- #1
Vitor Pimenta
- 10
- 1
Homework Statement
Below, two experiments (1 and 2) are described, in which the same quantity of solid carbon dioxide is completely sublimated, at 25ºC:
- The process is carried out in a hermetically sealed container, non-deformable with rigid walls;
- The process is carried out in a cilinder with a piston, whose mass is negligible and moves without friction
My problem is with the following statements:
a) ##\Delta {U_1} = \Delta {U_2}##
b) ##{q_1} = {q_2}##
Homework Equations
- First Law of Thermodynamics: ##\Delta U = q + w##
- Calorimetry equation for sublimation: ##q = n{C_s}## (in which ##n## is the number of moles of gas and ##{C_s}## is the molar latent heat of sublimation)
The Attempt at a Solution
[/B]
About process (1) , no work is done by the gas being generated (after all the container is rigid), so first law of thermodynamic reduces to ##\Delta {U_1} = {q_1}##.
Now, about process (2), work is done (since the piston can be moved by the gas being generated inside the cilinder), and the same law can only be written fully: ##\Delta {U_2} = {q_2} + {w_2}##
I further assume that, considering CO2 as ideal gas, the change in internal energy will be the same for both processes, since they start from the same state (solid, 25ºC) and end with gas at the same temperature (25ºC).
This way of thinking, however, would lead to ##{q_1} = {q_2} + {w_2}## , that is, the realization that heat exchanged by process (1) is not the same as by process (2). I reached the opposite conclusion of the exercise´s intended answer (since I think both changes in internal energies are the same and that the heat exchanged are different).
Lastly, I included calorimetry equation because that seems to be an argument for reaching the exercise´s intended answers: Assuming ##{C_s}## depends uniquely on temperature, its value would be the same for both processes (1 and 2), something that would lead to the conclusion that ##{q_1}={q_2}## (both processes employ the same ##n## of gas).