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: The sun is closer to the Earth is the summer and farther in the winter causing it to be hotter than it is colder in the summer and vise versa for the other question.
Explanation: I learned it ;)
Answer:
The elements in same period have same principle quantum number or energy shell.
The elements in same group shows similar chemical and properties.
Explanation:
Inn group:
The elements in same group i.e present in vertical column shows similar chemical properties.
The elements in same group having same number of valance electrons. while in chemical reaction bonds are break and formed and valance electrons are involved. That's why elements in same group having same number of valance electrons and shows similar chemical properties.
In period:
While as we move from left to right the number of valance electron increase by one in every element. But the electron is added in same shell which means that their physical and chemical properties are different but principal quantum number is same.
