Question

A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to change of volume of 0.8 cm$^3$. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was ______ litre. (Take 1 atm = $10^5$ Pa)

Hint:

- Bulk modulus relates pressure change to relative volume change: \(K = -\Delta P/(\Delta V/V_0)\) - 1 m\(^3\) = 1000 litres - Watch unit conversions (1 cm\(^3\) = 10\(^{-6}\) m\(^3\), 1 GPa = 10\(^9\) Pa)

Answer 1

Correct Answer: 4

Given: 

Initial pressure of the liquid \( P_i = 1 \, \text{atm} \)

Final pressure of the liquid \( P_f = 5 \, \text{atm} \)

The change in pressure (\( \Delta P \)) is:

\( \Delta P = P_f - P_i = 4 \, \text{atm} = 4 \times 10^5 \, \text{Pa} \)

The change in volume (\( \Delta V \)) is:

\( \Delta V = -0.8 \, \text{cm}^3 \)

The bulk modulus \( B \) is given as:

\( B = 2 \times 10^9 \, \text{Pa} \)

Now, using the formula for bulk modulus:

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

We can calculate the volume \( V \):

\( V = -B \times \left( \frac{\Delta V}{\Delta P} \right) \)

Substituting the given values:

\( V = -2 \times 10^9 \times \left( \frac{-0.8 \times 10^{-6}}{4 \times 10^5} \right) \)

Thus, the volume \( V \) is:

\( V = 4 \times 10^{-3} \, \text{m}^3 = 4 \, \text{litre} \)

Answer 2

Step 1: Calculate the pressure change (\(\Delta P\)). \[ \Delta P = P_{\text{final}} - P_{\text{initial}} = 5\,\text{atm} - 1\,\text{atm} = 4\,\text{atm} = 4 \times 10^5\,\text{Pa} \]
Step 2: Use the bulk modulus formula. The bulk modulus (\(K\)) is given by: \[ K = -\frac{\Delta P}{\Delta V/V_0} \] where: - \(K = 2\,\text{GPa} = 2 \times 10^9\,\text{Pa}\) - \(\Delta V = -0.8\,\text{cm}^3 = -0.8 \times 10^{-6}\,\text{m}^3\) (negative sign indicates volume decrease) - \(V_0\) is the initial volume in \(\text{m}^3\)
Step 3: Solve for initial volume (\(V_0\)). \[ 2 \times 10^9 = -\frac{4 \times 10^5}{-0.8 \times 10^{-6}/V_0} \] \[ 2 \times 10^9 = \frac{4 \times 10^5 \times V_0}{0.8 \times 10^{-6}} \] \[ V_0 = \frac{2 \times 10^9 \times 0.8 \times 10^{-6}}{4 \times 10^5} = \frac{1.6 \times 10^3}{4 \times 10^5} = 4 \times 10^{-3}\,\text{m}^3 = 4\,\text{litres} \]