Calculating Work in a Thermally Insulated Cylinder

In summary, the pressure is not constant for each compartment, and the pressure in each compartment is certainly not half the total pressure.
  • #1
SPhy
25
0

Homework Statement



A gas-tight frictionless piston of small thermal conductivity slides in a thermally insulated cylinder, dividing it into two compartments, A and B, each containing nA=n and nB=0.45n moles of ideal monatomic gas (n moles). Initially the temperature of the gas is To in compartment A and 4To in B. Assume that the system is in mechanical equilibrium at all times and that the mass of the piston and the effect of gravity are negligible. Consider the process by which the system reaches thermal equilibrium:

(a) What is the final temperature, Tf?

(b) What is the ratio of the volume of A to that of B initially, ri?

(c) What is the ratio of the volume of A to that of B after thermal equilibrium is reached, rf?

(d) Calculate the work done on the gas in compartment A, and then do the same for compartment B.I managed to answer (a), (b), and (c) correctly, getting values of 1.93To, 0.55, and 2.22 respectively.

Having a bit of difficulty with (d)

Homework Equations



dW = -PdV

W = -∫ Pdv = -∫ (nRT/V)dV

The Attempt at a Solution

The answer is to be in terms of nRT

I computed the work integral and I have,

W = -nRT ln (Vfa/Via)

Getting stuck as to what the ratio of Vfa to Via is.

I played around with the ideal gas law and have,

(Vfa/Via) = (1.93Pia/Pfa)

From here I am just running in circles.

Any help is appreciated!
 
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  • #2
SPhy said:
(Vfa/Via) = (1.93Pia/Pfa)
why?? I would rather use Charle's law .. constant Pressure
this is because the system is not affected by any external agent so system pressure remains constant, also its a frictionless piston which maintains equal pressure b/w the 2 compartments! so pressure is a constant for each compartment = ##\frac{ P_{total}}{2}##. Can you do the rest?
 
  • #3
deleted
 
  • #4
Suraj M said:
why?? I would rather use Charle's law .. constant Pressure
this is because the system is not affected by any external agent so system pressure remains constant, also its a frictionless piston which maintains equal pressure b/w the 2 compartments! so pressure is a constant for each compartment = ##\frac{ P_{total}}{2}##. Can you do the rest?
This is not correct. The pressure is not constant for each compartment, and the pressure in each compartment is certainly not half the total pressure.

Chet
 
  • #5
Let P represent the pressure within both chambers (the process is quasistatic and the piston is massless), and let V be the total volume of the two chambers. If you did part b, then you know the ratio of the volumes initially. In terms of V, what are the individual volumes? In terms of V and T0, what is the initial pressure P0.

You know that the sum of the internal energies of the gases in the two chambers is constant during the process. If, during the transition from the initial to the final state, the temperature of the gas in chamber is TA and the temperature in chamber B it TB, express TA in terms of TB. In terms of TA and TB, what is the ratio of the volumes in the two chambers? In terms of TA and TB and V, what are the individual volumes in the two chambers? In terms of TA, TB, and V, what is the pressure in the two chambers?

Chet
 
  • #6
Chestermiller said:
The pressure is not constant for each compartment,
There is a massless, frictionless piston between the 2 compartments, doesn't that maintain the constant pressure between the compartments? Yes but not half of total, i was wrong there!
 
  • #7
The system is isolated, so the internal energy is constant. It is ideal (monoatomic gas), so the total energy is U=3/2 nRTA+3/2 0.45 nRTB=const.

The total volume VA+VB is also constant, and the pressure is equal at both sides.
p(VA +VB) = nRTA+0.45 nRTB= 2/3 U . The pressure has to be constant during the process.
 
  • #8
Suraj M said:
There is a massless, frictionless piston between the 2 compartments, doesn't that maintain the constant pressure between the compartments? Yes but not half of total, i was wrong there!
The pressures in the two compartments are equal to one another, but your implication that the pressure remains constant at its initial value during the entire process was far from obvious (at least to me). ehild's analysis in post #7 elegently demonstrated this.

Chet
 
  • #9
Chestermiller said:
This is not correct. The pressure is not constant for each compartment,

Chet

That was a hasty statement ... :devil:
 
  • #10
ehild said:
That was a hasty statement ... :devil:
Yes. I made this statement before I had solved the problem to completion, but, even if the assumption were made that the pressure was not constant, the analysis would have led to the conclusion that it was. I doubt that the member I had responded to had obtained this result by analyzing the problem, given the remainder of his response. In retrospect, I probably should have said "The pressure is not necessarily constant for each compartment"

Chet
 
  • #11
Chestermiller said:
Yes. I made this statement before I had solved the problem to completion, but, even if the assumption were made that the pressure was not constant, the analysis would have led to the conclusion that it was. I doubt that the member I had responded to had obtained this result by analyzing the problem, given the remainder of his response. In retrospect, I probably should have said "The pressure is not necessarily constant for each compartment"

Chet

You can not read the mind of other people. Suray M guess was right, although not a complete proof.
Yes, it would have been more appropriate to say that " the pressure was not necessarily constant" during the process.
 
  • #12
SPhy said:
Having a bit of difficulty with (d)

Homework Equations



dW = -PdV

W = -∫ Pdv = -∫ (nRT/V)dV

The Attempt at a Solution

The answer is to be in terms of nRT

I computed the work integral and I have,

W = -nRT ln (Vfa/Via)
That is wrong, T changes during the process, you can not pull it out from the integral. Find how Va, Ta, Vb, Tb, and P change during the process.
Hint: The total volume of the compartment is constant. The total energy of the whole system is also constant. It is an ideal gas. How is the internal energy of the ideal gas related to the temperature?
 

What is the definition of work in thermodynamics?

Work in thermodynamics is defined as the transfer of energy from one system to another due to a difference in temperature or pressure. It is often represented as the area under the curve on a pressure-volume diagram.

How is work related to the first law of thermodynamics?

The first law of thermodynamics states that energy can neither be created nor destroyed, only transferred or converted. Work is one of the ways in which energy can be transferred between systems, and it is an important component of the first law of thermodynamics.

What is the difference between work and heat in thermodynamics?

In thermodynamics, work is the transfer of energy due to a difference in temperature or pressure, while heat is the transfer of energy due to a temperature difference. Work is a form of energy transfer that can be controlled, while heat is a spontaneous transfer of energy.

How is work calculated in thermodynamics?

The amount of work done in a thermodynamic process can be calculated by multiplying the force applied to a system by the distance the system moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

What are some real-world examples of work in thermodynamics?

Some examples of work in thermodynamics include the compression of a gas in an engine, the expansion of a gas in a turbine, and the movement of a piston in a steam engine. Work is also involved in everyday activities such as pumping gas, turning on a light switch, and using a refrigerator.

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