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Thermodynamics - Adiabatic Process

  • #1

Homework Statement



A cylinder, A, containing a monatomic ideal gas, is fitted with a frictionless leakfree
piston. The axis of the cylinder is oriented vertically, as in the figure below,
so that the weight of the piston maintains the gas at a constant pressure P0. The
cylinder is linked to an evacuated vessel, B, by a thin capillary tube. Initially the
valve, V, shown in the figure, is closed and the temperature of the gas is T0. Both
A and B are adiabatically isolated from their surroundings, and heat conduction
along the capillary can be ignored.
The value, V, is opened and the gas is admitted to B. The system slowly adjusts
to a new state of equilibrium, and the final pressure is P0. Show that the
equilibrium temperature of the gas in B is (5/3)T0.

Figure attached.

Homework Equations



If f is the quadratic degrees of freedom:

PV = nRT
PVk = constant where k = 1 + 2/f
[tex]\Delta[/tex]U = Q + W
U = (1/2)*fnRT
VTf/2 = constant

The Attempt at a Solution



I know f=3 because its monoatomic and Q=0 because its adiabatic.

I thought that the lowering of the piston in A is quasistatic so the above equations all apply but gas entering B is not quasistatic so we can only worry about its final equilibrium state.

The piston lowering is doing work on the gas increasing its energy by W= -P[tex]\Delta[/tex]V.

I'm thinking that the only way to get the final temp in B is by using U = (1/2)*fnRT but I cant seem to make the connection between what happens in A and B.

I thought that if I found the energy in A initially, UA,i = (1/2)*fnRT0, and added the work done I could have the final energy in the system and then somehow relate this to B.

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,584
346
You cannot use the adiabatic condition because the process is not quasi-static. This is a free expansion into B from A, at least initially.

What you can use is conservation of energy since the loss of gravitational potential energy is ultimately converted into internal energy of the gas in B. This energy increases the temperature of the gas in B.

Let [itex]\Delta V[/itex] be the volume of B (and the change in volume of A):

[tex]P_0\Delta V = \Delta U_B = n_BC_v\Delta T_B[/tex]

The trick is in finding n_B.

AM
 

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