Question

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107 N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.

Answer 1

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Step-by-step explanation:

The bulk modulus is represented by the following differential equation:

`K = - V\cdot (dP)/(dV)`

Where:

`K` - Bulk module, measured in pascals.

`V` - Sample volume, measured in cubic meters.

`P` - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

`-(K \,dV)/(V) = dP`

This resultant expression is solved by definite integration and algebraic handling:

`-K\int\limits^{V_(f)}_{V_(o)} {(dV)/(V) } = \int\limits^{P_(f)}_{P_(o)}\, dP`

`-K\cdot \ln \left |(V_(f))/(V_(o)) \right| = P_(f) - P_(o)`

`\ln \left| (V_(f))/(V_(o)) \right| = (P_(o)-P_(f))/(K)`

`(V_(f))/(V_(o)) = e^{(P_(o)-P_(f))/(K) }`

The final volume is predicted by:

`V_(f) = V_(o)\cdot e^{(P_(o)-P_(f))/(K) }`

If
`V_(o) = 1\,m^(3)`,
`P_(o) - P_(f) = -10132500\,Pa` and
`K = 2.3* 10^(9)\,Pa`, then:

`V_(f) = (1\,m^(3)) \cdot e^{(-10.1325* 10^(6)\,Pa)/(2.3 * 10^(9)\,Pa) }`

`V_(f) \approx 0.996\,m^(3)`

Change in volume due to increasure on pressure is:

`\Delta V = V_(o) - V_(f)`

`\Delta V = 1\,m^(3) - 0.996\,m^(3)`

`\Delta V = 0.004\,m^(3)`

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.