Fluids solutions

Convert volume: V = 3 L = 0.003 m³. Convert temperature: T = 35 + 273.15 = 308.15 K.

Why: The SI gas constant needs cubic meters and an absolute kelvin temperature, so convert both inputs first.

Use absolute pressure in PV = nRT, with R = 8.31446 J/(mol·K).

Why: The gas law counts molecular pressure relative to vacuum, which is why the stated absolute pressure belongs here.

Rearrange: n = PV/(RT) = (30 Pa)(0.003 m³)/[(8.31446 J/(mol·K))(308.15 K)].

Why: Solve the gas law for n by dividing pressure times volume by RT.

The diagram shows a downward tether. V = (4/3)πr³ = (4/3)π(0.15 m)³ = 0.0141372 m³.

Why: The whole sphere is submerged, so its entire geometric volume displaces water; the lower tether can pull only downward.

Buoyancy is upward: F_B = ρgV = 138.686 N. Weight is downward: mg = 98.1 N.

Why: Archimedes’ force equals displaced-water weight and points up, opposite the sphere’s own weight.

Before cutting, equilibrium gives F_B − mg − T = 0, so T = 40.5856 N.

Why: At rest, the downward tether supplies the extra downward force needed to balance buoyancy greater than weight.

Immediately after cutting, with upward positive, ma = F_B − mg, so a = 40.5856/10 = 4.05856 m/s² upward.

Why: Cutting the tether removes T but leaves buoyancy and weight; their upward difference accelerates the sphere in the stated no-drag model.

At constant pressure and fixed amount, V₁/T₁ = V₂/T₂, with temperature in kelvins.

Why: Fixed gas amount and pressure make volume proportional to absolute temperature, not the Celsius reading.

T₁ = 288.15 K. Therefore T₂ = 288.15(240/150) = 461.04 K.

Why: Use the 240/150 volume ratio to scale the first kelvin temperature and find the second one.

Convert back: T₂ = 461.04 − 273.15 = 187.89 °C.

Why: Subtract the kelvin offset only after the gas-law calculation to report the second Celsius temperature.

The numerical data describe warming, not cooling. An equilibrium ice-water bath at ordinary pressure is near 0 °C, and boiling fluorine is a cryogenic bath; the named baths cannot describe these volume data.

Why: The larger volume requires a hotter bath at fixed pressure; that result contradicts the two named baths rather than fixing them silently.

The water buoyant force is 10.4 − 5.6 = 4.8 N = ρ_water gV.

Why: The drop in scale reading is the upward buoyant force, which gives the weight of displaced water.

The block weight is ρ_block gV, so ρ_block = ρ_water(10.4/4.8) = 2166.67 kg/m³.

Why: Full immersion gives the block and displaced water the same volume, so dividing their weights removes g and V.

Gasoline buoyancy is 10.4 − 6.5 = 3.9 N. For the same submerged volume, ρ_gasoline/ρ_water = 3.9/4.8.

Why: The gasoline scale drop is its buoyant force; equal submerged volumes make the buoyancy ratio a liquid-density ratio.

Thus ρ_gasoline = 812.5 kg/m³.

Why: Multiply that buoyancy ratio by known water density to obtain the unknown gasoline density.

Let the common cross-sectional area be A = πR². The mass is A[(1500)(0.05) + (400)(0.07)] = 103A kg/m².

Why: Add the masses of the two rod segments using their own densities and the shared cross-sectional area.

The rod displaces Ah of water if its bottom is depth h below the surface. Floating equilibrium gives 1000Ahg = 103Ag.

Why: The submerged volume is area times bottom depth; a floating rod displaces water whose weight equals the rod’s total weight.

Cancel A and g: h = 103/1000 = 0.103 m.

Why: The shared area and gravity cancel, leaving the density-weighted segment lengths to determine the submerged depth.

Since 0.05 m < h < 0.12 m, the lower section is fully submerged and 0.053 m of the upper section is submerged, consistent with the assumed orientation.

Why: Check the calculated depth against both segment boundaries so the assumed partial immersion is physically consistent.

The source image states 40 cm, so Δz = 0.40 m. Continuity gives v₂ = v₁(d₁/d₂)² = 0.09(8.0/3.4)² = 0.498270 m/s.

Why: The rise is 40 cm as printed; incompressible continuity makes speed grow by the inverse cross-sectional area ratio.

Bernoulli gives P₂ = P₁ + (ρ/2)(v₁² − v₂²) − ρgΔz.

Why: Along the lossless streamline, increased speed and height both take energy from the pressure term.

Convert P₁ = 4.6(101325) = 466095 Pa. The speed term is −120.086 Pa and the elevation term is −3924 Pa.

Why: Convert atmospheres to pascals before combining pressure with the kinetic and elevation terms.

