Last modified on 28 October 2010, at 16:35

Problems In Highschool Chemistry/PhChem/thermo1/Problem1

Read the paragraph in "The Setup" section and answer the questions that follow it. This question is supposed to replace a theory part explaining obvious facts with an interactive session.

The SetupEdit

A horizontally aligned cylinder is divided into two sections by a vertical massless frictionless sliding adiabatic piston. Both the left and the right sections have one mole each of an ideal gas (whose molar specific heat is 20 J/mol K) at 300 K and 2.0 L. The ideal gas in the right section is maintained at a constant temperature by means of a diathermic contact with some external agent. The ideal gas in the left section, however, is in contact with a heater, which heats it slowly so that the left section expands. By the time the heater stops supplying heat, the volume of the right section has decreased to 1.0 L.

Now answer the accompanying questions.

  • Note - Before solving the problem, make sure that you understand the whole of it. This problem is of intermediate level. Like most questions in physical chemistry, it seems tough only because of its length. To understand the whole scenario in this (and every other) question, note down the necessary data and diagrams where you are solving the problem. Doing so will familiarize you with the problem, and give you some mental exercise.
  • So before you start to solve, draw a cylinder resting on its curved surface, divided into two parts. To each part assign the values of the gas parameters, and note which section will be heated and which one will be maintained at a constant temperature.

The QuestionsEdit

Question 1Edit

What process is the gas in the right section R undergoing?

Since it is maintained at a constant temperature, it is undergoing an isothermal process. Also, since the volume change is taking place slowly (quasi-statically), it is a reversible process. It is thus, a reversible isothermal process. The diathermic contact with an external agent just means heat can be exchanged between section R and the external agent. This is used to maintain the constant temperature in section R.

Question 2Edit

What is the significance of the adiabatic piston in between Sections L and R?

The adiabatic piston ensures that heat does not flow between the two sections. That is, the heat supplied to section L is utilized in heating the gas and doing PV work only, and is not passed on to section R.

Question 3Edit

What is then, the work done by the gas in the section R?

Use the expression/formulae for isothermal process ∫PextdV = nRTln(Vf/Vi). All of the values are known, n=1, T = 300 K constant, Vf = 1.0 L and Vi=2.0 L. The answer comes out to be -1.73 x 10^3 J. Since by IUPAC convention, work done by the gas has the same magnitude but opposite sign as ∫PextdV, the final answer is 1.73 x 10^3 J.

Question 4Edit

∆E and q for the gas in section R are ___J and ___J respectively.

Well, in an isothermal process, ∆E is always zero, because for an ideal gas, Internal energy depends only on temperature, and the temperature remains constant. q is to be found using the First Law Of Thermodynamics, and it comes out to be -1.73 x 10^3 J

Question 5Edit

When the heating stops, the piston stands still. What does this tell us about the pressures of the ideal gases in the two sections?

If the piston stands still, it means it is in a state of mechanical equilibrium. (Can you explain satisfactorily the difference between mechanical and thermodynamic equilibrium? Refer any textbook or wikipedia and note the answer just now if you cannot.) This makes us conclude that the net force acting on it is zero.

In other words, the force from the gas in section R and the force from the gas in section L are same in magnitude and opposite in sign. Equating their magnitudes, Pr x Piston Area = Pl x Piston Area. That is, the pressures of ideal gases in both the sections are same! This is a very important result, do remember how we concluded it.

Question 6Edit

If the piston had some finite mass, and the cylinder was aligned vertically, would the pressures be same in case of equilibrium?

Definitely not! You can draw a free body diagram of the piston and check. The ideal gas in the bottom section has to counterbalance the force due to the gas above, as well as the gravitational force acting downward. This means, the gas below will be at a higher pressure, by an amount (Piston Mass)g/(Piston Area) pascals. (Convince yourself about this point.)

Question 7Edit

By performing actual integration, find the value of ∫PextdV for the gas in Section L.

This is a very nice question, because it helps to clarify the significance of each and every variable. Pext for the gas in section L is nothing but the pressure against which it is expanding. This pressure is the pressure of the gas in section R.

∫PextdV = ∫PrdVl

Now we substitute Pr as nRTr/Vr.

∫PextdV = ∫PrdVl = ∫(nRTr/Vr)(dVl) = nRTr∫(dVl)/(Vr)

(The numerator was taken out of integration, because it is a constant)

Now comes the main step of the question. Can we integrate the variable Vr with respect to the variable Vl? Definitely not. So we substitute the variable Vr as [4.0 L - Vl] and carry out the integration from limits 2.0L to 3.0 L. This is because, the total volume of the cylinder will remain constant at 2.0 + 2.0 = 4.0 L, and if the volume of one section increases by some amount, the volume of the other section decreases by the same amount.

(The substitution Vr = 4.0L - Vl is similar to the case of evaluating ∫PextdV for an ideal gas in an isothermal process. There, since we could not integrate P wrt V, we substituted it as nRT/V)

The definite integration yields the value of the integral to be (nRTr)(-1)(ln[{4-3}/{4-2}]) = nRTrln2 = 1.73 x 10^3 J same in magnitude, but opposite in sign, as ∫PextdV for section R. Note this point carefully.

Question 8Edit

We substituted Pext = Pr = nRTr/Vr. = nRTr/(4.0 - Vl). But we do know that since this is a quasi static process, the piston is always in mechanical equilibrium and hence the pressures of gases in both the section are always same. So why did not we substitute Pext = Pr = Pl = nRT/Vl?

Definitely we could have done this. But take a look back at Q7. The numerator nRTr was constant an could be taken out of integration. Is this the case with nRTl as well? No. We would again need to substitute Tl as some function of Vl, so that we could carry on with the integration. All in all, what we did was correct.

Question 9Edit

Prove that the values of ∫PextdV for the two sections are always going to be the additive inverse of each other for this scenario.

Consider ∫PextdVr + ∫PextdVl. To prove our point, we will prove this expression to be zero, no matter what. The first point is, as we discussed above, the Pext for one section is the pressure of the ideal gas in the other section. Now, both are equal at any point of time in our case (because the process is quasi static, both mechanical and thermodynamic equilibrium are maintained at every stage. Convince yourself that this statement leads to our conclusion.}

Also, we have already proved before that the change in the volume of one section is the aditive inverse of the change in the volume of the other section. See Q7, and by the way, this is obvious. So,

dVr = -dVl.

Hence, ∫PextdVr + ∫PextdVl. = ∫PextdVr - ∫PextdVr = 0.

Question 10Edit

Find the work done w by the gas in section L, its ∆E and the heat supplied by the heater.

  • w = - ∫PextdV = -1.73 x 10^3 J
  • ∆E = nCv∆T. Now finding the change in temperature is a different task altogether. But use the skills that you derived from solving the table. The initial temperature is known to us. But for the final state, we only know the volume. The temperature and pressure are not known. We need two equations then. The first is the ideal gas equation. The second is obviously, Pl = Pr. = nRTr/Vr.
  • Substitute this value of Pl in the ideal gas equation (do not obtain the numerical value, it will just increase some calculation). After a lot of cancellation (to our satisfaction), we obtain Tl = 600K. n=1 moles, Cv = 20.0 J/K mol and ∆T = 300K. This gives the answer to be 1.20 x 10^4 J
  • The heat supplied by the heater is nothing but q for section l, which is to be found using the first law of thermodynamics and is found equal to, after rounding off, 1.4 x 10^4 J.