).
Since the volume is halved (1/21/2), the pressure must double. 40 psi×2=80 psi40 \text{ psi} \times 2 = 80 \text{ psi}.
In a compressible gas, pressure and volume are inversely proportional when temperature is held constant.
2
Determine the effect of the temperature change on the gas pressure using Gay-Lussac's Law.
Since the absolute temperature is doubled, the pressure must double again. 80 psi×2=160 psi80 \text{ psi} \times 2 = 160 \text{ psi}.
In a confined gas, absolute pressure is directly proportional to absolute temperature.
3
Calculate the force exerted by the actuator using the final pressure and the piston area (F=P×AF = P \times A).
160 psi×5 in2=800 lbs160 \text{ psi} \times 5 \text{ in}^2 = 800 \text{ lbs}.
Force is defined as pressure applied over a specific surface area.

Key Concept

The Combined Gas Law allows simultaneous evaluation of pressure, volume, and temperature changes in a pneumatic system, which is then used to calculate mechanical force.
Question 2Question →

An industrial pneumatic press system utilizes trapped gas to generate force. The gas is initially at an absolute pressure of 30 psi30 \text{ psi} and occupies a volume of 200 in3200 \text{ in}^3. During the operation cycle, a mechanical ram pushes down, reducing the gas volume to 50 in350 \text{ in}^3 while the temperature remains constant. If the face of the piston has a cross-sectional area of 5 in25 \text{ in}^2, what is the total upward force, in pounds, exerted by the compressed gas on the piston at this new volume?

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Answer: 600

Answer

600
Since the temperature is constant, Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2) applies. The volume is reduced from 200 in3200 \text{ in}^3 to 50 in350 \text{ in}^3, which is one-quarter of its original size. Because pressure and volume are inversely proportional for a fixed amount of gas at a constant temperature, the absolute pressure must increase by a factor of four (30 psi×4=120 psi30 \text{ psi} \times 4 = 120 \text{ psi}). The total force exerted by a pressurized fluid on a surface is the product of its pressure and the surface area (F=P×AF = P \times A). Multiplying the new pressure of 120 psi120 \text{ psi} by the piston area of 5 in25 \text{ in}^2 gives a total upward force of 600 lbs600 \text{ lbs}.

Step-by-Step Solution

1
Calculate the final pressure of the gas after compression.
P2=120 psiP_2 = 120 \text{ psi}
According to Boyle's Law, when temperature is constant, the product of absolute pressure and volume is constant (P1V1=P2V2P_1V_1 = P_2V_2).
2
Calculate the total upward force exerted by the gas.
F=600 lbsF = 600 \text{ lbs}
Force is equal to pressure multiplied by the area over which it is applied (F=P×AF = P \times A).

Key Concept

Combining Boyle's Law for isothermal compression with the basic relationship between pressure, force, and area.
Question 3Question →

A heavy-duty pneumatic shock absorber relies on a sealed chamber of air to dampen impacts. Before an impact, the air inside the chamber has an absolute pressure of 90 psi90 \text{ psi}, occupies a volume of 60 in360 \text{ in}^3, and is at a temperature of 300 K300 \text{ K}. During a severe impact, the piston rapidly compresses the air volume to 20 in320 \text{ in}^3, causing the internal temperature to spike to 400 K400 \text{ K}. Assuming no air escapes, what is the new absolute pressure inside the shock absorber?

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Answer: 360 psi360 \text{ psi}

Answer

360 psi360 \text{ psi}
The correct answer correctly applies the Combined Gas Law. The volume is reduced to one-third of its original size, which alone would triple the pressure to 270 psi270 \text{ psi}. Simultaneously, the absolute temperature increases by a factor of 4/34/3, which further scales the pressure to 270×43=360 psi270 \times \frac{4}{3} = 360 \text{ psi}.

Step-by-Step Solution

1
Identify the initial state (P1,V1,T1P_1, V_1, T_1) and final state (V2,T2V_2, T_2) variables from the problem.
P1=90 psiP_1 = 90 \text{ psi}, V1=60 in3V_1 = 60 \text{ in}^3, T1=300 KT_1 = 300 \text{ K}; V2=20 in3V_2 = 20 \text{ in}^3, T2=400 KT_2 = 400 \text{ K}.
Establishing known parameters is required to use the Combined Gas Law.
2
Set up the Combined Gas Law equation: P1×V1T1=P2×V2T2\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}.
90×60300=P2×20400\frac{90 \times 60}{300} = \frac{P_2 \times 20}{400}
Since both volume and temperature are changing simultaneously, the combined relationship must be used.
3
Rearrange the formula to isolate the final pressure (P2P_2).
P2=P1×(V1V2)×(T2T1)P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)
This setup groups the variables into easy-to-calculate ratios.
4
Substitute the values and calculate the final result.
P2=90×(6020)×(400300)=90×3×43=360 psiP_2 = 90 \times \left(\frac{60}{20}\right) \times \left(\frac{400}{300}\right) = 90 \times 3 \times \frac{4}{3} = 360 \text{ psi}.
Mathematical evaluation yields the final absolute pressure inside the shock absorber.

Key Concept

Combined Gas Law and Compressibility
Question 4Question →

Arrange the following events in the correct chronological order for the extension stroke of a basic single-acting pneumatic cylinder:

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Answer

The correct sequence is: 1) The control valve is opened, 2) Air flows into the cylinder, 3) Pressure builds against the piston, 4) The piston rod extends.
The chronological sequence follows the fundamental transfer of energy in a pneumatic system. It begins with control (valve opening), followed by fluid transmission (air flowing into the cylinder), followed by energy accumulation (pressure building against the piston due to gas compressibility), and ends with mechanical work (piston extension).

Step-by-Step Solution

1
Identify the initiating control action
The control valve is opened.
A pneumatic system requires a signal or mechanical actuation to release energy from the compressor or tank.
2
Trace the path of the working fluid
Air enters the cylinder chamber.
Air moves from a region of higher pressure (supply) to lower pressure (cylinder) through the open valve.
3
Determine the physical effect inside the actuator
Pressure builds against the piston.
As air fills the confined space of the cylinder, it exerts uniform pressure on the piston face.
4
Identify the resulting mechanical output
The piston extends.
Once the force of the air pressure is greater than the opposing load and friction, the piston moves outward.

Key Concept

Pneumatic System Operation and Force Generation
Question 5Question →

A mechanic pushes down on the handle of a manual air pump, sealing the valve and compressing the trapped air. If the volume of the air chamber is reduced to exactly one-half of its original size while the temperature remains constant, what happens to the air pressure inside the pump?

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Answer: The pressure doubles.

Answer

The pressure doubles.
According to Boyle's Law, the pressure of a given mass of an ideal gas is inversely proportional to its volume when the temperature remains constant. Because the mechanic forces the same amount of air into half the space, the air molecules strike the walls of the pump chamber twice as often, causing the pressure to double.

Step-by-Step Solution

1
Identify the relationship between gas volume and pressure at a constant temperature.
The relationship is inversely proportional, described by Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2).
Understanding this core gas law is necessary to determine the outcome of volume changes in a sealed pneumatic system.
2
Apply the given volume change to the inverse relationship.
Since the volume is multiplied by 1/21/2, the pressure must be multiplied by the reciprocal, which is 22.
To keep the product of pressure and volume constant, halving one variable strictly requires doubling the other.

Key Concept

Boyle's Law and Gas Compressibility
Estimated Time:45s
Question 6Question →

A backup breathing canister has a volume of 50 in350 \text{ in}^3 and stores gas at an absolute pressure of 2000 psi2000 \text{ psi}. During a transfer process, the gas is allowed to expand into a larger holding tank, reaching a total final volume of 250 in3250 \text{ in}^3. Assuming the temperature remains constant, what is the new absolute pressure of the gas in psi\text{psi}?

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Answer: 400

Answer

400
According to Boyle's Law, the pressure and volume of a gas are inversely proportional when the temperature is constant (P1V1=P2V2P_1V_1 = P_2V_2). Since the volume increased by a factor of 5 (from 50 to 250 cubic inches), the pressure must decrease by a factor of 5. Dividing the initial 2000 psi by 5 gives a final pressure of 400 psi.

Step-by-Step Solution

1
Identify the initial and final states of the gas.
P1=2000 psiP_1 = 2000 \text{ psi}, V1=50 in3V_1 = 50 \text{ in}^3, and V2=250 in3V_2 = 250 \text{ in}^3.
Organizing the known variables is the first step in solving any gas law problem.
2
Apply Boyle's Law to relate pressure and volume.
Equation to use: P1V1=P2V2P_1 V_1 = P_2 V_2.
Boyle's Law states that for a fixed mass of gas at constant temperature, pressure and volume are inversely proportional.
3
Substitute the known values into the equation and solve for the final pressure (P2P_2).
2000×50=P2×2502000 \times 50 = P_2 \times 250, which simplifies to 100000=250P2100000 = 250 P_2. Dividing both sides by 250 gives P2=400P_2 = 400.
This mathematical calculation yields the new absolute pressure after the gas has expanded.

