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## Tuesday, June 3, 2008

### AP Physics B- Answer to Free Response Question on Thermodynamics

In the post dated 31st May 2008, two free response practice questions on thermodynamics were given to you. As promised, I give below model answers along with the questions:

(1) One mole of an ideal mono atomic gas at standard temperature (T0) and standard pressure (P0) is contained in a cylinder fitted with a frictionless piston. The gas has volume V0 under these conditions. The following three operations are performed on the gas:

(i) The gas is heated at constant volume so that its pressure is doubled.

(ii) The gas is then made to undergo isothermal expansion so that the pressure is restored to the initial value.

(iii) The gas then undergoes an isobaric compression so that its volume is restored to the initial value.

(a) Draw a P-V diagram to represent the processes numbered (i), (ii) and (iii) above, marking the initial (as well as final) state of the gas by point A, and the intermediate states by points B and C.

(b) What is the final temperature of the gas? Justify your answer.

(c) During the process (i) what is the work done by the gas? Justify your answer.

(d) Determine the numerical value of the volume (V2) of the gas at the end of the isothermal expansion mentioned in process (ii).

(e) If the work done by the gas during its isothermal expansion is given to be 1.45P0V0, calculate the net work done during the entire cycle of operations involving the processes (i), (ii) and (iii). (a) The required P-V diagram is shown in the adjoining figure.

(b) As shown in the P-V diagram, the pressure and the volume of the gas are finally rstored to the initial values P0 and V0 respectively. Since PV/T is a constant for a given mass of ideal gas, the temperature of the gas must be equal to the initial temperature T0.

(c) In process (i) the volume is constant. The work done is the integral of PdV where dV is the change in volume at pressure P. Since there is no change in volume, no work is done in process (i).

[To explain this in more fundamental terms, the point of application of force does not move when the volume is constant and hence no work is done].

(d) We have PV = nRT where P is the pressure, V is the volume, R is the universal gas constant, T is the absolute (Kelvin) temperature and n is the number of moles of the gas.

Therefore, V = nRT/P

Substituting the values appropriate to the end of process (ii),

V2 = 1×RT/P0 where T is the temperature at which the isothermal expansion occurs.

From state A (P0, V0, T0) as shown in the P-V diagram the gas changes to state B (2P0,V0,T). Therefore, P0V0 /T0 = 2P0V0 /T from which T = 2T0.

Substituting for T in the expression for V2 we obtain

V2 = 1×RT/P0 = R×2T0 /P0 = (8.31×2×273)/(1.0×105) = 0.045 m2, nearly.

(e) The work done by the gas during the isothermal expansion is the area under the curve BC in the P-V diagram. The work done on the gas during the isobaric compression is the area under the curve CA. The curve AB represents the isochoric (constant volume) change which involves no work. Therefore, the net work (W) done (by the gas) during the entire cycle of operations involving the processes (i), (ii) and (iii) is given by

W = Area under the curve BC Area under the curve CA

= 1.45P0V0 – P0V0 = 0.45 P0V0

Note:

You may get simpler questions with variations. For instance, a P-V diagram similar to the one you have drawn may be given in the question itself and you may be asked to state whether heat energy is added to the gas or removed from the gas during the process indicated by curve AB. Indeed heat is added since the pressure is increased at constant volume and this can happen only if the temperature of the gas is increased. You may be required to calculate the heat absorbed by the gas during the process AB. The heat absorbed is nCv(T –T0) where n is the number of moles of the gas (n = 1 in the above question), Cv is the molar specific heat at constant volume and T is the temperature of the gas in state B. You can calculate T using 2P0V0/T = P0V0/T0. The value of Cv may be given in the question, but even if it is not given you should know the value for a mono atomic gas to be (3/2)R where R is the universal gas constant.

(2) Two moles of nitrogen are contained in a vertical conducting cylinder provided with a frictionless piston of negligible mass and area equal to 1.5×10–2 m2 (Fig.). Nitrogen behaves as an ideal gas. It is brought to the following states through a sequence of operations as follows: (i) Initially the temperature of nitrogen is the same as the laboratory temperature of 27º C and its volume is V1. This state of the gas is designated as state A

(ii) The piston is now pulled up with a gradually increasing force of final value 60 N, maintaining the temperature at 27º C itself. The volume of the gas now becomes V2. This state of the gas is designated as state B.

(iii) The temperature of the gas is now increased to 100º C by placing the cylinder in contact with a bath containing boiling water. The volume of the gas now becomes V3. This state of the gas is designated as state C.

(iv) With the cylinder still in contact with the boiling water bath, the pulling force of 60 N on the piston is now withdrawn and the gas is allowed to get compressed to state D.

(v) The gas in the cylinder is now subjected to a certain process in which the piston moves up through a distance of 0.1 m without any change in the pressure of the gas. The gas attains state E after the expansion. Now answer the following:

(a) Calculate the volume (V1) of the gas in state A

(b) Indicate by putting a tick (√) mark whether the process from state A to state B is adiabatic, isothermal or isobaric

(c) Calculate the pressure of the gas in state C.

(d) Between which states work is done on the gas? Indicate by tick mark:

A and B____ B and C____ C and D____ D and E____

(e) Calculate the work done when the gas changes its state from D to E as mentioned in (v) above.

(a) The pressure of the gas is initially equal to the atmospheric pressure P0 = 1.0×105 Nm–2.

We have PV = nRT, where P is the pressure, V is the volume, R is the universal gas constant, T is the absolute (Kelvin) temperature and n is the number of moles of gas in the given sample.Therefore we have

V = nRT/P

Substituting the values appropriate to the state A of the nitrogen sample, we obtain

V1 = 2×RT/P0 = (2×8.31×300)/(1.0×105) = 0.05m2, nearly.

(b) The process from state A to state B is isothermal since the temperature remains constant through out. The pressure of the gas changes and hence it is not isobaric. The cylinder is in not thermally insulated from the surroundings and there is enough time for the temperature of the gas to attain the value of 27º C through out the process since the full force is not applied abruptly.

(c) In state C, the pressure of the gas is the difference between the atmospheric pressure and the opposing pressure (force per unit area) produced by the pulling force.

Therefore, pressure = 1.0×105 – [60/(1.5×10–2)] = 0.96×105 Nm–2.

(d) Work is done on the gas when the gas is compressed. This happens when the pulling force is withdrawn, between states C and D

(e) The change from state D to state E is through an isobaric process and the work done (by the gas) is PdV = PAdx = (1.0×105)×(1.5×10–2) ×0.1 = 150 J.