What pressure is required to reduce the volume of 1kg of water from 100 to 99L? Bulk modulus for water is 2 x 10¹¹ N/m²
To solve this problem, we can use the equation relating pressure, volume, and bulk modulus:
ΔP = B × (ΔV / V)
Where:
ΔP is the change in pressure,
B is the bulk modulus,
ΔV is the change in volume, and
V is the initial volume.
In this case, we want to find the pressure required to reduce the volume of 1 kg of water from 100 L to 99 L. We know the bulk modulus of water is 2 x 10¹¹ N/m².
First, we need to calculate the change in volume:
ΔV = V_final - V_initial
= 99 L - 100 L
= -1 L
Next, we can plug the values into the equation:
ΔP = (2 x 10¹¹ N/m²) × (-1 L / 100 L)
Note that we have converted volume from liters to meters cubed (L to m³) for consistency with the units of bulk modulus.
Now, we can simplify the equation:
ΔP = (2 x 10¹¹ N/m²) × (-0.01 m³ / 1 m³)
= -2 x 10⁹ N/m²
Therefore, the pressure required to reduce the volume of 1 kg of water from 100 L to 99 L is approximately -2 x 10⁹ N/m² (negative because the volume is decreasing).