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$1 \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted into steam at $100^{\circ} \mathrm{C}$ by boiling at atmospheric pressure. The volume of water changes from $1.00 \times 10^{-3} \mathrm{~m}^3$ as a liquid to $1.671 \mathrm{~m}^3$ as steam. The change in internal energy of the system during the process will be
(Given latent heat of vaporisation $=2257 \mathrm{~kJ} / \mathrm{kg}$ , Atmospheric pressure $=1 \times 10^5 \mathrm{~Pa}$ )

Option: 1
$+2476 \mathrm{~kJ}$

Option: 2
$-2426 \mathrm{~kJ}$

Option: 3
$-2090 \mathrm{~kJ}$

Option: 4
$+2090 \mathrm{~kJ}$

Answers (1)

best_answer

Change in volume at constant pressure and temp $\rightarrow$

$\begin{aligned} \Delta \mathrm{V}= & \mathrm{V}_2-\mathrm{V}_1=1.671-0.001 \\ \end{aligned}$

$\begin{aligned} \Delta \mathrm{V}= & 1.67 \mathrm{~m}^3 \quad \ldots . .(1) \\ \end{aligned}$

$\begin{aligned} \Delta \mathrm{Q}= & \Delta \mathrm{U}+\mathrm{w} \\ \end{aligned}$

$\begin{aligned} \mathrm{mL}_{\mathrm{v}}= & \Delta \mathrm{U}+\left(1.013 \times 10^5\right)(1.67) \\ \end{aligned}$

$\begin{aligned} & \Delta \mathrm{U}=(2257-170) 10^3 \\ \end{aligned}$

$\begin{aligned} & \Delta \mathrm{U}=2090 \mathrm{~kJ} \text { (approx.) } \quad \end{aligned}$

Posted by

Suraj Bhandari

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