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 × 109 N/m2 and 1 ATM = 105 N/m2
Pressure-bulkmodulus*.001
why 0.001? 1/100 is 0.01. I thought that was 1 %. Also is that pressure minus (-) or pressure (equals)?
To find the pressure increase required to compress water by 1%, we can use the equation:
ΔP = -V/V * B
Where:
ΔP is the pressure increase,
V is the initial volume of water,
ΔV is the change in volume (given as 1% of the initial volume), and
B is the bulk modulus of water.
The bulk modulus (B) measures the resistance of a material to change in volume under an applied pressure.
Given:
ΔV = -0.01V (since it is given that the volume decreases by 1%)
B = 2 × 10^9 N/m²
1 ATM = 10^5 N/m²
Now, let's substitute the known values into the equation:
ΔP = (-0.01V / V) * (2 × 10^9 N/m²)
Simplifying the equation gives:
ΔP = (-0.01) * (2 × 10^9 N/m²)
ΔP = -2 × 10^7 N/m²
Finally, let's convert the pressure change from N/m² to ATM using the conversion given:
1 ATM = 10^5 N/m²
ΔP = (-2 × 10^7 N/m²) / (10^5 N/m²)
ΔP = -200 ATM
Since pressure cannot be negative, we take the magnitude of the pressure change:
ΔP = | -200 | = 200 ATM
Therefore, a pressure increase of 200 ATM must be applied to compress water by 1% of its initial volume.