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.

Compute the bulk modulus of water from the following data: Initial vol

1.	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.
Answer» <P> Solution :`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.