When 1, g of water at 0^{circ}C and 1 times 1 | vedclass

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(A) The work done by a system at constant pressure is given by $W = P \Delta V = P(V_f - V_i)$.Given:Pressure $P = 1 	imes 10^5\, N/m^2$.Initial volu

Vedclass · Accepted Answer

$0.0091$

Vedclass · Answer

(A) The work done by a system at constant pressure is given by $W = P \Delta V = P(V_f - V_i)$.Given:Pressure $P = 1 	imes 10^5\, N/m^2$.Initial volume of $1\, g$ of water at $0^{\circ}C$ is $V_i = 1\, cm^3 = 1 	imes 10^{-6}\, m^3$.Final volume of ice is $V_f = 1.091\, cm^3 = 1.091 	imes 10^{-6}\, m^3$.Change in volume $\Delta V = V_f - V_i = (1.091 - 1) 	imes 10^{-6}\, m^3 = 0.091 	imes 10^{-6}\, m^3$.Work done $W = 1 	imes 10^5 	imes 0.091 	imes 10^{-6} = 0.091 	imes 10^{-1} = 0.0091\, J$.

When $1\, g$ of water at $0^{\circ}C$ and $1 \times 10^5\, N/m^2$ pressure is converted into ice of volume $1.091\, cm^3$,the external work done will be (in $, J$)

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