One mole of H2 O(g) at 1.00 atm and 100 .^∘ C occupies a volume of 30.6 L. When one mole of H2

One mole of H2 O(g) at 1.00 atm and 100 .^∘ C occupies a...

One mole of $\mathrm{H}_{2} \mathrm{O}(g)$ at $1.00 \mathrm{~atm}$ and $100 .^{\circ} \mathrm{C}$ occupies a volume of $30.6 \mathrm{~L}$. When one mole of $\mathrm{H}_{2} \mathrm{O}(g)$ is condensed to one mole of $\mathrm{H}_{2} \mathrm{O}(l)$ at $1.00 \mathrm{~atm}$ and $100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}$ of heat is released. If the density of $\mathrm{H}_{2} \mathrm{O}(l)$ at this temperature and pressure is $0.996 \mathrm{~g} / \mathrm{cm}^{3}$, calculate $\Delta E$ for the condensation of one mole of water at $1.00 \mathrm{~atm}$ and $100 .{ }^{\circ} \mathrm{C}$.

One mole of $\mathrm{H}_{2} \mathrm{O}(g)$ at $1.00 \mathrm{~atm}$ and $100 .^{\circ} \mathrm{C}$ occupies a volume of $30.6 \mathrm{~L}$. When one mole of $\mathrm{H}_{2} \mathrm{O}(g)$ is condensed to one mole of $\mathrm{H}_{2} \mathrm{O}(l)$ at $1.00 \mathrm{~atm}$ and $100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}$ of heat is released.
If the density of $\mathrm{H}_{2} \mathrm{O}(l)$ at this temperature and pressure is $0.996 \mathrm{~g} / \mathrm{cm}^{3}$, calculate $\Delta E$ for the condensation of one mole of water at $1.00 \mathrm{~atm}$ and $100 .{ }^{\circ} \mathrm{C}$.

Key Concepts

First Law of Thermodynamics

This fundamental principle states that the change in internal energy of a system (?E) is equal to the heat added to the system (q) plus the work done on the system (w). It is crucial for analyzing energy changes in any physical or chemical process, including phase transitions.

Enthalpy and Internal Energy Relationship

At constant pressure, the change in enthalpy (?H) of a system includes not only the change in its internal energy (?E) but also the work done by the system due to expansion or compression (P?V). This relationship, expressed as ?H = ?E + P?V, is essential for converting between heat measurements and total energy changes in a process.

Work in Constant Pressure Processes

In processes that occur at constant pressure, work is performed as the system changes its volume. This work is calculated by the product of pressure and the change in volume (w = -P?V), where the negative sign indicates that work done by the system leads to a loss in its internal energy. Understanding this concept is key for analyzing energy changes during phase transitions where volume differences are significant.

Phase Transitions

Phase transitions, such as condensation or vaporization, involve significant changes in physical properties, including volume and heat exchange. During these processes, not only is heat transferred (as quantified by ?H) but work is also done due to the change in volume. Recognizing how these aspects interplay is important for a full thermodynamic analysis of the process.

Transcript

00:01 1 mole of steam at 1 .00 atmosphere and 100 point degrees celsius occupies a volume of 30 .6 liters.

00:37 When 1 mole of h2og is condensed to 1 mole of h2ol is condensed to 1 mole of h2ol, at the same temperature and pressure.

01:15 4 .66 kilojoules of heat is released.

01:29 If density of h2ol is 0 .996 grams per centimeter cubed at one atmosphere and 100 degrees c, calculate delta e for the condensation of one mole of h2o at, again, 1 .00 atmospheres, and 100 degrees c.

02:28 Okay, this shouldn't be too bad.

02:33 Let's see.

02:36 So let's do initial and final conditions.

02:45 I have one mole of each.

02:56 Pressure at each is one atmosphere.

03:08 My temperature at each is 100 degrees c and my volume for my initial was 30 .6 liters and my volume here will be 1 mole times 18 .02 grams per mole times my density, which is 0 .996 grams per cubic centimeter.

04:09 Do i really want to get that to moles, or where do i want to leave that? i should have stretched that out a little further.

04:18 I'm going to get that to liters.

04:23 So my volume, excuse me, volume will equal one mole times 18 .02 grams per mole.

04:40 Times 0 .996 grams per cubic centimeter times 1 ,000 cubic centimeters per liter.

04:56 Where is my calculator? i have lost my calc.

05:02 Oh, there it is.

05:03 Found it.

05:06 Hang on.

05:08 There we go.

05:09 So i'll have 18 .02 divided by .96...