# Problems In Highschool Chemistry/PhChem/thermo1/splmntry2

A gas contained in a cylinder fitted with a fractionless piston expands against a constant external pressure of 1atom from a volume of 5litres to a volume of 10 liters .During the process the system absorbs 400J of thermal energy from its surroundings.Determine ∆E for the process (R=8.314Jpre mole per kelvin and 1atom =101325NM^-2

### First thoughts[edit | edit source]

What is this process? Well, it is a phase change of water. Since heat transfer is involved, q, ∆E and ∆H are also associated with this process. How does work w come into being? Actually, the system in question is undergoing a change in volume as well against a pressure. Hence, some work must have been performed.

### The Solution[edit | edit source]

**q/∆H** - Note that this process has been carried out at constant atmospheric pressure. So, the heat involved q would be nothing but the enthalpy of the reaction. Please make a note that the standard enthalpy quoted here is for 1.00 moles, and for vaporization. Here, we are dealing with 10.00/18.00 moles of water vapour which has to condense. So, the ∆H is 5/9 times of the standard enthalpy in magnitude and opposite in sign.

That is, q = ∆H = (- 5/9)(∆H_{vap}) = -22.6 kJ. This is obvious, heat should be taken out of the system if it has to be condensed.

**W** - What is the work done? Consider ∫P_{ext}dV. Since the pressure is constant, it comes out of the integral, and we are left to evaluate P∆V. Now, this is another important point to note. The water vapour would occupy 17.01 L (according to the ideal gas equation), while the water droplets would occupy somewhere around 0.010 L. (assuming the density of water to be 1gm/ml at 100 ^{o}C). The change in volume is approximately equal to the volume of the water vapour itself.

Hence, in most condensation/vaporization problems, we never bother to find the change in volume. We assume liquids to have negligible volume in comparison to vapours. Even in our case, we will not take the volume of water into account because it is negligible, and because it is calculated through assumptions.

Hence, P∆V amounts to P (=1atm) x ∆V (=17.01 L) = 17.01 atm-L = 17.01 x 101000 x (1/1000) = 1718 J. w = -∫P_{ext}dV = -1718 J

**∆E** - Now we find ∆E. According to the first law of thermodynamic/Conservation of energy, ∆E = Q – ∫P_{ext}dV = Q + w = -22.6 kJ – 1718 J = -24.3 kJ

Please note that the value of w is in the order of joules, while that of ∆H is of the order of kilojoules. Do not make any mistakes while adding them.

### Synopsis[edit | edit source]

- q and ∆H were equal for this process because it was carried out at constant pressure.
- w for this process was calculated using –P∆V
- ∆E would have been trickier, if going by the popular misconceptions. However, it was calculated through the first law of thermodynamics, since the other two parameters are already known.

### Concluding Notes[edit | edit source]

## Question 2[edit | edit source]

**Water expands when it freezes. Determine the amount of work done in Joules when a system consisting of 1.0L of liquid water freezes under a constant pressure of 1.0 atm and forms 1.1L of ice.**

### Solution[edit | edit source]

There has to be no confusion with this problem. The work done is –∫P_{ext}dV = -P∆V = 1.0 atm x 0.1 L = -[1 x 101000 x 0.1 x (1/1000)] = -10.1 J

### Notes[edit | edit source]

Not many examiners will ask this question directly. Most spin-offs will involve the calculation of the volumes and the first law of thermodynamics.

## Question 3[edit | edit source]

When 1 mole of ice melts at 0 ^{o}C and at constant pressure of 1 atm. 6050 joules of heat are absorbed by the system. The density of ice and water are 0.92 gm/ml and 1.00 gm/ml respectively. Calculate ∆H and ∆E for the reaction in joules. Report the exact values.

### First Thoughts[edit | edit source]

If you have read the concluding notes of the previous question, then you will find that this is exactly what we were talking about. ∆H, as in question 1, is equal to the q of the reaction because this is a constant pressure reaction. ∆E has to be found using the first law of thermodynamics, for which we also require w.

Just as we discussed above, we have not been given the volumes of the two states directly. Rather, we have to find them out using their densities. This is not much of a big task, but complicates the problem nonetheless.

### Solution[edit | edit source]

**∆H** – Since the process is carried out at constant pressure, the enthalpy change of the reaction is the same as the q value. Q is 6050 J in magnitude, and since heat has been absorbed by the system, it has a +ve sign. So, ∆H = q = +6050 J

**∫P _{ext}dV** – The integral again simplifies to P∆V. All we have to do now is to find the change in volume. We are dealing with 1 mole of water, which is 18 grams of mass. So, the volume is 18 grams / 0.92 g/ml = 19.56 ml ice and 18/1.00 = 18 ml water. ∆V is then -1.56 ml.

