# A 4-L rigid tank contains 2 kg of saturated liquid- vapor mixture of w

A 4-L rigid tank contains 2 kg of saturated liquid- vapor mixture of water at ${50}^{\circ }C$. The water is now slowly heated until it exists in a single phase. At the final state, will the water be in the liquid phase or the vapor phase? What would your answer be if the volume of the tank were 400 L instead of 4 L?

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Jim Hunt
Step 1
In this problem we need to determine the final phase of the water. To do that we need to calculate the specific volume v and the critical specefic volume ${v}_{c}$ of the water.
The problem states that the tank in which the water is in is rigid. That means that the specific volume does not change during the process. We can calculate that specific volume by using the given volume $V=4L$ and mass $m=2kg$. Before the calculation we need to express the volume in ${m}^{3}$ units.
$V=4L\cdot \frac{1{m}^{3}}{1000L}=0.004{m}^{3}$
$v=\frac{V}{m}$
$v=\frac{0.004{m}^{3}}{2kg}$
$v=0.002\frac{{m}^{3}}{kg}$
The critical specific volume ${v}_{c}$ we can calculate using the molar mass and critical molar volume ${V}_{c}=0.0560{m}^{3}kmo{l}^{-1}$.
${v}_{c}=\frac{{V}_{c}}{M}$
${v}_{c}=\frac{0.0560\frac{{m}^{3}}{kmol}}{18.015\frac{kg}{kmol}}$
${v}_{c}=0.00311\frac{{m}^{3}}{kg}$
The calculated specific volume is lower than the critical specific volume which means that the final phase of water in the 4L volume tank is liquid.
Step 2
If the volume of the tank is $V=400L$ we would need to first calculate the specific volume v of the water again. Before the calculation we need to express the volume in ${m}^{3}$ units.
$V=400L\cdot \frac{1{m}^{3}}{1000L}=0.4{m}^{3}$
$v=\frac{V}{m}$
$v=\frac{0.4{m}^{3}}{2kg}$
$v=0.2\frac{{m}^{3}}{kg}$
This specific volume is greater than the critical meaning the phase of the water in the 400L tank is vapor.

Ella Williams
Step 1
The standard value of critical specific volume of water is,
${v}_{cr}=0.0031\frac{{m}^{3}}{k}g$
For 4L tank volume, the specific volume is,
$v=\frac{V}{m}$
$=\frac{\left(4L\right)\left(\frac{1×{10}^{-3}}{1L}\right)}{2kg}$
$=0.002\frac{{m}^{3}}{k}g$
The specific volume of water is less than critical value. Hence water will exit in liquid state.
Step 2
For 400L tank volume, the specific volume is,
$v=\frac{V}{m}$
$=\frac{\left(400L\right)\left(\frac{1×{10}^{-3}}{1L}\right)}{2kg}$
$=0.2\frac{{m}^{3}}{k}g$
The specific volume of water is greater than critical value. Hence water will exit in vapor.