Key Concept

Boyle's Law (Inverse relationship between pressure and volume)

Alternative Method

Instead of full algebraic substitution, recognize the proportionality: the volume increases by a factor of 5 (250/50=5250 / 50 = 5). Because pressure and volume are inversely proportional, the pressure must be divided by 5. Thus, 2000 psi/5=400 psi2000 \text{ psi} / 5 = 400 \text{ psi}.
Estimated Time:45s
Question 7Question →

A perfectly sealed, rigid steel air compressor tank is left outside in direct sunlight on a very hot summer day. As the temperature of the trapped air inside the tank increases, what will happen to the internal air pressure?

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Answer: The pressure will increase.

Answer

The pressure will increase.
According to Gay-Lussac's Law, the pressure of a fixed mass of gas is directly proportional to its temperature when the volume remains constant. As the air inside the rigid steel tank absorbs heat from the sun, the air molecules gain kinetic energy. They move faster and collide with the internal walls of the tank more frequently and with greater force, which results in a higher internal pressure.

Step-by-Step Solution

1
Identify the relationship between temperature and pressure for a trapped gas.
Gay-Lussac's Law states that pressure is directly proportional to absolute temperature when volume is constant.
The tank is described as 'rigid' and 'perfectly sealed', meaning the volume cannot change.
2
Apply the temperature change to the system.
The sunlight heats the tank, increasing the temperature of the air inside.
To determine the final state of the gas.
3
Determine the outcome based on the direct relationship.
Because temperature increases, pressure must also increase.
Higher temperature means higher kinetic energy; molecules strike the rigid walls harder and more frequently.

Key Concept

Gay-Lussac's Law (Temperature-Pressure Relationship)
Question 8Question →

Place the following operational stages of a basic reciprocating air compressor in their proper sequential order, starting with the introduction of fresh air into the system.

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Answer

The correct operational sequence begins with the intake of ambient air, followed by the sealing of the inlet valve, then the compression of the trapped air, and finally the discharge of the pressurized air into the tank.
A basic reciprocating compressor operates in a continuous four-stage loop: it draws air in (intake), seals the chamber, squeezes the air into a smaller volume to build pressure (compression), and then pushes that pressurized air out to a tank (discharge).

Step-by-Step Solution

1
Identify the starting point of the pneumatic cycle.
The cycle must begin by drawing ambient air into the empty cylinder.
An air compressor needs a fresh supply of air to perform any mechanical work.
2
Determine the necessary condition for compression.
The inlet valve must close to seal the chamber.
If the chamber is not sealed, upward movement of the piston would simply push the air back out instead of compressing it.
3
Identify the actual compression action.
The piston moves up to decrease the internal volume.
Decreasing the volume of the sealed gas increases its pressure.
4
Identify the final stage of the cycle.
The outlet valve opens to release the compressed air.
The system must transfer the high-pressure air into the storage tank for future use.

Key Concept

Reciprocating compressor cycle and basic application of Boyle's Law
Question 9Question →

An aircraft's emergency pneumatic blow-down bottle contains nitrogen gas at an absolute pressure of 3000 psi3000 \text{ psi} with an internal volume of 200 in3200 \text{ in}^3. During an emergency landing gear extension, a valve opens and the gas expands into the actuator lines and cylinders, increasing the total volume occupied by the gas to 600 in3600 \text{ in}^3. Assuming the temperature remains constant during this rapid expansion, what is the new absolute pressure in the system?

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Answer: 1000 psi1000 \text{ psi}

Answer

The new absolute pressure in the system is 1000 psi1000 \text{ psi}.
According to Boyle's Law, when the temperature of a gas is held constant, its pressure and volume are inversely proportional. The gas expanded from an initial volume of 200 in3200 \text{ in}^3 to a total final volume of 600 in3600 \text{ in}^3. Because the volume increased by a factor of 3 (600/200=3600 / 200 = 3), the pressure must decrease by a factor of 3. Dividing the initial 3000 psi3000 \text{ psi} by 3 results in 1000 psi1000 \text{ psi}.

