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At 100^{\circ} \mathrm{C} and 1 \mathrm{bar}, in a closed container 120 \mathrm{~g} water was used to immerse a heating will. It was observed that at constant pressure 65 \% of the liquid was changed to vapours. What will be the work done for the lame if you are given with the densities of water in the liquid and gaseous state under these conditions are 1000 \mathrm{~kg} / \mathrm{m}^3 and 0.60 \mathrm{~kg}^3, respectively.

Option: 1

\mathrm{4489\ J}


Option: 2

\mathrm{333\ J}


Option: 3

\mathrm{10824\ J}


Option: 4

\mathrm{9946\ J}


Answers (1)

best_answer

\begin{aligned} & \mathrm{ w=-P_{\text {ext }}\left(V_f-V_c\right)} \\ = & -10^5\left(\frac{65 \times 10^{-3}}{0.60}+\frac{35 \times 10^{-3}}{1000}-\frac{120 \times 10^{-3}}{1000}\right) \end{aligned}
=-10^5 \times 10^{-3}(108.33+0.035-0.12)
\begin{aligned} & =-10^2(108.245) \\ |\mathrm{W}| & =10824 \mathrm{~J} . \end{aligned}

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shivangi.shekhar

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