Thermodynamics - Adiabatic Process

AI Thread Summary
In the discussion on thermodynamics regarding an adiabatic process, a monatomic ideal gas in a cylinder is analyzed as it expands into an evacuated vessel. The system maintains a constant pressure, and the initial temperature is T0. When the valve is opened, the gas undergoes a free expansion into vessel B, where the final pressure remains P0. The key to determining the equilibrium temperature in vessel B is recognizing that the energy lost from the gravitational potential energy of the piston is converted into the internal energy of the gas, leading to an increase in temperature. Ultimately, the equilibrium temperature in vessel B is shown to be (5/3)T0.
catdog90210
Messages
1
Reaction score
0

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
\DeltaU = 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\DeltaV.

I'm thinking that the only way to get the final temp in B is by using U = (1/2)*fnRT but I can't 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.
 
Last edited:
Physics news on Phys.org
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 \Delta V be the volume of B (and the change in volume of A):

P_0\Delta V = \Delta U_B = n_BC_v\Delta T_B

The trick is in finding n_B.

AM
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top