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Tutor profile: Kevin H.

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Kevin H.
Former Chemical Engineering Professor who misses teaching
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Questions

Subject: Physics (Thermodynamics)

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Question:

Helium is pumped into a balloon. The balloon rises into the atmosphere and expands as the pressure lowers. Assuming the balloon is completely adiabatic, does the temperature increase, decrease, or stay the same? Explain why?

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Kevin H.
Answer:

The temperature lowers. One may think, because of Boyle's Law, the temperature lowers because pressure lowers. However, you did not account for volume expanding. In the following: PV=nRT Pressure is decreasing and volume is increasing, so temperature COULD be increasing, decreasing, or staying the same depending upon how pressure and volume decrease/increase, respectively. The explanation stems from examing the problem in the lens of the first law of thermodynamics: energy cannot be destroyed or created. Mathematically, the equation form of the first law simplifies to: $$ W=m(u_2-u_1) $$ Because the system is closed and adiabatic. Also, the time dependency has no role, and we can assume changes in kinetic and potential energy are negligible. For an ideal gas, internal energy ONLY depends upon temperature. If heat capacity is constant, the above simplifies to: $$ W=mC_v(T_2-T_1) $$ As such, to determine the change in temperature, we need to know the work. There is no spinning, rotating, or electrical, but there is boundary work. Now we ask, is the work done ON the system or BY the system. ON indicates outside force pushing. By indicates the gas pushing. As the balloon expands, the gas is pressing harder than the atmosphere, so the work is BY the system. Work is a transfer of energy, so the gas transfers energy to the environment as it does work. Because of the first law, that energy is no longer available, so the gas has lost energy. As an ideal gas, that means it loses temperature.

Subject: Chemistry

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Question:

If I burn 4 g of hydrogen gas in an oxygen rich environment, what mass of water vapor (g) will be formed?

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Kevin H.
Answer:

First, oxygen-rich indicates the hydrogen is the limiting reagent. Burn indicates combustion, which is mainly irreversible. As such, the first step is to write out a proper reaction and balance it: $$ H_2(g)+O_2 (g) \rightarrow H_2O(g) $$ is our reaction, now let's balance it: luckily, we only have two elements: H and O. Let's check H first. There are two on each side, so it is good. Now check O. There is two on the left, but only one on the right. Let's increase the water (H2O) to 2. O is balance, but now H isn't. If we put a 2 in front of H2, there are four on both sides, so we are good now. $$ 2H_2(g)+O_2 (g) \rightarrow 2H_2O(g) $$ This is an important step, because without it, our math will likely be wrong. When doing stoichiometric math, I like to do ratios. You start with the ratio you have and have (at the end) what you need. So I have g of H2 and I need g of H20: $$ \frac{20.0\,g(H_2)}{1.0}.........=\frac{x\, g(H_2O)}{1.0} $$ I need ratios to cancel out g of H2 to eventually get to g of water. I have to think about relationships I have at my disposal. These can include density, conversion factors, but also stoichiometric ratios and molecular weights. Density is not helpful, because I don't need volumes. Everything is in SI units, so conversion factors are also no help, but the latter two will be VERY helpful $$ \frac{2.016\, g(H_2)}{1.0\, mol(H_2)} \: \: \frac{18.016\, g(H_2O)}{1.0\, mol(H_2O)} \: and\: \frac{2.0\, mol(H_2)}{2.0\, mol(H_2O)} $$ In short, the strategy is to cancel out what is on top to GET to what's at the end. If I refer to the above ratios as A, B, and C, I first divide by A to get rid of grams of H2. This leaves moles of H2. SOOO, what should I use to get rid of moles of H2?? That's right, I use ratio C (the stoichiometric ratio). This leaves me with moles of water. But I need mass. To go from moles to mass, I use molecular weights, or ratio B in this case. This results in the following equation: $$ \frac{20.0 \, g(H_2)}{1.0} \frac{1.0\, mol(H_2)}{2.016\, g(H_2)} \frac{2.0\, mol(H_2O)}{2.0\, mol(H_2)} \frac{18.016\, g(H_2O)}{1.0\, mol(H_2O)} = 178.73\, g(H_2O) $$

Subject: Chemical Engineering

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Question:

For an equimolar mixture between water and ethanol entering a distillation column, which will be found in a higher concentration at the top of the column?

Inactive
Kevin H.
Answer:

A distillation column is used to separate components. It relies upon differences in boiling points and vapor pressures. While ethanol is .(has a larger molecular weight), water has a higher boiling point because of its extremely strong intermolecular bonding. As such, ethanol boils easier, so vapors will be ethanol rich. Next, let's consider what will be at the top of the column, liquid or vapor? Due to gravity and densities, the liquids will drain low and the vapors will rise. So the top will be vapor rich. If ethanol is likely to be vaporized versus water, ethanol will be the component in higher concentration.

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