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1 g of water, of volume $1 \mathrm{~cm}^3$ at $100^{\circ} \mathrm{C}$ is converted into steam at same temperature under normal atmospheric pressure $=\left(\simeq 1 \times 10^5 \mathrm{~Pa}\right)$. The volume of steam formed equals $1671 \mathrm{~cm}^3$. If the specific latent heat of vaporisation of water is $2256 \mathrm{~J} / \mathrm{g}$, the change in internal energy is
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The correct answer is:
$2089 \mathrm{~J}$
Given, mass of water, $\mathrm{m}=\mathrm{lg}$
Volume of $1 \mathrm{~g}$ of water $=1 \mathrm{~cm}^3=10^{-6} \mathrm{~m}^3$
Volume of $1 \mathrm{~g}$ of steam $=1671 \mathrm{~cm}^3=1671 \times 10^{-6} \mathrm{~m}^3$
Pressure, $\mathrm{p}=1 \times 10^5 \mathrm{~Pa}$
Latent heat of vaporization of water, $\mathrm{L}=2256 \mathrm{~J} / \mathrm{g}$
Change in volume, $\Delta V=(1671-1) \times 10^{-6} \mathrm{~m}^3$
$$
=1670 \times 10^{-6} \mathrm{~m}^3
$$
Heat supplied, $\Delta \mathrm{Q}=\mathrm{mL}=1 \times 2256=2256 \mathrm{~J}$
As the steam expands, so the work done in expansion is
$$
\begin{array}{rlr}
\Delta \mathrm{W} & =\mathrm{p} \Delta \mathrm{V}=1 \times 10^5 \times 1670 \times 10^{-6} \quad \text { [from Eq. (i)] } \\
& =167 \mathrm{~J} & \ldots \text {.(iii) }
\end{array}
$$
According to first law of thermodynamics,
$$
\begin{array}{rlrl}
\Delta \mathrm{Q} & =\Delta \mathrm{U}+\Delta \mathrm{W} \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W} \\
& =2256-167 & & \text { [from Eq. (ii) and (iii)] } \\
& =2089 \mathrm{~J} &
\end{array}
$$
Volume of $1 \mathrm{~g}$ of water $=1 \mathrm{~cm}^3=10^{-6} \mathrm{~m}^3$
Volume of $1 \mathrm{~g}$ of steam $=1671 \mathrm{~cm}^3=1671 \times 10^{-6} \mathrm{~m}^3$
Pressure, $\mathrm{p}=1 \times 10^5 \mathrm{~Pa}$
Latent heat of vaporization of water, $\mathrm{L}=2256 \mathrm{~J} / \mathrm{g}$
Change in volume, $\Delta V=(1671-1) \times 10^{-6} \mathrm{~m}^3$
$$
=1670 \times 10^{-6} \mathrm{~m}^3
$$
Heat supplied, $\Delta \mathrm{Q}=\mathrm{mL}=1 \times 2256=2256 \mathrm{~J}$
As the steam expands, so the work done in expansion is
$$
\begin{array}{rlr}
\Delta \mathrm{W} & =\mathrm{p} \Delta \mathrm{V}=1 \times 10^5 \times 1670 \times 10^{-6} \quad \text { [from Eq. (i)] } \\
& =167 \mathrm{~J} & \ldots \text {.(iii) }
\end{array}
$$
According to first law of thermodynamics,
$$
\begin{array}{rlrl}
\Delta \mathrm{Q} & =\Delta \mathrm{U}+\Delta \mathrm{W} \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W} \\
& =2256-167 & & \text { [from Eq. (ii) and (iii)] } \\
& =2089 \mathrm{~J} &
\end{array}
$$
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