# Problems In Highschool Chemistry/PhChem/thermo1/table

Fill in the missing values in the table. The table contains 63 blanks, so we expect a student to solve the table in 90 minutes.

Note: -

1.) Each of the processes are carried out on 1 mole of  ideal gas.

2.) The odd numbered processes are for monoatomic ideal gases, while the rest are for diatomic ideal gases. For a monoatomic ideal gas, Cv = (3/2)R and for a diatomic ideal gas, Cv = (5/2)R. Cp = Cv + R, so theyre (5/2)R and (7/2)R respectively; not considering the vibrational degrees of freedom which we neglect at relatively low temperatures.

3.) Throughout the entire page, ∫PextdV implies the definite integral over the initial and the final stages as the lower and upper limits respectively. It DOES NOT designate the indefinite integral.

## The Question

 Process Ti Vi Pi Tf Vf Pf Q ∆E w ∫PextdV ∆H 1 Rev. Adiabatic 400K 02.0 L 04.0 L 2 Rev. Adiabatic 300K 01.00 atm 02.00 atm 3 Isobaric Rev. 300K 01.00 atm 600K 4 Rev. Isothermal 300K 03.0 L 06.0 L 5 Rev. Isothermal 250K 03.0 L 02.00 atm 6 Rev. Isochoric 02.00 atm 400K 1000 J 7 Irrev. Adiabatic 350K 01.00 atm = ext pressure 1000 J 8 Irrev. Isothermal 01.0 L 10.00 atm = ext pressure 0500 J

&nbsp

If you feel stumped by the sheer lack of data in the table, know that this is not an impossible task. It is, actually, a mechanical, mindless work. This is mainly because the table doesn't require a proper understanding of a process and the related intricacies. All it demands is a harmony among the formulae in the student's mind. Filling up this table will give a strong practice of the mathematics involved, and will also benefit the student in simplifying the more complex numerical problems.

If you still do not gather any courage to fill the table, you should consider solving the introductory problems.

 Process Ti Vi Pi Tf Vf Pf Q ∆E W ∫PextdV ∆H 1 Rev. Adiabatic 400K 02.0 L 16.42 atm 252K 04.0 L 05.18 atm 0 -1846 J -1846J 1846J -3076J 2 Rev. Adiabatic 300K 24.63 L 01.00 atm 365K 29.96 L 02.00 atm 0 1351J -1351J 1351J 1891J 3 Isobaric Rev. 300K 24.63 L 01.00 atm 600K 49.26 L 01.00 atm 6236 J 3741J -2495J 2495J 6236J 4 Rev. Isothermal 300K 03.0 L 8.21 atm 300K 06.0 L 04.10 atm 1729 J 0 -1729J 1729J 0 5 Rev. Isothermal 250K 03.0 L 6.84 atm 250K 10.26 L 02.00 atm 2556 J 0 -2556J 2556J 0 6 Rev. Isochoric 352 K 14.44 L 02.00 atm 400K 14.44 L 02.27 atm 1000 J 1000J 0 0 1397J 7 Irrev. Adiabatic 270 K 38.63 L 0.57 atm 350K 28.73 L 01.00 atm = ext pressure 0 1000J 1000J - 1000J 1663J 8 Irrev. Isothermal 424 K 01.0 L 34.81 424K 3.48 L 10.00 atm = ext pressure 500 J 0 -500J 0500 J 0

## Hints

### Column 1

(a)Pi - Since three of the gas parameters are known, the fourth parameter can always be found using the ideal gas equation on the initial state.

(b)Tf - Ideal gas equation has four parameters, only two are known, so we need one more equation.

Since this is a reversible adiabatic process, the pressure P and Volume V satisfy the relation (P)(V^y) = constant throughout. Also, the ideal gas equation is satisfied throughout in any reversible process. PV = nrt. Devide the first equation by the second, and you'll find that you just eliminated one unknown - P. It is now left to one equation, Ti[Vi^(y-1)] = Tf[Vf^(y-1])  and one unknown, Tf.

The final calculations require a simple use of log tables. In competitve exams however, the paper setter will give all the values of the logarithms involved, so you need not concern yourself with the complexity of the calculations.

(c)Pf - Now that three quantities in the ideal gas equation are known, you can easily find Pf.

Alternatively, you could directly find Pf through using P(V^y) = constant on initial and final stages of the process, and then used the ideal gas equation to find Tf. Notice that we avoided finding Pf when we were asked to find only Tf, by eliminating the unknown.

