How large a pressure increase (in ATM) must be applied to water if it is to be compressed in volume by 1%? The bulk modulus of water is 2 �~ 10^9 N/m^2 and 1 ATM = 10^5 N/m^2
I think that it is
P = bulk modulus * 0.01
because 1/100 = 0.01, which is 1% of a volume.
B = ĢP/ [(ĢV/Vi)] < --- I found this equation.
Correct.
You are on the right track! The equation you have mentioned is the correct equation to determine the pressure increase required to compress water by a certain percentage. Let's break it down further.
The equation you mentioned is:
B = ΔP / (ΔV / Vi)
Where:
B = Bulk modulus of water
ΔP = Pressure increase
ΔV = Change in volume (in this case, it's 1%, so ΔV = 0.01Vi)
Vi = Initial volume
To find the pressure increase (ΔP), you can rearrange the equation as follows:
ΔP = B * (ΔV / Vi)
Substituting the given values into the equation, we have:
ΔP = (2 x 10^9 N/m^2) * (0.01 * Vi / Vi)
Simplifying the equation, the Vi cancels out:
ΔP = (2 x 10^9 N/m^2) * 0.01
ΔP = 2 x 10^7 N/m^2
Since 1 ATM is equal to 10^5 N/m^2, divide the pressure increase by 10^5 to convert it to ATM:
ΔP = (2 x 10^7 N/m^2) / (10^5 N/m^2/ATM)
ΔP = 200 ATM
So, a pressure increase of 200 ATM is required to compress water by 1%.
You are on the right track. To determine the pressure increase needed to compress water by 1%, you can use the equation B = ΔP/ [(ΔV/Vi)]. Let's break this equation down step by step:
1. B is the bulk modulus of water, which is given as 2 x 10^9 N/m^2.
2. ΔP represents the change in pressure. This is what we want to find.
3. ΔV/Vi represents the change in volume as a fraction of the initial volume. In this case, since we want to compress the water, ΔV/Vi would be equal to -0.01 (-1% change).
Now, let's rearrange the equation to solve for ΔP:
B = ΔP/ [(ΔV/Vi)]
Multiply both sides of the equation by [(ΔV/Vi)]:
B * [(ΔV/Vi)] = ΔP
Substitute the given values:
2 x 10^9 N/m^2 * (-0.01) = ΔP
Simplifying the equation:
-2 x 10^7 N/m^2 = ΔP
Finally, converting the pressure units from N/m^2 to ATM:
ΔP = (-2 x 10^7 N/m^2) * (1 ATM/10^5 N/m^2)
ΔP = -200 ATM
Therefore, a pressure increase of 200 ATM must be applied to water in order to compress its volume by 1%.