A hydraulic press contains 0.25#"m"^3# of oil of bulk modulus 50000000pa the change in volume of oil when subjected to increase pressure of 60000000pa is?

1 Answer
Feb 21, 2018

The volume decreases by approximately #0.04 \ "m"^3#.

Explanation:

Considering that there is no change in pressure or the number of moles, we can use Boyle's law here, which states that

#Pprop1/V#

or

#P_1V_1=P_2V_2#

Since we need to solve for the change in volume, we might as well find the new volume first.

#V_2=(color(red)(P_1)color(blue)(V_1))/color(green)(P_2)#

Plugging in the values:

#V_2=(color(red)(50,000,000 \ color(black)cancelcolor(red)"Pa")*color(blue)(0.25 \ "m"^3))/color(green)(60,000,000 \ color(black)cancelcolor(green)"Pa")~~0.21 \ "m"^3#

Now, we need to find the change in the volume, that is

#DeltaV=V_2-V_1#

We got:

#DeltaV=0.21 \ "m"^3-0.25 \ "m"^3#

#DeltaV=-0.04 \ "m"^3#

That means, that the volume has decreased by #0.04 \ "m"^3#.