(d)Q - In an adiabatic process, the heat exchanged with the sorroundings is ________

(e)∆E - Internal energy is fairly simple for an ideal gas, and depends only upon temperature of the ideal gas.  It is nCvTf at Tf, and nCvTi at Ti. So the change, then, is nCv(Tf-Ti). Try to relate it to the expression for heat gained and heat lost learnt in thermal physics in the eleventh and ninth standard in Indian Schools.

(f)w - It is the aditive inverse of ∫PextdV in Chemistry conventions. In Physics, it is the same as ∫PextdV. See below on how to find ∫PextdV, that is the work done by the gas during the change in volume.

(g)∫PextdV - Well, performing the whole integration by substituting P is a good idea. P = nrt/v will not work because even T will vary. We cannot integrate two variables in this way. Instead, substitute P as [Constant]/(V^y) and carry out the integration. After the integration is done, substitute the value of constant as Pf(Vf^y) and Pi(Vi^y) to get the result in terms of known quantities.

But, the same work can be done in an easy manner! Use the first law, Q = ∆E + ∫PextdV. You will find that in an adiabatic process, since the heat exchanged is 0, ∫PextdV becomes the aditive inverse of the change in the internal energy. In laymen's terms, since in an adiabatic process, the heat input is zero, whatever work that is done by the gas comes at the expence of its internal energy (and vice versa). This explaining subtly why the First Law Of Thermodynamics is actually the manifestation of the law of conservation of energy.

**NOTE :: The integral requires the value of Pext. In a reversible process, since the External Pressure is the same as the internal pressure of a gas, the whole substitution makes sence. Remember, Pext is not necessrily equal to nrt/v, but internal pressure = nrt/v.**

(h)∆H - Finally, the enthalpy change of the reaction. It is pleasing to know that this again, depends only on the change in temperature for an ideal gas. That is, ∆H = nCp∆T.

The First Column, Done!!!

### Column 2

(a)Vi - Use the ideal gas equation over the first state.

(b)Tf - Ideal gas equation, this time again, will have two unknowns. We need one more equation, which comes from (P)(V^y) = constant. Going exactly by the steps in column one, however, we won't be eliminating the unknown Vf, while eliminating the known P. Instead, to eliminate Vf, while keeping the knowns, raise both sides of the ideal gas equation by Y and devide the first equation by the second. You will end up with (T^y)[P^(1-Y)] = constant. Use it on the initial and final stages.

(c)Vf - Now that three quantities in the ideal gas equation are known, you can easily find Vf.

Alternatively, you could directly find Vf through using P(V^y) = constant on initial and final stages of the process, and then used the ideal gas equation to find Tf. Notice that we avoided finding Vf when we were asked to find only Tf, by eliminating the unknown.

(d)Q - What is the defining charachteristic of an adiabatic process?

(e)∆E - nCv∆T pretty simple. :)

(f)w - Tip::Throughout the table, this column is going to be the aditive inverse of ∫PextdV

(g)∫PextdV - Tip::Although not throughout the table, but in any kind of adiabatic process (reversible or irreversible), ∫PextdV is the aditive inverse of ∆E, according to the first law.

**This is because the first law of the thermodynamics is universal, and it doesnt matter if the process is reversible or irreversible to apply it.

However, the ideal gas equation and any results derived from it are applicable only in the case of a thermodynamic equilibrium. Since this is not possible in case of an irreversible process, we cannot use the mentioned equations over the course of the reactions. Even then, the ideal gas equation can be used before and after an irreversible process takes place.**

(h)∆H -  ∆H = nCp∆T.

Second column completed as well.

### The Third Column

(a)Vi - the _______ equation, fill in the blank mentally.

(b)Vf - Again, two unknowns for the ideal gas equation. But is Pf actually an unknown? Remember that this is an isobaric process, and that the initial pressure is known, so the final pressure is known as well.

(c)Pf - Done

(d)Q - Non zero for the first time!! We will use the first law, Q = ∆E + ∫PextdV. See it again as the law of conservation of energy, the heat supplied is used in heating the gas as well as doing work. Now find the other two terms.

(e)∆E - nCv∆T :)

(f)∫PextdV - Since this is a reversible process, the external pressure is the same as the internal pressure, and the internal pressure is constant. It comes out of the integration. So we are left with  (P)∫dV, which evaluates to P∆V. Do the needful!!

(h)∆H - nCp∆T !!!