Step-by-Step Solution

1
Identify the appropriate gas law for a scenario with changing pressure and volume at a constant temperature.
Boyle's Law applies, which is expressed as P1×V1=P2×V2P_1 \times V_1 = P_2 \times V_2.
Boyle's Law dictates that the pressure of a fixed mass of gas is inversely proportional to its volume, provided the temperature remains constant.
2
Identify the known initial and final variables from the problem statement.
Initial pressure (P1P_1) = 3000 psi3000 \text{ psi}, Initial volume (V1V_1) = 200 in3200 \text{ in}^3, and Final volume (V2V_2) = 600 in3600 \text{ in}^3.
These are the critical parameters needed to solve for the unknown final pressure (P2P_2).
3
Substitute the known values into Boyle's Law and solve for the final pressure (P2P_2).
3000×200=P2×600→600000=P2×600→P2=1000 psi3000 \times 200 = P_2 \times 600 \rightarrow 600000 = P_2 \times 600 \rightarrow P_2 = 1000 \text{ psi}.
Isolating P2P_2 algebraically provides the exact pressure after the gas has fully expanded into the larger volume.

Key Concept

Boyle's Law (Inverse relationship between pressure and volume)
Question 10Question →

During the operation of a pneumatic nailer, its internal firing reservoir holds a compressed air volume of 20 in320 \text{ in}^3 at an absolute pressure of 120 psi120 \text{ psi}. When the trigger is pulled, the air expands into the driver chamber at a constant temperature until the absolute pressure drops to 30 psi30 \text{ psi}. Based on Boyle's Law, what is the new total volume of the expanded air in cubic inches?

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Answer: 80

Answer

80
According to Boyle's Law, pressure and volume are inversely proportional at a constant temperature (P1×V1=P2×V2P_1 \times V_1 = P_2 \times V_2). The pressure drops from 120 psi to 30 psi, which is a factor of 4 decrease. Because the relationship is inversely proportional, the volume must increase by a factor of 4. Therefore, the initial volume of 20 cubic inches is multiplied by 4, resulting in a final volume of 80 cubic inches.

Step-by-Step Solution

1
Identify the initial and final states of the gas.
Initial pressure P1=120 psiP_1 = 120 \text{ psi}, Initial volume V1=20 in3V_1 = 20 \text{ in}^3. Final pressure P2=30 psiP_2 = 30 \text{ psi}.
Establishing the known parameters is required before applying the appropriate gas law.
2
Apply Boyle's Law formula.
P1×V1=P2×V2P_1 \times V_1 = P_2 \times V_2
Since the temperature is constant, Boyle's Law relates the pressure and volume of the trapped air.
3
Substitute values and solve for the unknown final volume (V2V_2).
120×20=30×V2→2400=30×V2→V2=80120 \times 20 = 30 \times V_2 \rightarrow 2400 = 30 \times V_2 \rightarrow V_2 = 80
Isolating V2V_2 provides the final expanded volume of the air.

Key Concept

Boyle's Law states that the absolute pressure of a confined gas is inversely proportional to its volume, provided the temperature remains constant.

Alternative Method

Use proportional reasoning: Observe that the final pressure (30 psi) is exactly one-quarter (1/4) of the initial pressure (120 psi). Because pressure and volume are inversely proportional under constant temperature, the final volume must be four times (4x) the initial volume. 4 × 20 = 80.
Estimated Time:1m 0s
Question 11Question →

A heavy transport vehicle uses a pneumatic (air) suspension system to maintain a constant ride height regardless of cargo weight. Place the following events in the operation of this pneumatic leveling system in the correct chronological order, starting from the moment a heavy load is placed on the vehicle.

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Answer

The correct sequence is: load compresses springs, valve detects drop, air flows into springs, increased pressure raises vehicle.
The correct sequence follows the logical physical response of a pneumatic system to an external load. First, the load mechanically compresses the air springs. This downward movement triggers the control valve. The open valve then allows high-pressure air to flow from the reservoir into the springs. Finally, the newly added air increases the internal pressure, expanding the springs to lift the load and reset the system.

Step-by-Step Solution

1
Identify the initial triggering event.
The addition of heavy cargo forces the vehicle down, compressing the pneumatic springs.
Air is a compressible fluid, so adding weight decreases the volume of the air in the springs until the pressure balances the new weight.
2
Determine the system's mechanical response.
The leveling valve detects this drop in height and opens.
Pneumatic systems require a mechanical or electrical trigger to allow high-pressure air to enter the working components.
3
Trace the flow of the compressed gas.
Air moves from the storage reservoir into the springs.
Gas naturally flows from an area of higher pressure (reservoir) to an area of lower pressure (compressed springs).
4
Identify the final mechanical work performed by the gas.
The added air increases the pressure inside the springs, forcing them to expand and lift the vehicle.
As more gas is introduced into the flexible space, pressure increases, generating enough mechanical force to lift the load.

