Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm `(1 atm = 1. 013xx 10^(5) Pa),` Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Explanation

`Delta P = 100 atm = 100 xx 1.013 xx 10 ^(5) Pa`
`V _(1) = 100` liter `= 100 xx 10 ^(-3) m ^(-3) =0.1 m ^(3)`
`V _(2) = 100.5 ` liter ` = 100.5 xx 10 ^(-3) m ^(-3)`
``therefore Delta l = 1.87 xx 10 ^(-3) m = 1.87 mm` _(2) = 100.5 ` liter `= 100. 5 xx 10 ^(-3) m ^(-3)`
`Delta V =V_(2) - V _(1) = (100.5 xx 10 ^(-3) - 100 xx 10 ^(-3))= 0.5 xx 10 ^(-3) m ^(-3)`
Bulk modulus for water,
`B _(w) = ( P)/((Delta V )/(V))= (PV)/(Delta V )= (100 xx 1 . 0.13 xx 10 ^(5) xx 100 xx 10 ^(-3))/( 0.5 xx 10 ^(-3))`
`= 2.026 xx 10 ^(9) Pa`
Bulk modulus for air `B _(a) = 1.0 xx 10 ^(5) Pa `
`therefore` Ratio `= ("Bulk modulus for water" (B _(w)))/("BUlk modulus for air" (B _(a)))`
`= (2.026 xx 10^(9))/( 1. 0 xx 10 ^(5)) = 2.026 xx 10 ^(4)`
The ratio is large. Becaue gaes are more compressive than liquid and intermolecular distance in liquid are very small compared to gases.