Note : - In evaluating ∫PextdV, make sure that you use the correct units of each quantity! Since we want the answer to be in joules, i.e. SI units, we need to substitute pressure and volume in Pascals and Cubic Meters.

Note : - The enthalpy change of a reaction is the heat exchanged in a reaction when a process is carried out at constant pressure. This, in our case, means that the Q and ∆H entries in column three should be same. [Have you convinced yourself why?] But there may be slight differences between the two entries, this resulting from the rounding off of conversion factors. e.g. 1 atm = 101325 Pascals, but we often substitute 1 atm as 1.01 lac pascals.

Third column done as well. :)

### Column Four

(a)Pi  - ;)

(b)Tf - This question is the answer, "what is an isothermal process?"

(c)Pf- The ideal gas equation is OK, but observe that the RHS of it is constant. So, even the LHS would be. Hence, if the volume doubles up, the pressure reduces to half.

(d)Q - ∆E + ∫PextdV, now we will find both the terms and substitute them.

(e)∆E - Well, if the internal energy of an ideal gas depends only on its temperature, and if this is an isothermal process, where does it lead us?

(f)∫PextdV - This is a reversible process. As explained above, Pext is the same as the pressure of the gas molecules at any stage. Substitute Pext as nrt/v. Since the numerator is constant, we take it out of integration, and evaluate nrT ∫(dV/V) which is nrTln(Vf/Vi). Now substitute the values.

(g)∆H - nCp∆T :)

### Column Five

No additional info. This is a clone actually.

### Column Six

We expect the student to be seasoned enough now to realize that the gas parameters are found through ideal gas equation, and then the thermodynamic Joule values are found simply.

Note : Since the volume remains constant, change in volume is 0. i.e. dV=0, so ∫PextdV = 0

### Column Seven

This is the first irreversible process that we are dealing with, in this table. Pf = 1 atm = external pressure in the above table means that the gas expands/contracts against a constant external pressure of 1 atm, and the final pressure of the gas too is 1 atm. [We have some numerical problems where the final pressure isnt equal to the external pressure against which expansion/contraction took place, but do not bother yourself with it at this moment.]

Note carefully that all three gas parameters are missing for the initial state. That means, apart from the gas equation, we will require two more equations to find them all. And since this is an irreversible adiabatic process, we cannot use P(V^y)=Constant over the initial and final states. That means we have lost an equation!

How are we going to deal with this situation? The first thing to keep in mind is that every problem asked in this book has a solution. And this table, like most of the highschool problems, is all about making equations to find unknowns.

(a)Vf - Use the ideal gas equation to get the value.

Now try to relate the final parameters to the initial parameters. Obviously, the heat gained, work done, change in internal energy put some constraints on the possible values of the parameters.

(b)Ti - ∆E is known to us, but if we write it as nCv∆T, and equate it to 1000 J, we actually end up with an equation. This gives us the value of ∆T, which can be use to find the initial temperature, because the final temperature is known.

(c)Vi/∫PextdV - Refer the two reversible adiabatic processes in column 1 and column 2 that we studied. We made a statement over there, that since the characteristic of any kind of adiabatic process is "zero energy exchange", ∫PextdV is always the additive inverse of ∆E for an adiabatic process. Hence ∫PextdV = -1000 J.

But this is not all!! If you are thorough with integration, look at the word ∫PextdV with some concentration. The parameter Pext would jump in and out of the integral in your mind's eye once you understand that it is constant for this process. This yields ∫PextdV =Pext(∆V). Just as you found Ti by finding ∆T, find Vi by finding ∆V.

Fill up the rest of the entries yourself.

Note: This is for you to ponder. In the reversible adiabatic processes,  ∫PextdV = -∆E and actually evaluating the integral gave the same equation. But in the case of an irreversible adiabatic process, evaluating the integral and using the first law to find its value gives two distinct equations! This can be used for our benefit in problems. Please keep this in mind!

### Column Eight

Just as in Column Seven, we lose the equation ∫PextdV = nrtln(Vf/Vi), but we can write it as ∫PextdV = Pext(∆V).

An irreversible isothermal process is much shorter than an irreversible adiabatic process to complete.

(a)Vf - ∫PextdV = Pext(∆V) use this equation to find the final volume of the gas molecules.

(b)Tf - Two gas parameters are known now, use the ideal gas equation to find Tf. And if this is an isothermal process, what does knowing the final temperature mean?

The rest of the entries require no new logic, everything has been covered in the above columns.