Entropy Changes of Physical Processes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Put a tray of ice at zero degrees Celsius on a gas flame. Heat keeps going in, but the thermometer stays stuck at zero until all the ice has melted. So, while the thermometer is stuck, is the entropy changing or not? Hmm, hold that thought. Today we calculate entropy changes for melting, boiling, heating, expanding and mixing. These come up often in JEE, so let's begin.

0:33Phase change: entropy jumps at one temperature

So, let's answer the puzzle. Look at this graph: S, the entropy, going up, and T, the temperature, going across. It starts at zero for a perfect crystal at zero kelvin, as the first note on the right says, and as the solid warms, its entropy rises slowly. But at T m, with a small m for melting, the line shoots straight up! Temperature stuck, entropy jumping! So yes, it is changing: S jumps at each phase change, just as the notes say. Okay? Now, how big is that jump? The heat of fusion goes in at one fixed temperature. So the entropy change is this heat, divided by what? By the temperature, yes! Delta S equals delta H over T, okay? The little triangle is the capital Greek letter delta, meaning change in; S is entropy, and H is enthalpy. And the condition: only at the transition temperature, the melting point or the boiling point. Right? For melting, delta fus S equals delta fus H over T m. The small fus below means fusion, or melting. It is the first orange arrow on the graph. With me? For boiling, delta vap S equals delta vap H over T b, with vap for vaporisation and b for boiling. That's the second orange arrow, and look, it is much the larger jump, as the last note says. Clear?

1:56Trouton's rule: a quick check for boiling

Now, a beautiful shortcut. Trouton's rule: for many liquids, delta vap S is approximately eighty five to eighty eight joules per kelvin per mole. The wavy equals sign means approximately, and the units, with their powers of minus one, read joules per kelvin per mole. Interesting, right? But water's value is higher. Why? What is special about liquid water? Hydrogen bonding, exactly! In water and alcohols, hydrogen bonds partly order the liquid, so boiling gives an unusually large gain in disorder. Okay? Now, a topper's trick: turn the rule around. Delta vap H is about eighty eight times T b, in joules per mole. Chloroform boils at three hundred and thirty four kelvin, so its delta vap H is about twenty nine kilojoules per mole. Quick, isn't it?

2:46Ideal gas at constant temperature

Now, a gas. An ideal gas expands at constant temperature, and its molecules get more room. Does its entropy go up, or down? Up, of course! For n moles, delta S equals n R l n of V two over V one. R is the gas constant, l n is the natural log, and V one and V two, with small one and two below, are the starting and final volumes. Okay? Now careful with pressure, because it flips! At constant temperature, pressure falls as volume rises, so it is n R l n of P one over P two. Write P two over P one, and your sign comes out wrong. Clear? Now a classic trap. An ideal gas expands into a vacuum, from V to two V, the arrow on screen, in an insulated vessel. No heat flows, so q, the heat, is zero. So delta S is zero too, right? No! For an ideal gas, the temperature does not change here, and entropy is a state function: it depends only on the start and the end. So the formula holds whether the change is reversible or not, and delta S equals n R l n two, greater than zero. Got it?

3:57Heating: the temperature changes

Next, heating. Once the ice is water, its temperature rises. Can we still use delta H over T here? No, because T is not fixed any more. For heating n moles from T one to T two, delta S equals n C l n of T two over T one, where C is the molar heat capacity. Okay? Now, the C depends on the condition. At constant pressure, use C P, with P written small below; at constant volume, use C V instead. The two give different answers, so read the question carefully. Right?

4:33Mixing, and the reversible adiabatic change

Next, mixing. Two different ideal gases mix at the same T and P, that is, the same temperature and pressure. Delta S equals minus R times the sum of n i l n x i. The big zigzag is the capital Greek letter sigma, meaning sum of; n i, with a small i below, is the moles of each gas, and x i its mole fraction. Okay? See the minus sign in front. Can the entropy of mixing ever be negative? Never! Each mole fraction is less than one, so its log is negative, and minus times negative is positive. With me? Last case: a reversible adiabatic change. Adiabatic means no heat flows, so q rev, the reversible heat, is zero, and delta S is zero. So the entropy stays constant: isentropic. But only for a reversible change! Our free expansion was adiabatic too, but not reversible, and there the entropy went up. Got it?

5:31Example 1: benzene's entropy of vaporisation, Trouton's rule

Alright, Example one. Benzene boils at three hundred and fifty three kelvin, with delta vap H equal to thirty point eight kilojoules per mole. Find delta vap S, and compare it with Trouton's rule. Pause the video, and try it first. Done? Okay, let's check. At the boiling point, delta S equals delta H over T, okay? Thirty point eight over three hundred and fifty three: zero point zero eight seven three. That's the answer, right? Oh no, careful! Look at the units. Delta H is in kilojoules, but entropy is in joules per kelvin. So convert first: thirty point eight kilojoules per mole is thirty thousand eight hundred joules per mole. Okay? Now divide by three hundred and fifty three: eighty seven point three joules per kelvin per mole. That sits right inside eighty five to eighty eight, just as Trouton's rule predicts! Clear?

6:29Example 2: one mole of ice at 0°C to water at 100°C

Now Example two, and it is our puzzle from the start! One mole of ice at zero degrees Celsius is melted, and the water warmed to one hundred degrees Celsius. Find the total entropy change. Delta fus H is six point zero one kilojoules per mole, and C P of water is seventy five point three joules per kelvin per mole. Pause and try. Hint: how many steps? Done? Okay, let's check. Two steps: the ice melts at a fixed temperature, then the water warms up. For the warming, can I put one hundred over zero inside the log? No, no! One hundred over zero is not even defined. Always use kelvin: zero degrees Celsius is two hundred and seventy three kelvin, and one hundred degrees Celsius is three hundred and seventy three kelvin. Clear? Step one, melting. Delta S one, with a small one below, is delta H over T, okay? Six point zero one kilojoules is six thousand and ten joules, and over two hundred and seventy three that gives twenty two point zero joules per kelvin. Right? Step two, warming, with n equal to one. Seventy five point three times l n of three seventy three over two seventy three: twenty three point five joules per kelvin. With me? Add them: forty five point five joules per kelvin. So the entropy rises while the ice melts, thermometer stuck, and rises again as the water warms. Puzzle solved! The trap here is forgetting the melting step. Don't fall for it!

7:59Quick revision: phase change, gas, heating and mixing

So, quick revision. Phase change, at the transition temperature: delta S equals delta H over T, okay? Ideal gas at constant temperature: n R l n of V two over V one, and heating: n C l n of T two over T one. Mixing is always positive, and a reversible adiabatic change gives zero. Now go and try the practice problems on this page. All the best!