**P∆V** = -(1 atm x 0.00156 L) = -0.00156 atm-l = -0.16 J

**∆E** – According to the first law of thermodynamics, it comes out to be 6050.16 J. However, we do not report such values because of the use of significant figures.

### Synopsis[edit | edit source]

- Found ∆H
- Calculated ∫P
_{ext}dV for use in the first law expression to ultimately find ∆E.

### Concluding Notes[edit | edit source]

Physical chemistry problems, generally, are easy and use the same concept/logic. In most of the cases, if they are complicated, it is just because of the presented data. Also, this question wasn’t the best example of how this is done. Be prepared for more.

## Question 4[edit | edit source]

**A cylinder contains 10 litres of an ideal at 25 atm and 25 ^{o}C. However, the gas leaks out due to some malfunctioning. The atmospheric pressure is exactly 1 atm and temperature 25 ^{o}C during the entire leak. Assuming that the process is isothermal, how much work is done on the atmosphere due to the leaking gas?**

### First thoughts[edit | edit source]

This is a nice variation of the questions we have been doing. When the gas leaks out at constant temperature, it actually undergoes an isothermal expansion. A part of our system has vanished in the atmosphere, while the remaining is still inside the cylinder. This is intriguing, as we have been dealing with disciplined processes, where, unlike here, the system under question was at one place.

So, all we are left to do is solve the problem with the numerical values.

### Solution[edit | edit source]

The gas will leak out of the cylinder till the pressure in the cylinder has reduced to 1 atm. Since the temperature of the gas is constant, PV = constant for the gas. Now, note that, the gas in the cylinder will be at 1 atm and the gas leaked out will also be at 1 atm. That means, the pressure has reduced 10 times. The volume, therefore, has increased ten times and has become 250 litres. The work done by the gas is then -∫P_{ext}dV = -(1 atm) x (250 L - 10 L) = -240atm-L

The work done on the atmosphere because of the leaking gas will have exactly the same magnitude but opposite sign as above. (To do – Explain this point properly, and in detail) Hence, the final answer is 240 atm-litres = 24240 J

### Synopsis[edit | edit source]

- Found the volume of the gas after the leaking has stopped.
- Evaluated ∫P
_{ext}dV for the gas - The last point.

### Concluding Notes[edit | edit source]

## Question 5[edit | edit source]

**A magnesium strip of mass 20 gm with 20 percent inert impurities is dropped into a beaker of dilute hydrochloric acid. Calculate the work done by the system as a result of the reaction that takes place. The atmospheric pressure is 1.0 atm and temperature 25 ^{o}C. The molar mass of magnesium is 24.3 grams.**

### First Thoughts[edit | edit source]

Again, this is a different situation. Until now, we have been dealing with processes over ideal gas, but this is something different. Actually, the reaction that will take place is a redox reaction. The magnesium in the strip will be oxidized to magnesium ions, while some of the hydrogen ions in the acid will be reduced to molecular hydrogen. This gas will then occupy some volume, which is different from the previous volume occupied by the system. Hence, ∫P_{ext}dV is involved yet again.

Also, note that the amount of acid taken is assumed to be in excess, so that all of the magnesium is oxidized, while still leaving some hydrogen ions.

### Solution[edit | edit source]

For each mole of magnesium ion produced, one mole of molecular hydrogen H_{2} is produced. (DIY – write down the redox reaction. This is a fairly simple one, and after the statement above, no complexity is involved whatsoever.)

The question is, how many moles of magnesium are actually reacting? The mass of the strip taken is 20.00 grams, which contains 20 percent inert impurities. So, the mass of magnesium reacting would be 16 grams, which amount to 16/24.3 = 0.66 moles of magnesium. The impurities are inert, and hence do nothing in the reaction. They have to be ignored completely.

In this condition, 0.66 moles of hydrogen will occupy 16.15 L. This volume is in fact, the increase in volume of the acid beaker + strip system. Hence, ∆V = 16.15 L. P∆V = 16.15 atm-ltres = 1631 J. w = -∫P_{ext}dV = -1631 J

### Synopsis[edit | edit source]

- Found the mass of magnesium reacting
- Found the moles of hydrogen gas produced, and the volume it will occupy.
- This volume is equal to ∆V.
- Evaluated ∫P
_{ext}dV as usual, and reported with appropriate signs