Answer:
V2 = 90.7 mL
Explanation:
pressure and volume are inversely proportional, if the pressure is increased, the volume will decrease. In an isothermal process:
p1V1 = p2V2
V2 = p1V1/p2 = (277 torr×187 mL)/571 torr
V2 = 90.7 mL
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:
MgO- magnesium oxide
Cu(NO3)2- copper(11)nitrate
Li2CO3- lithium carbonate
Answer:
It is better to do chemistry
Explanation:
So that you will learn more of chemicals
Answer:
Conduction
Explanation:
Conduction is the transfer of heat energy from one substance to another or within a substance.
