The bulk modulus of a liquid is 4.0 10^10 N/m2. Suppose a volume of 0.6 m3 of the liquid is taken from the surface where the pressure is approximately that of the atmosphere (1.00 10^5 N/m2) and is lowered into a pool of liquid to a depth where the pressure is 2.00 10^7 N/m2. By how much does the volume of the liquid change?
use: V= (Vzero)(change in pressure)/B
To find the change in volume of the liquid, we can use the formula:
ΔV = (V0 * ΔP) / B
where:
ΔV is the change in volume
V0 is the initial volume of the liquid
ΔP is the change in pressure
B is the bulk modulus of the liquid
Given:
V0 = 0.6 m^3
ΔP = (2.00 * 10^7 N/m^2) - (1.00 * 10^5 N/m^2) = 1.99 * 10^7 N/m^2
B = 4.0 * 10^10 N/m^2
Now we can plug these values into the formula to find the change in volume:
ΔV = (0.6 m^3 * 1.99 * 10^7 N/m^2) / (4.0 * 10^10 N/m^2)
Calculating this value:
ΔV = (0.6 m^3 * 1.99 * 10^7 N/m^2) / (4.0 * 10^10 N/m^2)
≈ 2.985 * 10^-5 m^3
Therefore, the volume of the liquid changes by approximately 2.985 * 10^-5 m^3.
To find the change in volume of the liquid, we can use the formula:
ΔV = V₀ * (ΔP / B)
Where:
ΔV is the change in volume
V₀ is the initial volume of the liquid (0.6 m³ in this case)
ΔP is the change in pressure (the difference between the final and initial pressures)
B is the bulk modulus of the liquid (4.0 * 10¹⁰ N/m² in this case)
Given the initial pressure (P₁) is approximately equal to the atmospheric pressure (1.00 * 10⁵ N/m²) and the final pressure (P₂) is 2.00 * 10⁷ N/m², we can calculate the change in pressure:
ΔP = P₂ - P₁
= (2.00 * 10⁷ N/m²) - (1.00 * 10⁵ N/m²)
= 1.99 * 10⁷ N/m²
Now, we can substitute the values into the formula:
ΔV = (V₀ * ΔP) / B
= (0.6 m³ * 1.99 * 10⁷ N/m²) / (4.0 * 10¹⁰ N/m²)
By canceling the units and calculating the expression, we can find the change in volume.