3rd law of thermodynamics.. 2 Different Examples((Thnks))

Chemistry

1 answer:

VLD [36.1K]2 years ago

5 0

Answer:

See explanation

Explanation:

The third law of thermodynamics states that "the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero" (Wikipedia).

One example of the third law of thermodynamics has to do with steam. Steam is gaseous water. Since it is a gas, its molecules are free to move around therefore its entropy is high. When the temperature of the steam is decreased below 100 degrees, the molecules of steam loose energy and turn into liquid water and do not move as freely as they did in the gaseous state. If the temperature is further decreased to yield ice at zero degrees, the molecules of water are "frozen" in their positions and the entropy of the system decreases to zero.

Also, the ions in ionic crystal solids move around when the substance is in solution or in molten state hence the substance conducts electricity. When the ionic substance is in solid state, the ions do not move about and the entropy of the solid system tends towards zero.

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You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

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3 years ago

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Natali5045456 [20]

Cohesion and adhesion

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Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$