Thus P₂ = 462050.914 Pa = 4.560 atm. If the intended rise were 40 m, the alternative pressure would be 73574.914 Pa = 0.7261 atm.

Why: Add the two pressure changes for the stated 0.40 m rise; the 40 m alternative illustrates why that source dimension matters.

Convert temperature: T = 243.15 K. The ideal gas gives P = nRT/V = 34.9164 atm.

Why: Absolute kelvin temperature with the chosen liter-atmosphere R gives the ideal comparison pressure directly.

For the van der Waals model, rearrange [P + a(n/V)²](V − nb) = nRT to P = nRT/(V − nb) − a(n/V)².

Why: Isolate P in the real-gas equation to separate excluded-volume and attraction corrections.

V − nb = 0.4000 − (0.7000)(0.024) = 0.3832 L, which is positive.

Why: Check the volume remaining after nb is positive before using it as a denominator.

P = (0.7000)(0.0820574)(243.15)/0.3832 − (0.034)(0.7000/0.4000)² = 36.3431 atm.

Why: Use the reduced available volume for the kinetic term, then subtract the attraction term to obtain the corrected pressure.

Water supplied at a pressure above the surrounding air flows through the sprinkler. Pressure energy becomes kinetic energy in the jets, with some energy lost to friction.

Why: The supply-to-air pressure difference can accelerate water through the nozzles, while friction uses some of that energy.

For a given volume flow rate, a smaller nozzle area requires a larger outlet speed, Q = Av. The actual flow rate also depends on supply pressure and resistance.

Why: Conservation of flowing volume relates nozzle area to speed, but it does not fix Q without a supply-flow model.

A jet carries momentum away. The equal and opposite force on an angled nozzle produces torque about the sprinkler axis and makes the arms rotate.

Why: Turning the outgoing water changes its momentum; the reaction on an angled arm has a lever arm and therefore produces torque.

After leaving, the water follows projectile paths and breaks into droplets, distributing water across an area.

Why: Once outside, the water is no longer confined by the nozzle, so gravity shapes its path and breakup spreads the spray.

The two volume rates add: Q₁ = (5.6)(4.5)(24) = 604.8 m³/s and Q₂ = (8.9)(7.8)(15) = 1041.3 m³/s.

Why: For each rectangular cross-section, width times depth gives area, and multiplying by mean speed gives volume per second.

Continuity at the junction gives Q_river = Q₁ + Q₂ = 1646.1 m³/s.

Why: No storage or loss at the junction means the downstream volume rate must equal the sum of the two incoming rates.

Divide by downstream cross-sectional area: v_river = 1646.1/65 = 25.3246 m/s.

Why: Divide the conserved combined volume rate by the downstream area to find its mean speed.

The surface and outlet are both at atmospheric pressure. Neglecting surface speed, Bernoulli gives v = √(2gh) = √[(2)(9.81)(10)] = 14.0071 m/s.

Why: The equal atmospheric pressures cancel, and the large tank makes surface speed negligible, so the 10 m head sets the initial jet speed.

Opening area: A = π(0.9/2)² = 0.636173 m². The initial ideal rate is Q = Av = 8.91096 m³/s.

Why: Use the opening’s radius to find its area, then multiply by outlet speed for the initial volume rate.

Neglect air drag. The horizontal speed stays constant after departure, so x = vt = (14.0071)(3) = 42.0214 m.

Why: After launch, no horizontal force changes the jet’s horizontal speed in the no-drag model.

The specified flight would include a vertical drop (1/2)gt² = 44.145 m. No landing height is supplied; the 3 s flight is a given condition.

Why: A horizontal launch still falls under gravity; checking that drop keeps the given flight time from implying an unstated landing height.

V = 0.00180 m³ and T = 295.15 K.

Why: Convert liters to cubic meters and Celsius to kelvins to match the SI gas constant.

n = PV/(RT) = (182000)(0.00180)/[(8.31446)(295.15)] = 0.133496 mol.

Why: Use the stated absolute pressure in PV = nRT and solve for the number of moles.

At fixed n and T, P₁V₁ = P₂V₂.

Why: Keeping both temperature and gas amount fixed makes the ideal-gas pressure-volume product constant.

P₂ = (125 kPa)(2.40/0.850) = 352.941 kPa.

Why: Divide the original PV product by the smaller final volume, so compression raises the absolute pressure.

Gas laws use absolute pressure: P₁ = 220 + 101.3 = 321.3 kPa. Temperatures are 288.15 K and 328.15 K.

Why: Gauge pressure excludes atmospheric pressure, so add it back before using a gas law and convert both temperatures to kelvins.

At constant V and n, P₂ = P₁T₂/T₁ = 365.902 kPa absolute.

Why: With fixed tire volume and no gas loss, absolute pressure scales by the absolute-temperature ratio.

Subtract atmospheric pressure: P₂,gauge = 365.902 − 101.3 = 264.602 kPa.

