Answer:
at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.
Explanation:
Give brainliest
The pressure inside the can upon cooling is 0.4 atm.
<u>Explanation:</u>
Given -
Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K
Final Temperature, T2 = 208°C = 208 + 273 K = 481 K
Pressure upon cooling, P2 = ?
Using Gay Lussac's law:
P1/T1 = P2/T2
P2 = P1 X T2 / T1
P2 = 1 atm X 481 / 1181
P2 = 0.4 atm
Therefore, the pressure inside the can upon cooling is 0.4 atm.
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
