Question
Question: A sample of gas absorbs \[4000{\text{ kJ}}\] of heat. If volume remains constant, what is \(\Delta {...
A sample of gas absorbs 4000 kJ of heat. If volume remains constant, what is ΔU?
Solution
We are given that heat of the system is 4000 kJ. Also, we are given that the volume of the system is constant. To solve this we must know the equation that gives the relationship between heat flow and the resulting change in the internal energy of the system at constant volume. ΔU is the change in internal energy of the system.
Complete solution:
We know that the equation for the first law of thermodynamics is as follows:
Δq=ΔU+ΔW
Where Δq is the change in the heat of the system,
ΔU is the change in the internal energy of the system,
ΔW is the work done.
The equation for work done is as follows:
W=PΔV
Where W is the work done,
P is the pressure on the system,
ΔV is the change in volume.
We are given that the volume of the system is constant i.e. ΔV=0 Thus,
W=0
Thus, when the volume is constant the work done is equal to zero. Thus,
Δq=ΔU
Thus, the change in the heat of the system is equal to the change in internal energy of the system.
We are given that the heat of the system is 4000 kJ. Thus,
ΔU=4000 kJ
Thus, the internal energy i.e. ΔU is 4000 kJ.
Note: The energy of a system that arises due to the molecular state of motion of matter is known as the internal energy of the system. As the temperature increases the phase of the matter changes from solid to liquid or from liquid to gas and thus, the internal energy increases. Internal energy is an extensive property of a system i.e. it depends on the mass of the system and it is a state function. Internal energy is denoted by symbol U.