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.
Answer:
B: Increasing the volume inside the reaction chamber
Explanation:
Explanation:
the lone pairs will be negatively charged. these have a repulsion effect on other negatively charged electrons in the shells of atoms. picture a water molecule: the lone electron pair on the top of the oxygen will have a repulsion force on the 2 hydrogen atom's orbiting electrons to cause a bent molecular geometry.
High temperature and low pressure<--Most likely
Low temperature and high pressure<----Less likely.
So the answer to this is Low temperature and high pressure.
