Question

At 20 oC the densities of fresh water and ethyl alcohol are, respectively, 998 and 789 kg/m3. Find the ratio of the adiabatic bulk modulus of fresh water to the adiabatic bulk modulus of ethyl alcohol at 20 oC.

Answer 1

The ratio is
`(B_1)/(B_2) = 1.265`

Step-by-step explanation:

From the question we are told that

The density of fresh water is
`\rho__(f)} = 998 \ kg/m^3`

The density of ethanol is
`\rho_(e) = 789 \ kg /m^3`

Generally speed of a wave in a substance is mathematically represented as

`v = \sqrt{(B)/(\rho) }`

Here B is the adiabatic bulk modulus of the substance while
`\rho` is the density of the substance

So at constant wave speed

`\sqrt{(B_1)/(\rho_1) } = \sqrt{(B_2)/(\rho_2) }`

=>
`(B_1)/(\rho_1) = (B_2)/(\rho_2)`

=>
`B_1 \rho_2 = B_2\rho_1`

=>
`(B_1)/(B_2) = (\rho_1)/(\rho_2)`

Here
`\rho_1 =\rho__(f)} = 998 \ kg/m^3` and
`\rho_2 = \rho_(e) = 789 \ kg /m^3`

So

=>
`(B_1)/(B_2) = (998)/(789)`

=>
`(B_1)/(B_2) = 1.265`