Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
D. An electron may acr with either particle like or wave like
Answer:
hello your question is incomplete attached below is missing part
answer : Pauli's exclusion principle
Explanation:
The principle that was disobeyed is <em>Pauli's exclusion principle</em> this is because the Pauli's exclusion principle states that no two electrons can have the same set of quantum number.
In Box A the two electrons , have the same spin which means they have the same quantum number ( disobeys Pauli's exclusion principle )
The nucleus contains protons and nuetrons