Question

The increase in pressure required to decrease the volume of 200 L of water by 0.004 percent is (Bulk modulus of water is \( 2.1 \times 10^9 \) N/m\(^2\)):

• A. \( 8.4 \times 10^4 \) N/m\(^2\)
• B. \( 8.4 \times 10^3 \) N/m\(^2\)
• C. \( 8.4 \times 10^5 \) N/m\(^2\)
• D. \( 8.4 \times 10^6 \) N/m\(^2\)

Hint:

The bulk modulus describes how much pressure is needed to compress a substance. The greater the bulk modulus, the more pressure is required for a given volume change.

Answer 1

The Correct Option is A

To solve this problem, we need to use the concept of bulk modulus (K), which is defined by the formula:

\( K = -\frac{\Delta P}{\frac{\Delta V}{V}} \)

where:

• \(\Delta P\) is the change in pressure. 
• \(\Delta V\) is the change in volume.
• \(V\) is the original volume.
• The negative sign indicates that pressure increases when volume decreases.

We are given:

• Bulk modulus, \( K = 2.1 \times 10^9 \text{ N/m}^2 \)
• Original volume, \( V = 200 \text{ L} \)
• Volume change percentage, \( 0.004\% \)

Calculate the change in volume (\(\Delta V\)):

\( \frac{\Delta V}{V} = \frac{0.004}{100} = 0.00004 \)

Now, substitute the values back into the bulk modulus formula:

\( K = -\frac{\Delta P}{0.00004} \)

Solve for \(\Delta P\):

\( \Delta P = K \times 0.00004 \)

Substitute \( K \) into the equation:

\( \Delta P = 2.1 \times 10^9 \times 0.00004 \)

\( \Delta P = 8.4 \times 10^4 \text{ N/m}^2 \)

Thus, the increase in pressure required is \( \boxed{8.4 \times 10^4 \text{ N/m}^2} \).

Answer 2

The bulk modulus \( B \) is given by the relation: \[ B = \frac{-\Delta P}{\frac{\Delta V}{V}} \] Rearranging the equation to find the pressure change \( \Delta P \): \[ \Delta P = -B \times \frac{\Delta V}{V} \] Given: - \( B = 2.1 \times 10^9 \) N/m\(^2\), - \( \Delta V/V = 0.004\% = 0.00004 \), - Volume \( V = 200 \) L. Substitute the values: \[ \Delta P = -2.1 \times 10^9 \times 0.00004 = 8.4 \times 10^4 \, \text{N/m}^2 \] Thus, the required increase in pressure is \( 8.4 \times 10^4 \) N/m\(^2\).

Answer 3

The bulk modulus relates a fractional change in volume to the pressure that causes it: \( K = \dfrac{\Delta P}{|\Delta V|/V} \), so \( \Delta P = K \times \dfrac{|\Delta V|}{V} \). Notice that only the fractional change \( |\Delta V|/V \) enters this formula, so the absolute volume of 200 L never actually appears in the final number.

The fractional change here is \( 0.004\% = \dfrac{0.004}{100} = 4\times10^{-5} \). Multiplying by the bulk modulus:

\[ \Delta P = (2.1\times10^{9})\times(4\times10^{-5}) = 8.4\times10^{4}\ \text{N/m}^2 \]

Checking each option against this order of magnitude:

1. Option A, \( 8.4\times10^{4} \): matches the computed value exactly and is correct.
2. Option B, \( 8.4\times10^{3} \): is smaller by a factor of 10, which would only arise from mistakenly using \( 4\times10^{-6} \) for the fractional change instead of \( 4\times10^{-5} \).
3. Option C, \( 8.4\times10^{5} \): is larger by a factor of 10, consistent with mistakenly treating the percentage as \( 0.04\% \) instead of \( 0.004\% \).
4. Option D, \( 8.4\times10^{6} \): is larger by a factor of 100 and would only come from a two-decade error in the percentage conversion.

The correct answer is \( 8.4\times10^{4}\ \text{N/m}^2 \).