Question

# The pressure at certain depth in a sea is 80 atm. If the density of water at the surface of the sea is $$1.03 \times 10^3 kg/m^3$$ and the compressibility of water is $$45.8 \times 10^{-11}\,Pa^{-1}$$, calculate the density of water at that depth.$$1\,atm=1.013 \times 10^5\,Pa$$

Solution

## $$\Delta P=80atm=80\times 1.013\times { 10 }^{ 5 }Pa$$Let $$m$$ be the mass and $$\rho$$ be the density of water at depth.Volume at the surface of water $$V=\cfrac { m }{ \rho }$$And at depth $${ V }^{ \prime }=\cfrac { m }{ \rho ^{ \prime } }$$      $$\Delta V=V-{ V }^{ \prime }=m\left( \cfrac { 1 }{ \rho } -\cfrac { 1 }{ \rho ^{ \prime } } \right)$$Bulk modulus$$\left( K \right) =\cfrac { \Delta P }{ \cfrac { \Delta V }{ V } }$$Compressibility factor $$=\cfrac { 1 }{ K } =\cfrac { \Delta V }{ \Delta PV }$$        $$\cfrac { 1 }{ K } =m\left( \cfrac { 1 }{ \rho } -\cfrac { 1 }{ \rho ^{ \prime } } \right) \times \cfrac { \rho }{ m } \times \cfrac { 1 }{ \Delta P } =1-\left( \cfrac { \rho }{ \rho ^{ \prime } } \right) \times \cfrac { 1 }{ \Delta P } \\ P=\cfrac { 1.03\times { 10 }^{ 3 } }{ 1-0.0037 } \\ P^{ \prime }=1.034\times { 10 }^{ 3 }kg/{ m }^{ 3 }$$Physics

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