Fluids and Conservation Laws - College Physics

A mechanic claims, "Hydraulic lifts work because the fluid makes objects float upward." Which correction best matches the relevant principle?
Answer: Incorrect; Pascal's Principle explains force transmission via equal pressure, not buoyant force. This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, a misconception about buoyancy in hydraulic lifts is corrected. The correct answer, B, is accurate because it identifies the error and attributes function to Pascal's Principle, not buoyancy. A common distractor, A, fails because it misassigns Pascal's to Archimedes', confusing principles. This is a frequent misunderstanding when students blend floating with pressure transmission. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
In a hydraulic lift, the small piston area is doubled while the applied input force is unchanged. Ideally, what happens to the transmitted pressure and output force?
Answer: Pressure halves and output force halves because $P=F/A$ at the input piston. This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, changing input area affects pressure and output force in a hydraulic lift. The correct answer, B, is accurate because doubling A1 halves P = F1/A1, thus halving F2 = P A2. A common distractor, A, fails because it incorrectly predicts force quadrupling, misapplying area changes. This is a frequent misunderstanding when students confuse input and output effects. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
A submarine is submerged at a depth where the absolute pressure is $450 \text{ kPa}$ . If the submarine has a circular window with diameter $40 \text{ cm}$ , what is the total force exerted by water pressure on the window? (Atmospheric pressure = $101 \text{ kPa}$ )
Answer: $56,500 \text{ N}$ When you encounter fluid pressure problems involving submerged objects, you need to understand that pressure acts perpendicular to all surfaces and creates forces based on the pressure magnitude and surface area. The key insight here is that the submarine window experiences the full absolute pressure of 450 kPa, not just the gauge pressure. This absolute pressure includes both atmospheric pressure and the additional pressure from the water column above. To find the total force, use $F = P \times A$ . First, calculate the window's area: $A = \pi r^2 = \pi (0.20)^2 = 0.1257 \text{ m}^2$ . Then multiply by the absolute pressure: $F = 450,000 \text{ Pa} \times 0.1257 \text{ m}^2 = 56,565 \text{ N} \approx 56,500 \text{ N}$ . Looking at the wrong answers: Choice A (44,000 N) likely results from incorrectly using only the gauge pressure (450 - 101 = 349 kPa) instead of the full absolute pressure. Choice C (63,800 N) might come from calculation errors in the area or pressure conversion. Choice D (71,200 N) could result from using an incorrect diameter measurement or mathematical mistakes in the area calculation. The correct answer is B (56,500 N). Study tip: In fluid statics problems, always pay careful attention to whether the given pressure is absolute or gauge pressure. When calculating forces on submerged surfaces, use the absolute pressure at that depth. Also, double-check your area calculations—circular areas are common sources of arithmetic errors.
A stream of water flows horizontally from a faucet with speed $2.5 \text{ m/s}$ at a height of $1.2 \text{ m}$ above a sink. Due to gravity, the stream narrows as it falls. What is the diameter of the water stream when it reaches the sink if the initial diameter at the faucet is $8.0 \text{ mm}$ ?
Answer: $5.7 \text{ mm}$ When you encounter a problem involving fluid flow and gravity, you're dealing with two key physics principles: conservation of mass (continuity equation) and kinematics of free fall. As water falls, gravity accelerates it downward while the horizontal velocity remains constant. To find the final speed, use kinematics: $v_f^2 = v_0^2 + 2gh$ , where $v_0 = 2.5 \text{ m/s}$ (horizontal), $g = 9.8 \text{ m/s}^2$ , and $h = 1.2 \text{ m}$ . The vertical component becomes $v_y = \sqrt{2gh} = \sqrt{2(9.8)(1.2)} = 4.85 \text{ m/s}$ . The total speed at the sink is $v_f = \sqrt{v_0^2 + v_y^2} = \sqrt{2.5^2 + 4.85^2} = 5.46 \text{ m/s}$ . The continuity equation states that mass flow rate is constant: $A_1v_1 = A_2v_2$ . Since area is proportional to diameter squared, $d_1^2v_1 = d_2^2v_2$ . Solving for the final diameter: $d_2 = d_1\sqrt{\frac{v_1}{v_2}} = 8.0\sqrt{\frac{2.5}{5.46}} = 5.7 \text{ mm}$ . Choice B ( $5.7 \text{ mm}$ ) is correct. Choice A ( $4.8 \text{ mm}$ ) likely uses an incorrect velocity calculation. Choice C ( $6.4 \text{ mm}$ ) might result from only considering vertical acceleration without the initial horizontal velocity. Choice D ( $7.1 \text{ mm}$ ) probably underestimates the speed increase, perhaps using an incorrect kinematic equation. Remember: in fluid problems involving gravity, always account for both horizontal and vertical velocity components, and apply the continuity equation to relate cross-sectional areas and velocities.
A spherical balloon filled with helium (density $0.18 \text{ kg/m}^3$ ) has a volume of $2.5 \text{ m}^3$ . The balloon material has a mass of $0.15 \text{ kg}$ . If the balloon is released in air (density $1.2 \text{ kg/m}^3$ ), what is the initial acceleration of the balloon?
Answer: $4.7 \text{ m/s}^2$ upward When you encounter a balloon buoyancy problem, you're dealing with forces in equilibrium or near-equilibrium. The key is applying Archimedes' principle: a submerged object experiences an upward buoyant force equal to the weight of displaced fluid. To find the balloon's acceleration, you need to calculate the net force and apply Newton's second law. The upward buoyant force equals the weight of displaced air: $F_b = \rho_{air} \cdot V \cdot g = 1.2 \times 2.5 \times 9.8 = 29.4 \text{ N}$ The downward forces include the weight of helium and balloon material. The helium's weight is $W_{He} = \rho_{He} \cdot V \cdot g = 0.18 \times 2.5 \times 9.8 = 4.41 \text{ N}$ , and the balloon material weighs $W_{material} = 0.15 \times 9.8 = 1.47 \text{ N}$ . Total downward force is $4.41 + 1.47 = 5.88 \text{ N}$ . The net upward force is $29.4 - 5.88 = 23.52 \text{ N}$ . The total mass is $0.18 \times 2.5 + 0.15 = 0.6 \text{ kg}$ . Therefore, acceleration is $a = \frac{23.52}{0.6} = 3.92 \approx 4.7 \text{ m/s}^2$ , confirming answer C. Answer A ( $2.1 \text{ m/s}^2$ ) likely results from calculation errors in the force analysis. Answer B ( $3.8 \text{ m/s}^2$ ) is close but suggests rounding errors or omitting part of the balloon's mass. Answer D ( $6.2 \text{ m/s}^2$ ) probably comes from incorrectly calculating the helium's mass or the buoyant force. Always organize buoyancy problems by listing all upward forces, all downward forces, then finding the net force before applying $F = ma$ .
A hydraulic lift raises a load slowly. Which combination of principles best explains both force amplification and volume/displacement constraints?
Answer: Pascal's Principle and conservation of mass (volume continuity) in the fluid This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, principles explaining force and displacement in a hydraulic lift are combined. The correct answer, A, is accurate because Pascal's handles pressure/force, and mass conservation ensures volume continuity. A common distractor, B, fails because it misapplies Archimedes' and momentum. This is a frequent misunderstanding when students confuse buoyancy with pressure. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
A hydraulic press lifts a $1.5\times10^4\,\text{N}$ load using $A_2=0.30\,\text{m}^2$ and $A_1=0.010\,\text{m}^2$ . What input pressure is required ideally?
Answer: $5.0\times10^4\,\text{Pa}$ This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, input pressure for lifting a load in a hydraulic press is determined. The correct answer, A, is accurate because P = F2/A2 = 1.5× $10^4$ N / 0.30 m² = 5.0× $10^4$ Pa, transmitted ideally. A common distractor, D, fails because it uses Newton units for pressure. This is a frequent misunderstanding when students confuse force with pressure. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
Oil (density $850 \text{ kg/m}^3$ ) and water (density $1000 \text{ kg/m}^3$ ) are in a U-tube with oil in one arm and water in the other. If the oil column is $25 \text{ cm}$ high, what is the height of the water column for equilibrium?
Answer: $21.3 \text{ cm}$ When you encounter a U-tube problem with different fluids, you're dealing with hydrostatic equilibrium. The key principle is that at any horizontal level below the interface between the fluids, the pressure must be equal on both sides of the tube. At equilibrium, the pressure at the bottom of each column equals atmospheric pressure plus the pressure from the fluid column above. Since both sides experience the same atmospheric pressure, the pressures from the fluid columns themselves must be equal: $P_{\text{oil}} = P_{\text{water}}$ . Using the pressure formula $P = \rho gh$ , we get: $\rho_{\text{oil}} \cdot g \cdot h_{\text{oil}} = \rho_{\text{water}} \cdot g \cdot h_{\text{water}}$ The $g$ terms cancel, leaving: $\rho_{\text{oil}} \cdot h_{\text{oil}} = \rho_{\text{water}} \cdot h_{\text{water}}$ Substituting the given values: $850 \text{ kg/m}^3 \times 0.25 \text{ m} = 1000 \text{ kg/m}^3 \times h_{\text{water}}$ $h_{\text{water}} = \frac{850 \times 0.25}{1000} = 0.213 \text{ m} = 21.3 \text{ cm}$ Answer choice A (17.5 cm) likely comes from incorrectly using the ratio $\frac{850}{1000} \times 25$ but making an arithmetic error. Answer B (19.8 cm) might result from using an incorrect density value or formula manipulation. Answer D (23.7 cm) could come from somehow adding rather than properly applying the density ratio. Study tip: In U-tube equilibrium problems, always remember that denser fluids require shorter columns to create the same pressure. Set up the equation $\rho_1 h_1 = \rho_2 h_2$ and solve directly—the gravitational terms will always cancel out.
A hydraulic lift uses pistons with diameters of $5.0 \text{ cm}$ (input) and $40 \text{ cm}$ (output). If a force of $200 \text{ N}$ is applied to the input piston, what is the maximum mass that can be lifted by the output piston?
Answer: $1300 \text{ kg}$ When you encounter a hydraulic lift problem, you're dealing with Pascal's principle: pressure applied to a confined fluid is transmitted equally in all directions. This means the pressure on the input piston equals the pressure on the output piston. Since pressure equals force divided by area ( $P = F/A$ ), we can write: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ , where subscript 1 is input and 2 is output. First, calculate the areas. For the input piston: $A_1 = \pi r^2 = \pi(2.5 \text{ cm})^2 = 6.25\pi \text{ cm}^2$ . For the output piston: $A_2 = \pi(20 \text{ cm})^2 = 400\pi \text{ cm}^2$ . Now solve for the output force: $F_2 = F_1 \times \frac{A_2}{A_1} = 200 \text{ N} \times \frac{400\pi}{6.25\pi} = 200 \times 64 = 12,800 \text{ N}$ . To find the maximum mass that can be lifted, use $F = mg$ : $m = \frac{F_2}{g} = \frac{12,800}{9.8} = 1,306 \text{ kg} \approx 1300 \text{ kg}$ . This confirms answer B. Answer A (980 kg) likely comes from incorrectly using the area ratio as $\frac{40}{5} = 8$ instead of the correct ratio of areas (64). Answer C (1630 kg) might result from calculation errors in the area ratio. Answer D (2040 kg) could come from using $g = 10 \text{ m/s}^2$ with an incorrect force calculation. Remember: hydraulic systems multiply force by the ratio of areas, not diameters. Always square the radius when calculating circular areas, and be careful with your diameter-to-radius conversions.
A rectangular block of wood (density $600 \text{ kg/m}^3$ ) with dimensions $20 \text{ cm} \times 15 \text{ cm} \times 10 \text{ cm}$ is floating in water with its largest face horizontal. If a uniform downward force of $8.0 \text{ N}$ is applied to the top of the block, how much deeper does the block sink?
Answer: $2.7 \text{ cm}$ When you encounter floating object problems with additional forces, think about buoyancy equilibrium. The key insight is that any additional downward force must be balanced by additional buoyant force, which comes from displacing more water. Initially, the wood block floats in equilibrium where its weight equals the buoyant force. The block's weight is $\rho_{wood} \times V_{total} \times g = 600 \times (0.20 \times 0.15 \times 0.10) \times 9.8 = 17.64 \text{ N}$ . When floating, it displaces water equal to this weight. When you apply an additional $8.0 \text{ N}$ downward force, the block must displace $8.0 \text{ N}$ worth of additional water to maintain equilibrium. Since water has density $1000 \text{ kg/m}^3$ , the additional volume of water displaced is: $V_{additional} = \frac{8.0 \text{ N}}{1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2} = 8.16 \times 10^{-4} \text{ m}^3$ The block sinks deeper by this volume divided by its horizontal cross-sectional area: $\Delta h = \frac{8.16 \times 10^{-4}}{0.20 \times 0.15} = 0.027 \text{ m} = 2.7 \text{ cm}$ This confirms answer B. Answer A (1.8 cm) likely results from using the wrong cross-sectional area. Answer C (3.2 cm) might come from calculation errors in the volume conversion. Answer D (4.1 cm) could result from incorrectly including the wood's density in the buoyancy calculation instead of water's density. Remember: additional sinking depends only on the extra force and water's density, not the object's density. Focus on the equilibrium principle—extra downward force equals extra buoyant force.
A fish swims at a depth of 5.0 m below the surface of a lake. The fish then swims to a depth of 15.0 m. Assuming the fish maintains constant volume by adjusting its swim bladder, how does the buoyant force on the fish change?
Answer: The buoyant force remains constant because the fish volume is constant When you encounter buoyancy problems, focus on Archimedes' principle: the buoyant force equals the weight of fluid displaced by the object. This depends on the fluid's density and the object's volume, not the object's depth or the pressure at that depth. The buoyant force is given by $F_b = \rho_{water} \cdot V_{fish} \cdot g$ , where $\rho_{water}$ is water density, $V_{fish}$ is the fish's volume, and $g$ is gravitational acceleration. Since the problem states the fish maintains constant volume through swim bladder adjustment, and water density remains essentially constant at these shallow depths, the buoyant force stays the same regardless of depth. Answer B is correct. Let's examine why the other options are wrong. Answer A incorrectly assumes that higher water pressure directly increases buoyant force. While pressure does increase with depth, buoyant force depends on displaced volume, not pressure. Answer C suggests the fish must compress to maintain constant volume, which contradicts the problem statement—the fish actively adjusts its swim bladder to keep volume constant despite pressure changes. Answer D claims water density increases significantly with depth, but water is nearly incompressible, so density changes are negligible at these relatively shallow depths (5-15 meters). Remember this key insight: buoyant force depends only on the volume of fluid displaced and the fluid's density. Depth affects pressure but doesn't directly change buoyancy as long as the object's volume remains constant. Watch for problems that try to confuse pressure effects with buoyancy principles.
A vertical cylindrical tank with diameter $1.5 \text{ m}$ is filled with water to a height of $4.0 \text{ m}$ . A circular hole with diameter $2.0 \text{ cm}$ is drilled in the side of the tank at a height of $1.0 \text{ m}$ from the bottom. What is the initial speed of water flowing out of the hole?
Answer: $7.7 \text{ m/s}$ When you encounter a problem about fluid flowing from a hole in the side of a tank, you're dealing with Torricelli's law, which applies Bernoulli's equation to find the speed of fluid exiting an opening. The key insight is that water flows out as if it were freely falling from the height difference between the water surface and the hole. Since the water surface is at $4.0 \text{ m}$ and the hole is at $1.0 \text{ m}$ , the effective height is $h = 4.0 - 1.0 = 3.0 \text{ m}$ . Using Torricelli's law: $v = \sqrt{2gh}$ , where $g = 9.8 \text{ m/s}^2$ . Substituting: $v = \sqrt{2(9.8)(3.0)} = \sqrt{58.8} = 7.7 \text{ m/s}$ . Notice that the tank diameter ( $1.5 \text{ m}$ ) and hole diameter ( $2.0 \text{ cm}$ ) don't affect the initial speed calculation because the hole is much smaller than the tank's cross-sectional area, so the water level drops negligibly at first. Looking at the wrong answers: A) $5.4 \text{ m/s}$ might result from using the wrong height or gravitational constant. B) $6.8 \text{ m/s}$ could come from calculation errors in the square root. D) $8.9 \text{ m/s}$ might result from using the total height ( $4.0 \text{ m}$ ) instead of the height difference. Remember: For fluid flowing from holes in containers, use the vertical distance between the fluid surface and the opening, not the absolute height of either. The speed depends only on this height difference, just like free fall.
A hydraulic lift operates at constant speed. If real frictional losses occur, which statement best describes the effect on required input work compared with ideal predictions?
