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%.