• catdog90210
In summary, the problem involves a cylinder A containing a monatomic ideal gas, with a piston maintaining a constant pressure P0. The cylinder is connected to an evacuated vessel B by a capillary tube, and when a valve is opened, the gas is allowed to enter B and eventually reaches a new equilibrium state at pressure P0. The problem asks to show that the equilibrium temperature of the gas in B is (5/3)T0. Using conservation of energy, the final temperature can be calculated by equating the loss of gravitational potential energy to the increase in internal energy of the gas in B. The final temperature in B can then be found using the ideal gas law and the specific heat capacity of the gas.

## 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
$$\Delta$$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$$\Delta$$V.

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:
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

## What is an adiabatic process in thermodynamics?

An adiabatic process is a thermodynamic process in which no heat is transferred between the system and its surroundings. This means that the system is completely isolated and there is no exchange of thermal energy with the surroundings.

## What are some examples of adiabatic processes?

Some examples of adiabatic processes include the expansion or compression of a gas in a piston, the flow of air over an airplane wing, and the compression of a gas in a refrigerator.

## How does an adiabatic process affect the temperature of a system?

In an adiabatic process, the temperature of a system can change due to work being done on or by the system. For example, when a gas is compressed, its temperature will increase due to the work being done on it. Similarly, when a gas expands, its temperature will decrease as it does work on its surroundings.

## What is the first law of thermodynamics in relation to adiabatic processes?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In an adiabatic process, the change in internal energy of the system is equal to the work done on or by the system.

## How is an adiabatic process different from an isothermal process?

An adiabatic process does not allow for any heat transfer, while an isothermal process maintains a constant temperature by allowing for heat transfer. In an adiabatic process, the change in temperature is a result of work being done on or by the system, whereas in an isothermal process, the change in temperature is due to heat transfer.