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(B) Given: Initial pressure $(P_i) = 1 \ atm$,Final pressure $(P_f) = 5 \ atm$.Change in pressure $(dP) = P_f - P_i = 4 \ atm = 4 	imes 10^5 \ Pa$.Ch

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(B) Given: Initial pressure $(P_i) = 1 \ atm$,Final pressure $(P_f) = 5 \ atm$.Change in pressure $(dP) = P_f - P_i = 4 \ atm = 4 	imes 10^5 \ Pa$.Change in volume $(dV) = -0.8 \ cm^3 = -0.8 	imes 10^{-6} \ m^3$.Bulk modulus $(B) = 2 \ GPa = 2 	imes 10^9 \ Pa$.The formula for bulk modulus is $B = -\frac{dP}{dV/V}$,which implies $V = -\frac{B \cdot dV}{dP}$.Substituting the values: $V = -\frac{2 	imes 10^9 	imes (-0.8 	imes 10^{-6})}{4 	imes 10^5}$.$V = \frac{1.6 	imes 10^3}{4 	imes 10^5} = 0.4 	imes 10^{-2} \ m^3 = 4 	imes 10^{-3} \ m^3$.Since $1 \ m^3 = 1000 \ litres$,$V = 4 	imes 10^{-3} 	imes 1000 = 4 \ litres$.

$A$ sample of a liquid is kept at $1 \ atm$. It is compressed to $5 \ atm$,which leads to a 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 $1 \ atm = 10^5 \ Pa$)

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