Last modified on 18 July 2009, at 19:01

Problems In Highschool Chemistry/PhChem/thermo1/Splmntry

Question 1Edit

1.00 mole of helium gas is allowed to expand from 22.4L to 44.8L isothermally at 273.15 K. The expansion is free expansion type. The external pressure against which the gas is expanding is zero. In other words, the gas is expanding in vacuum. Find the values of q, w, ∆E and ∆H for this process. Assume ideal gas behaviour.

The solutionEdit

Although it might seem tiring to find the values of four parameters, a thorough practice of filling up the table we discussed before might make this an obvious mechanical task. Since the process is an isothermal process, the temperature change is zero. And since we are to assume ideal gas behavior, ∆E and ∆H are zero. This is because, for an ideal gas, these two parameters depend solely upon the temperature. Secondly, since the gas is expanding against zero pressure, it is offered zero resistance. Hence, no work is one by or upon the gas. If this sounds unconvincing, consider ∫PextdV. Since the external pressure is constant, we take it out of the integral, and since the external pressure is zero, the whole term is zero by itself.

w, then, too becomes zero in value.

Q, that is, the heat exchange in the process also has to have the value 0, according to the first law of thermodynamics. Have you noted that it never mattered what the gas was, or what its gas parameters were, so long as we considered it to be ideal? In the question, we have mentioned various data regarding the gas, none of which were used in the solution.

SynopsisEdit

  • P=0, so ∫PextdV = 0
  • Using the first law of thermodynamics, we showed that q = 0 as well.

Concluding NotesEdit

Note the whole discussion as, “An isothermal free expansion of an ideal gas is also adiabatic”.

Is the converse true? Is an adiabatic free expansion of an ideal gas isothermal too? Think in terms of conservation of energy. If the idea doesn’t seem obvious, use the expression for the first law of thermodynamics.

Question 2Edit

Two moles of a monoatomic ideal gas occupy a volume of 30 L at 300 K. The gas is allowed to expand to 50L at a constant external pressure of 550 mm Hg adiabatically. Find the work done by the gas and the final temperature of the gas. Keep in mind that the final pressure of the gas is not necessarily equal to the external pressure it expands against.

A little adviceEdit

While solving Physical Chemistry, a problem either seems familiar or it does not. And to make sure it does seem familiar in an examination, one should be able to dissect the data given and the scenario into fragments which have been previously dealt with.

Working on these grounds, we realize that we have already solved some problems dealing with irreversible expansion. The only difference between them and this one is that in the latter, we do not know that final pressure of the gas. Isn’t it? Where does it lead to? A simple thought says that we have another unknown, finding which would require an extra piece of data. Indeed, compare this problem with the previous ones from the table, and you’ll realize we have more data here.

The SolutionEdit

Here we go. The work done by the gas is nothing but -∫PextdV when we are in the IUPAC domain.

We now evaluate the integral. Since the pressure is constant, we take it out of the integral sign and the remaining integral amounts to ∆V. We know the change in volume of the gas, and we know the external pressure against which it has expanded. Hence, the work done by the gas is P∆V = (550/760 atm) x (20 L) atm-litres = (550x101000/760) x (20/1000) J =. 1462 J. (Check your level – are you thorough with unit conversions?). w is then equal to -1462 J.

Also, since it is an adiabatic process, ∫PextdV = - ∆E = w. This is the second step we’re performing. (Can you explain this using the conservation of energy? Since the heat supplied is zero, whatever work is done by the gas is done at the cost of its internal energy.) Equate the change in internal energy to nCv∆T and you’ll find the change in temperature. The number of moles are two, and the Cv is 1.5R (monoatomic ideal gas). This is enough to find the final temperature. The mistake you could have made was finding the final temperature using the ideal gas equation with 550/760 atm as the final pressure.

The final answers are -1462J and 242.4K.

SynopsisEdit

  • Evaluated the ∫PextdV as P∆V and found one answer, w
  • Equated ∫PextdV to –nCv∆T (w = nCv∆T) and found Tf

Concluding NoteEdit

Make it a general approach to assume the final pressure not being equal to the external pressure against which the system has expanded in cases of irreversible processes, unless specifically mentioned otherwise. This will help in most questions. Since these questions are easy going by the concepts, and still, mistakes can be made, they are among the deciding questions in exams like AIEEE or even the JEE.

Question 3Edit

One mole of a monoatomic ideal gas occupies a volume of 11.2 L at 273K. The gas is allowed to expand by 22.4L at a constant external pressure of 1 atm isothermally. Find the work done by the gas. The final pressure of the gas is not necessarily equal to the external pressure it expands against.

The SolutionEdit

This is basically the same question as above. There may be a few on-the-surface differences, but nothing more than that. The work done, as we know, is nothing but -∫PextdV = -P∆V = -[1 atm x 22.4 L atm-litres] = -(101000 x 22.4 x 10^-3) Joules = -2262.4 J. The final pressure, can be found using the ideal gas equation, of course, the final temperature is 273K (this is an isothermal process).

SynopsisEdit

  • Evaluated P∆V, and answered the value.

NotesEdit

One mistake that people may make is that the gas expands by 22.4 L, not that the final volume is 22.4L. In other words, the data is ∆V = 22.4 L. This mistake should be avoided, because in most competitive objective exams, an option leading from this described mistake will also be put up. Falling prey to that option means that you failed to gain easy marks, and at the same time losing marks which a person not attempting that question wont. This leads, ultimately, to losing ranks.

Question 4Edit

n moles of an ideal gas with molar specific heat Cv is taken from initial state T1/ V1 to final final volume V2 through a reversible adiabatic process. Show that the value of ∫PextdV is

nCvT_1\left [ 1 - \left ( \frac{V_1}{V_2} \right ) ^ \frac{R}{Cv} \right ]

First ThoughtsEdit

To many students, this problem might seem too obscure to induce any desire to solve it. Others might think it is familiar but would not attempt the problem just because of an unknown fear. Relax. The value of ∫PextdV is nothing but –nCv∆T = nCv(T1 - T2).

Our approach would be substituting each and every unknown we encounter, in the hope that if we do not make any mistakes, nothing can go wrong.

The SolutionEdit

Write down the required quantity, nCv(T1 - T2), and start manipulating it step by step. Take T1 out of the bracket first, because that is exactly the term that we are supposed to generate.

Now, we are left the bracketed term beside nCvT1. As we have mentioned before, if we do not make any mistake, we are sure to go home. This means that the complex looking bracket in the question, simplifies to 1 - (T2)/(T1). But how? We have done this several times before, while dealing with reversible adiabatic processes: the relation between the Temperature and Volume of the ideal gas at any stage in an adiabatic process, which gives us (T2)/(T1) = [(V1)/(V2)]^(y-1), where y stands for the adiabatic exponent of the gas.

(Do you remember this? PV^y = constant, and PV = nRT. On eliminating pressure from these two equations, we arrive at the relation TV^{y-1) = constant, giving T2)/(T1) = [(V1)/(V2)]^(y-1) when applied over the initial and final states. Please refer back to the tables, if you do not know this.)

But, the adiabatic exponent does not appear anywhere in the final term. So? Eliminate it, obviously. We know the molar heat capacity, and can report Universal Gas Constant as well. Y-1 is nothing but (Cp/Cv) -1 which simplifies to R/Cv, just what we need.

SynopsysEdit

  • Used (T2)/(T1) = [(V1)/(V2)]^(y-1) to eliminate T2
  • Used y = Cp/Cv to eliminate the adiabatic exponent, y itself.

Concluding NotesEdit

Do not be bamboozled by the perplexity of any problem, just try to make it seem easy.