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Q.
How much pressure (in atm) is needed to compress a sample of water by $0.4 \%$ ? (Assume, Bulk modulus of water $\left.\approx 2.0 \times 10^{9}\, Pa \right)$
TS EAMCET 2019
Solution:
Given, percentage change in volume $=0.4 \%$
Bulk modulus of water, $B=2 \times 10^{9} Pa$
The change in volume of water is given by
$\Delta V=-\frac{p \cdot V}{B}$
where, $p=$ pressure applied on the sample of the matter.
Given,
$\frac{\Delta V}{V}=-0.4 \%=\frac{0.4}{100}$
Pressure required, $p=B\left(-\frac{\Delta V}{V}\right)$
$=\frac{0.4}{100} \times 2 \times 10^{9} \,Pa =\frac{0.4}{100} \times \frac{2 \times 10^{9}}{101325} atm$
$=79\, atm \approx 80 \,atm$
Hence, $80\, atm$ pressure is needed to compress a sample of water by $0.4 \%$.