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#ethyl alcohol

4 public questions tagged with this topic.

The pressure at a depth of 3m in ethyl alcohol (ρ\=806kg/m3) is measured with atmospheric pressure as 1.01×105Pa. What i

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 806kg/m3, g = 10m/s2, h = 3m. P = 1.01×105+806×10×3 = 1.01×105+24180 = 1.2518×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.25 × 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.

What is the gauge pressure at a depth of 1.9m in ethyl alcohol (ρ\=806kg/m3)? (Take g\=10m/s2)

Gauge pressure: Pg = ρgh. ρ = 806kg/m3, g = 10m/s2, h = 1.9m. Pg = 806×10×1.9 = 15314Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.5 × 10⁴ 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.

A manometer with ethyl alcohol (ρ\=806kg/m3) shows a height difference of 0.3m. What is the pressure difference? (Take g

ΔP = ρgh. ρ = 806kg/m3, g = 10m/s2, h = 0.3m. ΔP = 806×10×0.3 = 2418Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2418 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.

What is the gauge pressure at a depth of 4.5m in ethyl alcohol (ρ\=806kg/m3)? (Take g\=10m/s2)

Gauge pressure: Pg = ρgh. ρ = 806kg/m3, g = 10m/s2, h = 4.5m. Pg = 806×10×4.5 = 36270Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.6 × 10⁴ 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.