Problem 923

Question

The compressibility of water \(4 \times 10^{-5}\) per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be.......... (A) \(4 \times 10^{-5} \mathrm{CC}\) (B) \(4 \times 10^{-5} \mathrm{CC}\) (C) \(0.025 \mathrm{CC}\) (D) \(0.004 \mathrm{CC}\)

Step-by-Step Solution

Verified
Answer
The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere is \(0.4 \, \mathrm{CC}\). The given options appear to be incorrect, as none of them match the calculated answer.
1Step 1: Identify the variables
We have the following information given: - Compressibility of water = \(4 \times 10^{-5}\) per unit atmospheric pressure - Volume of water = 100 cubic centimeters (CC) - Applied pressure = 100 atmospheres (atm)
2Step 2: Use the formula for finding the decrease in volume
The formula for finding the decrease in volume using compressibility, volume, and applied pressure is: Decrease in Volume = Compressibility × Volume × Applied Pressure
3Step 3: Substitute the given values into the formula
Substitute the given values into the formula: Decrease in Volume = \((4 \times 10^{-5})(100)(100)\)
4Step 4: Calculate the decrease in volume
Now, we can calculate the decrease in volume: Decrease in Volume = \((4 \times 10^{-5})(100)(100) = 0.4\, \mathrm{CC}\)
5Step 5: Choose the correct answer from the options
Comparing the calculated decrease in volume with the given options, we can conclude that the correct answer is: (A) \(4\times10^{-5}\,\mathrm{CC}\) (Incorrect) (B) \(4\times10^{-5}\,\mathrm{CC}\) (Incorrect) (C) \(0.025\,\mathrm{CC}\) (Incorrect) (D) \(0.004\,\mathrm{CC}\) (Incorrect) However, the calculated decrease in volume is not among the given options. The closest option is \(0.004\,\mathrm{CC}\), but the correct answer should be \(0.4\,\mathrm{CC}\). There seems to be an error in the given answer choices.

Key Concepts

Volume Change Under PressureAtmospheric Pressure EffectsCompressibility Formula
Volume Change Under Pressure
When we apply pressure to a liquid like water, its volume decreases. This phenomenon occurs because pressure forces the molecules closer together. But, is water highly compressible? Not really. Water is known for its low compressibility. Just a tiny volume change occurs even under high pressure.
For example, in our exercise, we're applying 100 atmospheres of pressure to water. We use the formula to calculate volume change:
  • Decrease in Volume = Compressibility \(\times\) Volume \(\times\) Applied Pressure
Substituting the values, the calculation results in a small volume change. Despite the significant pressure, water's volume changes very little due to its low compressibility. This characteristic is what makes liquids like water quite stable under various conditions.
Atmospheric Pressure Effects
Atmospheric pressure serves as the baseline pressure at which water and most substances are commonly subjected. However, when provided with extra pressure, as in the case of the exercise, we essentially shift from our everyday experiences to those that require more substantial calculation considerations.
Normal atmospheric conditions involve one atmosphere of pressure. If this pressure increases, as when considering our exercise's 100 atmospheres, its effects become more pronounced. For example, under these increased pressures,
  • Water molecules, with their closely packed nature, experience further forced proximity
  • Volume change gets highlighted, although minor due to water's nature
In simple terms, atmospheric pressure changes alter how we predict and calculate compressive forces on fluids, showing just how important it is to consider atmospheric influences when examining compressibility outcomes.
Compressibility Formula
The compressibility of a substance tells us how much it can be compacted under pressure. For water, the compressibility is given as a constant value, showing its resistance to volume changes. Let's break down the formula used in such calculations:
  • Compressibility: A measure of the fractional change in volume per unit pressure.
  • Volume: The initial amount of water we're considering.
  • Applied Pressure: The pressure exerted on the water.
This leads us to the key formula used in our exercise:\[\text{Decrease in Volume} = \text{Compressibility} \times \text{Volume} \times \text{Applied Pressure}\]Substituting the known values into this formula allows us to calculate how much water's volume decreases under given pressures. Understanding this formula is vital for explaining why water maintains such stability against pressure changes more effectively than gases.