I've read a solution of a problem, in which there are two different gases in a container, initally at equilibrium and separated by an adiabatic fix wall. At some time, this wall is changed by a diathermic mobile wall, so the equilibrium point changes. You have to find the final state of the...
Ok. What about this:
Lower chamber:
\Delta U_1 = Q_{stirrer} - Q_{up} + p\Delta V
So using the expression for the internal energy and the constant volume:
2R\Delta T = 500 - Q_{up}
Upper chamber:
\Delta U_2 = Q_{up} - p\Delta V
So:
4R\Delta T = Q_{up} - p_0\Delta V
Adding up these...
Homework Statement
We have a cylindrical container filled with a gas: the container has an upper compartment with 2 moles of this gas, and another compartment below with 1 mole of the same gas, separated by a diathermic boundary, and also has an adiabatic mobile plunger over the upper...
Well, I'm not sure of the initial condition, that's why I asked ;)
I guess it must be zero internal energy when temperature is zero, so then we can drop de deltas.
Observation: I changed a little the statement of the problem, as I noticed I had additional constraints.
Now to the problem: I have found the equation (which I didn't remember...)
dU =C_{V}dT +\left[T\left(\frac{\partial p}{\partial T}\right)_{V} - p\right]dV
So with the equation of...
Given the following equation of state and C_p = 3R, find the equation of the internal energy in terms of T and V
p(V-b) = RT
Any clues? I can use also the specific heat at constant pressure or volume. Thanks!
Hello everybody! This is my first thread and I'm just kind of discovering the forum. If I should make a presentation or should have posted this thread somwhere else, let me know please!
I have plenty of books and notes about a first course in Thermodynamics, but I'm having a bit of trouble when...