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#pressure calculation

34 public questions tagged with this topic.

Ideal Gas Laws and Thermal Properties

A gas at 1atm and 300K occupies 2L. If it is compressed to 0.5L while temperature rises to 450K, what is the final press

Given: P1 = 1atm, V1 = 2L, T1 = 300K, V2 = 0.5L, T2 = 450K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×20.5×450300 = 1×4×1.5 = 6atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 atm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Fluid Pressure, Pascal's Law and Manometers

Water flows at 2.9m/s at 3.2m height with pressure 1.9×105Pa. What is the pressure at 1.8m height with speed 4.0m/s? (ρ\

Bernoulli’s: P1+12ρv12+ρgh1 = P2+12ρv22+ρgh2. Left: 1.9×105+12×1000×8.41+1000×10×3.2 = 1.9×105+4205+32000 = 2.26205×105. Right: P2+12×1000×16+1000×10×1.8 = P2+8000+18000 = P2+26000. 2.26205×105 = P2+26000, P2 = 2.00205×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.00 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Universal Law of Gravitation

A 60kg person stands on two bones, each with area 25cm2. What is the average pressure on the bones? (Take g\=10m/s2)

Pav = FA, F = mg = 60×10 = 600N. Total area A = 2×25×10−4 = 5.0×10−3m2. Pav = 6005.0×10−3 = 1.2×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.2 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Surface Tension, Capillarity and Excess Pressure

A bubble of radius 4mm is formed at 50cm depth in a soap solution (ρ\=1.2×103kg/m3, S\=0.025N/m). What is the total pres

Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1.2×103×9.8×0.5 = 5880Pa. 2Sr = 2×0.0254×10−3 = 12.5Pa. Pi = 1.01×105+5880+12.5 = 1.06892×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.069 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

What is the total pressure at a depth of 3.2m in water (ρ\=1000kg/m3) with atmospheric pressure 1.01×105Pa? (Take g\=9.8

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1000kg/m3, g = 9.8m/s2, h = 3.2m. P = 1.01×105+1000×9.8×3.2 = 1.01×105+31360 = 1.3236×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.32 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

A hydraulic lift raises a 2100kg load using a small piston of radius 3.0cm and a large piston of radius 12cm. What force

F1 = A1A2F2, F2 = 2100×9.8 = 20580N. A1 = π(0.03)2, A2 = π(0.12)2, A1A2 = 0.00090.0144 = 116. F1 = 2058016 = 1286.25N≈1286N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1286 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Universal Law of Gravitation

A 45kg person stands on two bones, each with area 18cm2. What is the average pressure on the bones? (Take g\=9.8m/s2)

Pav = FA, F = mg = 45×9.8 = 441N. Total area A = 2×18×10−4 = 3.6×10−3m2. Pav = 4413.6×10−3 = 1.225×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.225 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

What is the excess pressure inside a water drop of radius 2.5mm at 100∘C? (Surface tension = 0.3×10−3N/m)

Excess pressure: ΔP = 2Sr. S = 0.3×10−3N/m, r = 2.5×10−3m. ΔP = 2×0.3×10−32.5×10−3 = 0.24Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.24 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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