9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

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9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

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Vishal Baghel | Contributor-Level 10

5 months ago

Let us assume the depth = h, pressure at depth, $P_{h}$ = 80 atm = 80 $\times 1.013 \times 10^{5}$ Pa

Density of water at the surface, $ρ_{s}$ = 1.03 $\times 10^{3}$ kg/ $m^{3}$

Let density of water at depth h be $ρ_{h}$

Let $V_{s}$ be the volume at the surface and $V_{h}$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_{s} - V_{h}$ From the relation m = $ρ V,$ we get

ΔV = m ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ ) = $ρ_{s} V_{s}$ ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ )

$\frac{Δ V}{V_{s}}$ = 1- $\frac{ρ_{s}}{ρ_{h}}$ ……(i)

Bulk modulus of water, k = $V_{1} \frac{Δ P}{Δ V}$ = $V_{s} \frac{Δ P}{Δ V}$

$\frac{Δ V}{V_{s}}$ = $\frac{Δ P}{k}$ …….(ii)

Bulk modulus of water, k = 2.

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Let us assume the depth = h, pressure at depth, $P_{h}$ = 80 atm = 80 $\times 1.013 \times 10^{5}$ Pa

Density of water at the surface, $ρ_{s}$ = 1.03 $\times 10^{3}$ kg/ $m^{3}$

Let density of water at depth h be $ρ_{h}$

Let $V_{s}$ be the volume at the surface and $V_{h}$ be the volume at depth h and ΔV be the change in volume. Let m be the mass of water.

ΔV = $V_{s} - V_{h}$ From the relation m = $ρ V,$ we get

ΔV = m ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ ) = $ρ_{s} V_{s}$ ( $\frac{1}{ρ_{s}}$ - $\frac{1}{ρ_{h}}$ )

$\frac{Δ V}{V_{s}}$ = 1- $\frac{ρ_{s}}{ρ_{h}}$ ……(i)

Bulk modulus of water, k = $V_{1} \frac{Δ P}{Δ V}$ = $V_{s} \frac{Δ P}{Δ V}$

$\frac{Δ V}{V_{s}}$ = $\frac{Δ P}{k}$ …….(ii)

Bulk modulus of water, k = 2.1 $\times 10^{9}$ Pa

Hence $\frac{Δ V}{V_{s}} = \frac{80 \times 1.013 \times 10^{5}}{2.1 \times 10^{9}}$ = 3.86 $\times 10^{- 3}$

From equation (i) $\frac{ρ_{s}}{ρ_{h}}$ = 0.99614

$ρ_{h} = 1.03 \times 10^{3} / 0.99614$ = 1.034 $\times 10^{3}$ kg/ $m^{3}$

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9.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

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