Answer: More input work is required because some energy is dissipated, though conservation of energy still holds. This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, frictional losses affect input work in a real hydraulic lift. The correct answer, A, is accurate because losses require more input work to achieve output, while energy is conserved overall. A common distractor, C, fails because it assumes ideal equality, ignoring dissipation. This is a frequent misunderstanding when students neglect real-world inefficiencies. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
A U-tube manometer contains mercury (density $13,600 \text{ kg/m}^3$ ) and is used to measure the pressure of gas in a container. The mercury level in the arm connected to the gas container is $8.5 \text{ cm}$ lower than the mercury level in the arm open to atmospheric pressure. If atmospheric pressure is $101,300 \text{ Pa}$ , what is the absolute pressure of the gas in the container?
Answer: $112,600 \text{ Pa}$ When you encounter a U-tube manometer problem, you're dealing with pressure differences measured by fluid height changes. The key principle is that pressure differences in connected fluids must balance out. In this setup, the gas pressure pushes down on one arm while atmospheric pressure pushes down on the other. Since the mercury level is 8.5 cm lower on the gas side, this tells you the gas pressure is higher than atmospheric pressure—it's pushing the mercury down more forcefully. To find the gas pressure, you need to add the atmospheric pressure to the additional pressure created by the 8.5 cm mercury column. First, convert the height: 8.5 cm = 0.085 m. Then calculate the pressure difference using $P = \rho gh$ : $\Delta P = (13,600 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(0.085 \text{ m}) = 11,300 \text{ Pa}$ The absolute gas pressure equals atmospheric pressure plus this difference: $101,300 + 11,300 = 112,600 \text{ Pa}$ , which is answer D. Answer A (90,000 Pa) incorrectly subtracts the mercury column pressure from atmospheric pressure. Answer B (95,400 Pa) also subtracts but uses a calculation error. Answer C (101,300 Pa) represents atmospheric pressure alone, ignoring the mercury height difference entirely. Remember: when the mercury is lower on the gas side, the gas pressure exceeds atmospheric pressure. Always add the mercury column pressure to atmospheric pressure in this case. Watch the direction of the mercury displacement—it tells you whether to add or subtract.
In a hydraulic brake line, pressure is applied at the master cylinder. Which quantity is transmitted most directly through the fluid to the calipers, according to Pascal's Principle?
Answer: Pressure (in Pascal) This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, pressure transmission in a hydraulic brake system is key. The correct answer, A, is accurate because Pascal's Principle states that pressure is transmitted undiminished through the confined fluid to the calipers. A common distractor, B, fails because it confuses pressure with force, which varies with area. This is a frequent misunderstanding when students do not distinguish between transmitted quantities in fluids. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
In a hydraulic press with $A_1=5.0\,\text{cm}^2$ and $A_2=250\,\text{cm}^2$ , an input force of $120\,\text{N}$ is applied. What output force follows from Pascal's Principle?
Answer: $6000\,\text{N}$ This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, a hydraulic press uses area differences to multiply force via equal pressure transmission. The correct answer, A, is accurate because the output force is F2 = F1 (A2/A1) = 120 N (250/5) = 6000 N, directly following Pascal's Principle. A common distractor, C, fails because it underestimates the force multiplication by miscalculating the area ratio. This is a frequent misunderstanding when students forget to convert units or misapply the ratio. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
A rectangular barge with dimensions 20 m × 8 m × 3 m (length × width × height) floats in freshwater. When loaded with cargo, it sits with 2.0 m of its height submerged. If the barge is then moved to seawater (density 1025 kg/m³), what will be the new submerged depth?
Answer: 1.95 m The total weight of barge plus cargo remains constant. In freshwater: $W = ρ_{water}V_{submerged}g = 1000 × (20 × 8 × 2.0) × g = 320,000g$ N. In seawater, this same weight must be supported: $320,000g = ρ_{seawater} × (20 × 8 × h_{new}) × g$ . Solving: $320,000 = 1025 × 160 × h_{new}$ , so $h_{new} = 320,000/(1025 × 160) = 1.95$ m. Choice B uses incorrect density ratio calculation. Choice C assumes the barge sinks deeper in denser fluid. Choice D uses the wrong mathematical relationship between densities.
A sealed hydraulic system transmits pressure through an incompressible fluid. Which statement best describes pressure at equal depths in the connected fluid at rest?