Why: The requested tire-gauge reading is the final absolute pressure minus the unchanged atmospheric pressure.

T₁ = 291.15 K and T₂ = 251.15 K.

Why: The gas-law temperature ratio must use absolute temperatures even though both given readings are Celsius.

For fixed n, V₂ = V₁(P₁/P₂)(T₂/T₁).

Why: No leakage fixes n; the flexible equilibrium balloon changes volume with both pressure and temperature.

V₂ = 3.60(92.0/68.0)(251.15/291.15) = 4.20144 L.

Why: Apply both ratios together: the pressure reduction more than offsets cooling, giving a larger balloon volume.

V = 0.0280 m³, T = 315.15 K, and M = 0.0280 kg/mol.

Why: Convert volume, temperature and molar mass to SI units so the resulting mass comes out in kilograms.

n = PV/(RT); m = nM = PVM/(RT) = 0.0433843 kg.

Why: The gas law gives moles, and multiplying by nitrogen’s molar mass turns that amount into mass.

Convert to grams: m = 43.3843 g.

Why: The question asks for grams, so convert the calculated kilogram mass at the end.

Combine PV = nRT with n = m/M: ρ = m/V = PM/(RT).

Why: Replacing n with m/M and dividing by V isolates density without needing a particular sample volume.

T = 281.15 K, M = 0.0440 kg/mol.

Why: Use absolute temperature and kilograms per mole so the density expression returns kg/m³.

ρ = (86000)(0.0440)/[(8.31446)(281.15)] = 1.61875 kg/m³.

Why: Insert the pressure and molecular mass into PM/(RT) to obtain carbon dioxide’s ideal-gas density.

Ideal: P = nRT/V = 10.1751 atm.

Why: The ideal calculation supplies a baseline that omits both finite molecular size and attractions.

Excluded volume: V − nb = 1.47438 L, giving nRT/(V − nb) = 10.3519 atm.

Why: Finite molecular size reduces available volume, raising the pressure term when the denominator becomes smaller.

Attraction correction: a(n/V)² = 0.5744 atm. Thus P_vdW = 10.3519 − 0.5744 = 9.77752 atm.

Why: Molecular attraction reduces pressure, so subtract its positive correction from the excluded-volume term.

Excluded volume raises the kinetic pressure term; attraction reduces the final pressure.

Why: Compare each correction’s direction separately; the total here lies below the ideal pressure because attraction dominates.

Buoyancy is 31.5 − 24.0 = 7.50 N.

Why: The immersed scale reading is lower because buoyancy supports part of the sample’s weight.

V = F_B/(ρg) = 7.50/(1000 × 9.81) = 0.000764526 m³ = 764.526 cm³.

Why: The sample is fully immersed, so its volume equals the displaced-water volume inferred from buoyancy.

ρ_sample = ρ_water W_air/F_B = (1000)(31.5)/7.50 = 4200 kg/m³.

Why: Divide the sample’s weight by displaced-water weight; the common volume and g cancel, leaving the density ratio.

For floating equilibrium, ρ_liquid V_sub g = ρ_ice V_total g.

Why: Floating at rest means upward displaced-liquid weight balances the weight of the whole ice block.

The submerged fraction is ρ_ice/ρ_liquid.

Why: Cancel gravity and total volume to express the submerged fraction as ice density divided by liquid density.

Fresh water: 917/1000 = 0.917. Seawater: 917/1025 = 0.894634.

Why: Evaluate the ratio for each liquid; denser seawater supports the ice with a smaller submerged fraction.

At the limit of full submersion, maximum displaced-water mass = ρV = (1000)(0.180) = 180 kg.

Why: Full submersion uses the whole external volume, setting the largest mass of water the pontoon can displace.

The total supported mass equals 180 kg, including the pontoon.

Why: At the floating limit, displaced-water weight must support the combined pontoon and payload weight.

Payload = 180 − 42.0 = 138 kg.

Why: Subtract the pontoon’s own mass so the answer counts only the extra supported load.

V = 620 × 10⁻⁶ = 0.000620 m³.

Why: Convert cubic centimeters to cubic meters before multiplying by the densities in kg/m³.

Equilibrium gives T + ρ_water Vg − ρ_steel Vg = 0.

Why: The stationary tool has two upward forces, cable tension and buoyancy, balancing its downward weight.

T = (ρ_steel − ρ_water)Vg = (6800)(0.000620)(9.81) = 41.3590 N.

Why: Subtract water’s displaced-weight contribution from the steel’s weight to obtain the required upward cable tension.

Pascal’s principle gives equal added pressure: F_small/A_small = F_large/A_large.

Why: At equal height in the connected static fluid, the two pistons transmit the same added pressure.

The area ratio can use the common cm² unit.

