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?

Question

Elena Albarran · Accepted Answer

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

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#.

Adam Arnold · Answer

undefined To find the change in volume of the oil when subjected to an increased pressure, we can use the formula for bulk modulus:

$$ \text{Bulk modulus (K)} = \dfrac{\Delta P}{\frac{\Delta V}{V}} $$

Where:
- $ \Delta P $ is the change in pressure.
- $ \Delta V $ is the change in volume.
- $ V $ is the initial volume.

Rearranging the formula to solve for $ \Delta V $, we get:

$$ \Delta V = \dfrac{V \cdot \Delta P}{K} $$

Given:
- $ V = 0.25 \, \text{m}^3 $
- $ \Delta P = 60000000 \, \text{Pa} $
- $ K = 50000000 \, \text{Pa} $

Substituting the values into the formula:

$$ \Delta V = \dfrac{0.25 \, \text{m}^3 \cdot 60000000 \, \text{Pa}}{50000000 \, \text{Pa}} $$

$$ \Delta V = 0.3 \, \text{m}^3 $$

Therefore, the change in volume of the oil when subjected to an increased pressure of 60000000 Pa is $ 0.3 \, \text{m}^3 $.