Key Concept

Operation of pneumatic control systems and the relationship between air pressure, volume, and mechanical force.
Question 12Question →

A pneumatic lifting system uses compressed air to operate a main lifting cylinder. Initially, the cylinder contains 2.0 m32.0 \text{ m}^3 of air at an absolute pressure of 100 kPa100 \text{ kPa}. To prepare for a heavy lift, the system's compressor forces the piston inward, reducing the trapped air's volume to 0.5 m30.5 \text{ m}^3. Assuming the temperature of the air remains constant during this process, what is the new absolute pressure inside the cylinder?

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Answer: 400 kPa400 \text{ kPa}

Answer

The new absolute pressure is 400 kPa400 \text{ kPa}.
According to Boyle's Law, the pressure of a fixed mass of gas is inversely proportional to its volume, provided the temperature remains constant. The equation is P1V1=P2V2P_1V_1 = P_2V_2. By substituting the given values (100 kPa×2.0 m3=P2×0.5 m3100 \text{ kPa} \times 2.0 \text{ m}^3 = P_2 \times 0.5 \text{ m}^3), we find that reducing the volume to one-quarter of its original size forces the pressure to quadruple, resulting in a final pressure of 400 kPa400 \text{ kPa}.

Step-by-Step Solution

1
Identify the relevant gas law for a situation with changing pressure and volume at a constant temperature.
Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2) applies to this scenario.
The problem specifies a trapped gas undergoing volume compression while maintaining a constant temperature.
2
Extract the known initial and final values from the problem statement.
Initial pressure (P1P_1) = 100 kPa100 \text{ kPa}. Initial volume (V1V_1) = 2.0 m32.0 \text{ m}^3. Final volume (V2V_2) = 0.5 m30.5 \text{ m}^3.
Organizing known variables is necessary before attempting to solve for the unknown final pressure (P2P_2).
3
Substitute the known values into Boyle's Law equation and solve for the unknown final pressure.
100×2.0=P2×0.5100 \times 2.0 = P_2 \times 0.5. This simplifies to 200=0.5P2200 = 0.5P_2, which yields P2=400 kPaP_2 = 400 \text{ kPa}.
Calculating the algebraic solution provides the final absolute pressure required by the question.

Key Concept

Boyle's Law (Inverse Relationship of Pressure and Volume)
Question 13Question →

A heavy-duty vehicle's pneumatic braking system relies on an accumulator tank to store potential energy. Initially, a volume of 150 in3150 \text{ in}^3 of air is compressed at an absolute pressure of 90 psi90 \text{ psi} inside the accumulator. When the operator engages the brakes, the air expands isothermally (at a constant temperature) into the connected brake chambers, causing the system's absolute pressure to drop to 30 psi30 \text{ psi}. What is the total volume that the air now occupies, in cubic inches?

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Answer: 450

Answer

450
According to Boyle's Law, the pressure and volume of a confined gas are inversely proportional when the temperature is held constant. The initial state is 90 psi90 \text{ psi} and 150 in3150 \text{ in}^3. The final pressure drops to 30 psi30 \text{ psi}, which is one-third of the original pressure. Therefore, to balance the equation and conserve the quantity of gas, the volume must expand to three times its original size. 150×3=450 in3150 \times 3 = 450 \text{ in}^3.

Step-by-Step Solution

1
Identify the known variables from the scenario.
P1=90 psiP_1 = 90 \text{ psi}, V1=150 in3V_1 = 150 \text{ in}^3, P2=30 psiP_2 = 30 \text{ psi}.
Establishing the initial and final states of the gas is necessary to select the correct formula.
2
Select the appropriate gas law.
Use Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2.
The problem states the expansion happens isothermally (constant temperature) and involves inversely proportional pressure and volume changes.
3
Substitute the values and solve for V2V_2.
90×150=30×V290 \times 150 = 30 \times V_2, so V2=13500/30=450 in3V_2 = 13500 / 30 = 450 \text{ in}^3.
Mathematical calculation is required to find the unknown final volume.

