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)

The bulk modulus $$B$$ of a liquid is defined as
$$B = -\,\frac{\Delta P}{\dfrac{\Delta V}{V}}$$
where $$\Delta P$$ is the change in pressure, $$\Delta V$$ is the change in volume (negative for compression), and $$V$$ is the initial volume.

We are given:

• Initial pressure $$P_1 = 1$$ atm, final pressure $$P_2 = 5$$ atm.
Hence the pressure rise is
$$\Delta P = P_2 - P_1 = 5 - 1 = 4$$ atm.

• Take $$1$$ atm $$= 10^{5}$$ Pa, so
$$\Delta P = 4 \times 10^{5}\ \text{Pa}$$.

• Bulk modulus $$B = 2$$ GPa $$= 2 \times 10^{9}$$ Pa.

• Magnitude of volume decrease $$|\Delta V| = 0.8\ \text{cm}^3 = 0.8 \times 10^{-6}\ \text{m}^3$$.

Using the definition of $$B$$ and taking magnitudes (since we know the sign corresponds to compression),
$$B = \frac{\Delta P \, V}{|\Delta V|} \quad\Longrightarrow\quad V = \frac{B\,|\Delta V|}{\Delta P}\,.$$

Substituting the numerical values:
$$V = \frac{2 \times 10^{9}\ \text{Pa} \;\times\; 0.8 \times 10^{-6}\ \text{m}^3}{4 \times 10^{5}\ \text{Pa}}$$

Step-wise calculation:
$$2 \times 0.8 = 1.6$$
$$10^{9} \times 10^{-6} = 10^{3}$$
Numerator $$= 1.6 \times 10^{3} = 1600$$
Denominator $$= 4 \times 10^{5}$$
Therefore,
$$V = \frac{1600}{4 \times 10^{5}} = 0.4 \times 10^{-2}\ \text{m}^3 = 0.004\ \text{m}^3.$$

Conversion to litres (as $$1\ \text{m}^3 = 1000\ \text{L}$$):
$$V = 0.004\ \text{m}^3 \times 1000\ \frac{\text{L}}{\text{m}^3} = 4\ \text{L}.$$

Hence, the initial volume of the liquid was $$4$$ litres.