2. E and Q for a rising ballon: A room has a high ceiling and a sky light, and o

Question

2. E and Q for a rising ballon:

A room has a high ceiling and a sky light, and on a sunny day the bottom and the top of the room are at different temperatures, TB and TT respectively. Take the pressure to be constant throughout the room with p = patm. A balloon filled with a lighter-than-air ideal gas is held near the floor where it has volume VB. Then the balloon is released and it rises to the top of the room, where it comes to equilibrium with volume VT .

a) What is the work done by the air in the balloon? Give your answer in terms of the pressure and volumes.

b) Rewrite the result of (a) in terms of the the temperatures and the number of moles n of gas in the balloon.

c) Assume that the gas inside the balloon is a monatomic ideal gas. Find: the change in the internal energy of the gas, and the amount of heat absorbed or emitted As in part (b), give your answer in terms of n and TT , TB.

d) From (c) you can read off a specific heat. This is the specific heat at constant what? And the value is?

3. Computing specific heats:

Suppose that the energy and pressure of a particular gas are described by the following relations, E = 3/2 nRT , pV = nR(T + (alpha)T2) Note that hese are the same as for an ideal gas except for the (alpha)T2 in the equation of state. Here a is a constant.

a. What are the dimensions of alpha?

b. Find Cv , the specific heat at constant volume.

c. Find Cp, the specific heat at constant pressure.

PLEASE PLEASE HELP!

Explanation / Answer

Second question

A. Work done by air = Pressure *( change in volume)

      =Patm*(VT-VB)

B.Ideal gas law, PV=nRT. Therfore Work done by air = nR(TT-TB)

C.Change in internal energy = Cv*dt = Cv*(TT-TB).

    According to first law of thermodynamics the heat is,Q=Change in Internal energy+work done

     Therfore Q=Cv*(TT-TB)+ Patm*(VT-VB)

D.Cv= specific heat at constant volme = 0.718Kj/Kg K

Third question

a. Since the AlphaT2 has to add with T both should have same dimention. T has the same dimention with T2 alpha have zero dimnention.

b.E=W+change in internal energy

   E=W+Cv(T2-T1)

3/2nRT=nR(T+alphaT2)+ Cv dt, Therfore Cv= (3/2nRT-nR(T+alphaT2))/dt

c. Cp=R+Cv

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