Key Concept

Boyle's Law (Pressure-Volume relationship at constant temperature)

Alternative Method

Instead of using the formula P1V1=P2V2P_1V_1 = P_2V_2 algebraically, you can use proportional reasoning. The pressure drops from 90 psi90 \text{ psi} to 30 psi30 \text{ psi}, which is a reduction by a factor of 3 (90/30=390 / 30 = 3). Because pressure and volume are inversely proportional at a constant temperature, the volume must increase by that exact same factor. Therefore, simply multiply the initial volume by 3: 150×3=450 in3150 \times 3 = 450 \text{ in}^3.
Estimated Time:1m 0s
Question 14Question →

A sealed pneumatic cylinder in a robotic assembly arm contains air at an absolute pressure of 20 psi20 \text{ psi}. The cylinder has an initial internal volume of 60 cm360 \text{ cm}^3. During a rapid compression stroke at a constant temperature, the piston reduces the volume to 15 cm315 \text{ cm}^3. What is the new absolute pressure inside the cylinder?

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Answer: 80 psi80 \text{ psi}

Answer

The absolute pressure increases to 80 psi80 \text{ psi} because reducing the volume by a factor of 4 causes the pressure to multiply by 4.
According to Boyle's Law, pressure and volume are inversely proportional when temperature is constant (P1V1=P2V2P_1V_1 = P_2V_2). The initial pressure-volume product is 20×60=120020 \times 60 = 1200. To maintain this constant with a new volume of 15 cm315 \text{ cm}^3, the new pressure must be 1200÷15=80 psi1200 \div 15 = 80 \text{ psi}. Alternatively, recognizing that the volume is reduced by a factor of 4 (60÷15=460 \div 15 = 4) means the pressure must increase by a factor of 4 (20×4=80 psi20 \times 4 = 80 \text{ psi}).

Step-by-Step Solution

1
Identify the initial conditions and the relevant gas law.
Initial pressure (P1P_1) is 20 psi20 \text{ psi} and initial volume (V1V_1) is 60 cm360 \text{ cm}^3. Since temperature is constant, use Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2.
Boyle's Law defines the inversely proportional relationship between the pressure and volume of a gas when temperature remains unchanged.
2
Set up the equation with the known values.
20 psi×60 cm3=P2×15 cm320 \text{ psi} \times 60 \text{ cm}^3 = P_2 \times 15 \text{ cm}^3
Substituting the given parameters prepares the equation to isolate the unknown final pressure (P2P_2).
3
Solve for the final pressure (P2P_2).
1200=P2×151200 = P_2 \times 15, which simplifies to P2=120015=80 psiP_2 = \frac{1200}{15} = 80 \text{ psi}.
Isolating P2P_2 reveals the new absolute pressure after the compression stroke.

Key Concept

Boyle's Law (Pressure-Volume Inverse Relationship)
Question 15Question →

A mechanical pressure relief valve is installed on a pneumatic air receiver tank to prevent over-pressurization. Arrange the following events in the correct chronological sequence, from the moment the tank begins to over-pressurize until the system stabilizes.

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Answer

The sequence begins with system pressure exceeding the setpoint. This creates a pneumatic force that overcomes the spring tension, unseating the valve. The compressed air then vents, lowering system pressure until the spring reseats the valve.
The correct operational sequence of a pneumatic relief valve relies on the balance between gas pressure and mechanical spring force. Initially, an over-pressure condition occurs in the tank. Because force equals pressure multiplied by area, this high pressure exerts an upward force on the valve plug. When this pneumatic force exceeds the downward force of the calibrated spring, the plug lifts. This physical unseating creates a pathway, allowing the highly compressed gas to rapidly expand and vent to the lower-pressure atmosphere. As the gas escapes, the internal tank pressure drops until the pneumatic force is weaker than the spring tension, allowing the spring to snap the valve closed again.

Step-by-Step Solution

1
Identify the initial condition that triggers the safety system.
The air pressure inside the tank rises above the safe setpoint limit.
A pneumatic relief sequence must begin with an over-pressure event.
2
Determine the mechanical interaction between the compressed air and the valve.
The pneumatic force overcomes the mechanical spring tension.
For the valve to operate, fluid force (Pressure × Area) must defeat the physical mechanism keeping it closed.
3
Establish the physical movement resulting from the force imbalance.
The valve plug lifts off its seat to expose the exhaust port.
The net upward pneumatic force pushes the valve open.
4
Identify the behavior of the gas once the pathway is open.
Compressed air escapes into the atmosphere, reducing tank pressure.
Gas naturally expands and flows from areas of high pressure to low pressure.
5
Determine the condition that ends the sequence.
The spring pushes the valve closed as the pneumatic pressure decreases.
When tank pressure drops sufficiently, the pneumatic force is no longer strong enough to resist the spring.

Key Concept

Pneumatic System Control and Pressure Relief Operations
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