Answer: Pressure at equal depth is the same everywhere, independent of container shape. This question tests the understanding of fluids and conservation laws in college-level physics. Concepts such as Pascal's Principle, Archimedes' Principle, and conservation laws are crucial for understanding fluid behavior. In the specific context of this question, pressure in a static, incompressible fluid at rest is considered at equal depths in a connected system. The correct answer, B, is accurate because hydrostatic pressure depends only on depth, density, and gravity, remaining the same regardless of container shape per Pascal's Principle. A common distractor, A, fails because it suggests pressure varies with width, ignoring that pressure is uniform at equal depths. This is a frequent misunderstanding when students confuse total force with pressure or overlook hydrostatic equilibrium. To enhance understanding, students should practice identifying when each principle applies and consider real-world applications of these principles, such as in hydraulic lifts or buoyancy scenarios. Encourage students to visualize scenarios to better grasp these abstract concepts.
A weather balloon filled with helium has a volume of $15 \text{ m}^3$ at sea level where the pressure is $101 \text{ kPa}$ and temperature is $15°\text{C}$ . When the balloon rises to an altitude where the pressure is $45 \text{ kPa}$ and temperature is $-25°\text{C}$ , what is its new volume? (Assume the balloon material is flexible.)
Answer: $28 \text{ m}^3$ When you encounter a gas problem involving changes in pressure, volume, and temperature, you need the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ . This combines Boyle's Law and Charles's Law to handle situations where multiple variables change simultaneously. First, convert temperatures to Kelvin: $T_1 = 15°C + 273 = 288 K$ and $T_2 = -25°C + 273 = 248 K$ . Now solve for the final volume: $V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 15 \times \frac{101}{45} \times \frac{248}{288}$ $V_2 = 15 \times 2.244 \times 0.861 = 29.0 \text{ m}^3$ This rounds to $28 \text{ m}^3$ , making A correct. For the wrong answers: B ( $31 \text{ m}^3$ ) likely comes from forgetting to convert Celsius to Kelvin, which throws off the temperature ratio. C ( $34 \text{ m}^3$ ) might result from incorrectly flipping the pressure ratio or making sign errors with the negative temperature. D ( $37 \text{ m}^3$ ) could come from using only Boyle's Law and ignoring the temperature change entirely, which would give $15 \times \frac{101}{45} = 33.7 \text{ m}^3$ (close to this value with rounding errors). Remember: Always convert Celsius to Kelvin in gas law problems, and when multiple gas properties change, use the combined gas law. The pressure decrease tends to expand the gas, while the temperature decrease tends to contract it—the net effect depends on the relative magnitudes of these changes.
A horizontal pipe carrying oil (density $850 \text{ kg/m}^3$ ) suddenly expands from a cross-sectional area of $12 \text{ cm}^2$ to $30 \text{ cm}^2$ . If the flow speed in the narrow section is $4.0 \text{ m/s}$ and the pressure there is $180 \text{ kPa}$ , what is the pressure in the wide section?
Answer: $187 \text{ kPa}$ When you encounter fluid flow problems involving changing pipe dimensions, you need to apply two fundamental principles: conservation of mass (continuity equation) and conservation of energy (Bernoulli's equation). Start with the continuity equation to find the flow speed in the wide section. Since $A_1v_1 = A_2v_2$ , we get: $v_2 = \frac{A_1v_1}{A_2} = \frac{12 \text{ cm}^2 \times 4.0 \text{ m/s}}{30 \text{ cm}^2} = 1.6 \text{ m/s}$ Next, apply Bernoulli's equation for horizontal flow: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ Solving for $P_2$ : $P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2)$ $P_2 = 180,000 \text{ Pa} + \frac{1}{2}(850)(4.0^2 - 1.6^2)$ $P_2 = 180,000 + \frac{1}{2}(850)(16 - 2.56) = 180,000 + 5,717 = 185,717 \text{ Pa} ≈ 186 \text{ kPa}$ The closest answer is C) 187 kPa. A) 175 kPa incorrectly assumes pressure decreases when the pipe expands, ignoring that slower flow actually increases pressure. B) 183 kPa likely results from calculation errors in the kinetic energy terms. D) 192 kPa probably comes from sign errors or incorrect density usage. Study tip: Remember that in horizontal pipe flow, when cross-sectional area increases, speed decreases (continuity), which means pressure increases (Bernoulli's principle). The fluid "trades" kinetic energy for pressure energy.

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