Question 244: Properties of Matter

Question

The increase in pressure required to decrease the volume of a water sample by $0.2 \%$ is $\mathrm{P} 	imes 10^5 \mathrm{Nm}^{-2}$. Bulk modulus of water is $2.15 	imes 10^9 \mathrm{Nm}^{-2}$. The value of P is _________ .

Accepted Answer

The bulk modulus of a material is defined as:
$ B = \frac{-\Delta P}{\left(\frac{\Delta V}{V}ight)} $
Given:
Bulk Modulus, $ B = 2.15 	imes 10^9 \, 	ext{Nm}^{-2} $
Change in volume percentage, $ \frac{\Delta V}{V} = 0.2\% = \frac{0.2}{100} = 0.002 $
To find the required increase in pressure, $ \Delta P $, we rearrange the formula for the bulk modulus:
$ B = \frac{-\Delta P}{- \left(\frac{\Delta V}{V}ight)} $
Plug in the known values:
$ 2.15 	imes 10^9 = \frac{\Delta P}{0.002} $
Solving for $ \Delta P $:
$ \Delta P = 2.15 	imes 10^9 	imes 0.002 $
$ \Delta P = 4.3 	imes 10^6 \, 	ext{Nm}^{-2} $
Since the problem specifies $ \Delta P = P 	imes 10^5 \, 	ext{Nm}^{-2} $, we have:
$ 4.3 	imes 10^6 = P 	imes 10^5 $
$ P = 43 $
The value of $ P $ is therefore 43.

Properties of Matter

The increase in pressure required to decrease the volume of a water sample by 0.2 \% is \mathrm{P} \times 10^5 \mathrm{Nm}^{-2}. Bulk modulus of water is 2.15 \times 10^9 \mathrm{Nm}^{-2}. The value of P is _________ .

Integer / Numeric Answer Type