Why: Both areas use cm², so their ratio is dimensionless without a separate conversion.

F_small = (14500 N)(25.0/850) = 426.471 N.

Why: Multiply the large load by the smaller-to-larger area ratio to find the smaller input force.

Each static liquid layer contributes ρgh. Atmospheric pressure cancels in the gauge value.

Why: Pressure gains accumulate through successive static liquid layers; gauge pressure subtracts the shared atmospheric surface pressure.

P_gauge = (850)(9.81)(0.350) + (1000)(9.81)(1.20) = 14690.5 Pa.

Why: Add the oil and water density-height contributions, using each layer’s own density.

At the interface level, the pressure from the alcohol arm equals the pressure from the water arm.

Why: The connected water is static, so pressure must agree at the same horizontal interface level in both arms.

P_atm + ρ_alcohol g h = P_atm + ρ_water g(0.0800 m).

Why: Both free surfaces share atmospheric pressure, leaving each column’s density times depth to determine the interface pressure.

h = (1000/790)(0.0800) = 0.101266 m.

Why: Cancel atmospheric pressure and g, then solve for the taller alcohol column needed with its lower density.

Q = Av = π(0.00600 m)²(1.80 m/s) = 0.000203575 m³/s.

Why: The radius sets the hose area, and area times mean speed is the steady delivered volume per second.

Required volume is 0.0750 m³.

Why: Convert the target barrel volume to cubic meters so it matches the calculated volume-rate units.

t = V/Q = 368.414 s = 6.14024 min.

Why: Divide required volume by steady flow rate to get the fill time, then convert seconds to minutes.

Continuity gives v_up = (2.00/5.00)²v_throat = 0.16v_throat.

Why: Equal incompressible volume rates make the wider upstream pipe’s speed smaller by the squared diameter ratio.

Bernoulli gives ΔP = (ρ/2)(v_throat² − v_up²) = (ρ/2)(1 − 0.16²)v_throat².

Why: The pipe is horizontal, so the pressure drop supplies the increase in kinetic energy per unit volume.

v_throat = √[2(5400)/(1000(1 − 0.0256))] = 3.32923 m/s.

Why: Substitute the continuity ratio and solve the pressure equation for the throat’s speed.

Q = π(0.0100 m)²v_throat = 0.00104588 m³/s.

Why: Multiply throat speed by its area to find the volume rate shared by both pipe sections.

Bernoulli: P₂ = P₁ + (ρ/2)(v₁² − v₂²) − ρgΔz.

Why: Without pumps or losses, Bernoulli balances pressure, kinetic and gravitational energy per unit volume between the two points.

The speed term is −3200 Pa and the height term is −63765 Pa.

Why: Both the higher speed and the increased height reduce the upper pressure, giving two negative corrections.

P₂ = 310000 − 3200 − 63765 = 243035 Pa.

Why: Subtract both corrections from the starting gauge pressure; the same atmospheric offset cancels between points.

Mass flow rate is ṁ = ρQ = (1000)(0.00340) = 3.40 kg/s.

Why: Multiply volume per second by water density to obtain the kilograms arriving at the wall each second.

Normal water momentum decreases at rate ṁv = (3.40)(17.0) = 57.8 N.

Why: The wall removes all incoming normal velocity, so the normal momentum change per second sets the force magnitude.

The wall exerts this force opposite the incoming flow on the water; the water exerts the equal opposite force on the wall.

Why: The water and wall forces form an action-reaction pair; the force on the wall points along the incoming jet.

Torricelli’s result gives v = √(2gh) = √(2 × 9.81 × 3.20) = 7.92364 m/s.

Why: Equal atmospheric pressures and negligible surface speed make the water depth set outlet speed through Bernoulli.

For the 1.25 m fall, t = √(2H/g).

Why: The jet starts horizontally, so its vertical fall from the opening determines the flight time.

Range x = vt = √(2gh)√(2H/g) = 2√(hH) = 4.00 m.

Why: Horizontal range is outlet speed times fall time; substituting both expressions cancels g for this ideal geometry.

h_f = (0.0260)(45.0/0.0600)(2.20²)/(2 × 9.81) = 4.81040 m.

Why: Use the supplied Darcy relation to express pipe friction as an extra equivalent height the pump must overcome.

Total pump head = 12.0 + 4.81040 = 16.8104 m; equal reservoir pressures and negligible surface speeds cancel.

Why: The pump supplies both the elevation rise and friction head; equal open-reservoir pressures and small surface speeds add no net terms.

Q = π(0.0300)²(2.20) = 0.00622035 m³/s.

Why: The pipe’s circular area and mean speed give the volume lifted each second.

Hydraulic power P = ρgQH = 1025.80 W. Actual input power would be higher if efficiency were below 100%.

Why: Multiply weight flow rate by total head for hydraulic power; an inefficient pump needs more input power than it gives the water.