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Question: The compressibility of water is \(4 \times 10^{-5}\) per unit atmosphere pressure. The decrease in v...

The compressibility of water is 4×10−54 \times 10^{-5} per unit atmosphere pressure. The decrease in volume of 100  cm3100\;cm^3 water under a pressure of 100  atm100\; atm will be:
A). 0.4  cm30.4\;cm^3
B). 4×10−5  cm34 \times 10^{-5}\;cm^3
C). 0.025  cm30.025\;cm^3
D). 0.04  cm30.04\;cm^3

Explanation

Solution

Recall that compressibility measures the relative volume change in response to pressure change. Thus, arrive at a relation between compressibility, volume, change in volume, and pressure using the definition of compressibility and determine the decrease in volume. Remember that the volume decreases as pressure increases for an isothermal system.

Formula Used: Compressibility of a fluid β=−1VΔVΔP\beta = -\dfrac{1}{V}\dfrac{\Delta V}{\Delta P}, where V is the volume of the fluid, ΔV\Delta V is the change in volume of the fluid, ΔP\Delta P is the change in pressure that causes the change in volume.

Complete step-by-step solution:
Let us begin by establishing an understanding of compressibility.
The compressibility of a fluid characterizes the change in volume that will be produced in a fluid by a change in pressure. This change of bulk (volume) is brought about by the particles of the fluid being brought closer together by the pressure.
Quantitatively, it is a measure of the relative volume change ΔVV\dfrac{\Delta V}{V} of a fluid as a response to pressure change ΔP{\Delta P}, and is given as:
Compressibility β=−1VΔVΔP\beta = -\dfrac{1}{V}\dfrac{\Delta V}{\Delta P}
The negative sign is introduced to make compressibility positive in the case where there is a decrease in volume induced by an increase in pressure, which is what compressibility means in a literal sense anyway.
Now, in the context of the question, let us begin by listing out the parameters given to us:
We have compressibility of water β=4×10−5  atm−1\beta = 4 \times 10^{-5}\;atm^{-1} ,
Volume of water V=100  cm3V= 100\;cm^3
Change in pressure of water ΔP=100  atm\Delta P = 100\;atm
From the relation for compressibility:
β=−1VΔVΔP\beta = -\dfrac{1}{V}\dfrac{\Delta V}{\Delta P}
⇒ΔV=−βVΔP\Rightarrow \Delta V = - \beta V \Delta P
Plugging in the values in the above relation:
ΔV=−(4×10−5)(100)(100)=−4×10−1=−0.4  cm3\Delta V = - (4 \times 10^{-5})(100)(100) = - 4 \times 10^{-1} = -0.4\;cm^3
The negative sign indicates that there is a decrease in volume upon exerting the pressure.
Therefore, the correct option would be A. 0.4  cm30.4\;cm^3.

Note: The specification of the compressibility in the question is actually ill-defined because for any system, the magnitude of compressibility depends heavily on whether the process is isothermal or isentropic.
Isothermal compressibility: βT=−1V(∂V∂P)T\beta_T = -\dfrac{1}{V}\left(\dfrac{\partial V}{\partial P}\right)_T
Where the partial differential (change in quantities) is taken at a constant temperature.
Isentropic compressibility: βS=−1V(∂V∂P)S\beta_S = -\dfrac{1}{V}\left(\dfrac{\partial V}{\partial P}\right)_S
Where the partial differential (change in quantities) is taken at constant system entropy.
Note that, for a solid, the distinction between the two is usually negligible but for fluids, the difference is remarkable since molecules of ‘a’ fluid are more loosely bound and are more likely impacted by variations in entropy and temperatures than solids.