The change of phase from a solid to a gas is called _____. evaporation melting sublimation vaporization

2 answers:

4 0

Solid to gas stage changes are known as "sublimation." In many cases, solids transform into gases simply after an intermediate liquid state. It is an endothermic procedure (happens with the absorption of heat from the environment).

Some example:

Freeze drying - Water can be sublimated in sustenance items by utilizing a vacuum.

"Dry ice" or, in other words carbon dioxide and as the name proposes it experiences stage change from solid state to liquid state straightforwardly without wetting or going into liquid state.

monitta3 years ago

4 0

The correct answer to the question is sublimation i.e the change of phase from solid to gas is called sublimation.

EXPLANATION:

Before answering the question, first we have to understand various changes of states of matter.

There are various state changes or phase changes in nature.

Melting: Melting is the heat phenomenon in which substance in solid phase is converted into liquid at its melting point. For instance, ice converted into water.

Sublimation: Sublimation is the type of phase change in which solid substance is directly converted into gas without melting. For instance, the vaporization of camphor.

Vaporization: Vaporization is the type of phase change of matter in which a liquid substance is converted into gas at its boiling point. For instance, boiling of water.

As per our question, the change is occurring from solid to gas.

Hence, the phase change will be sublimation.

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A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

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Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: