How would the volume of an $18 \cdot g$ $18*g$ mass of water change if its temperature changed by over $100$ $100$ $^{\circ} C$ $""^@C$ ?

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How would the volume of an $18 \cdot g$ $18*g$ mass of water change if its temperature changed by over $100$ $100$ $^{\circ} C$ $""^@C$ ?

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Answer 1 · 2017-09-08T17:53:05

Well, there is a PHASE CHANGE associated with your question......

Explanation:

$18 \cdot g$ $18*g$ clearly constitutes a mole of water. However, the water can be condensed or in the gaseous phase. Water is quite involatile, and $18 \cdot g$ $18*g$ of ice or water indeed contains a molar quantity of $O H_{2}$ $OH_2$ molecules, i.e. $2 \times N_{A}$ $2xxN_A$ $H$ $H$ atoms, and $N_{A}$ $N_A$ $O$ $O$ atoms; $18 \cdot g$ $18*g$ in total.

At normal pressures, if we heat the water up over $100$ $100$ $^{\circ} C$ $""^@C$ , a phase change occurs, and the water enters the gaseous phase. And $18 \cdot g$ $18*g$ of GASEOUS water would indeed occupy $22.4 \cdot L$ $22.4*L$ ....of course the temperature of the gas phase would also change the volume.

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How would the volume of an $18 \cdot g$ $18*g$ mass of water change if its temperature changed by over $100$ $100$ $^{\circ} C$ $""^@C$ ?

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How would the volume of an 18*g18⋅g18*g mass of water change if its temperature changed by over 100100100 ""^@C∘C""^@C?

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How would the volume of an $18 \cdot g$ $18*g$ mass of water change if its temperature changed by over $100$ $100$ $^{\circ} C$ $""^@C$ ?