Respuesta :
Answer:
See explanation.
Explanation:
In thermodynamic we used to use the expression [tex] \varDelta U=Q+W[/tex] for conservation of energy that is the first law of thermodynamics with [tex] \varDelta U [/tex] the internal energy of the system (U is directly proportional to temperature), Q the net heat and W the net work. In our case Q=0 because the gas is insulated so there's no exchange of heat with the environment:
[tex]\varDelta U=W [/tex]
Work for perfect gases is approximtely [tex] -P\Delta V[/tex]
with P pressure (is constant in our case and always positive) and V volume.
[tex] \varDelta U=-P\Delta V=-P(V_{final}-V_{initial})[/tex]
Note if the gas is compressed [tex] V_{final}-V_{initial}[/tex] is negative (because [tex] V_{final}<V_{initial}[/tex]) so [tex] \varDelta U[/tex] is positive that means U is going to increase so the temperature too because they are proportional.
In the other case if the gas expands [tex] V_{final}-V_{initial}[/tex] is positive (because [tex] V_{final}>V_{initial}[/tex]) and [tex]\varDelta U [/tex] is negative so U is going to decrease and temperature too.
From the principle of conservation of energy which is used in first law of thermodynamics, it has been established that;
When gas is compressed, temperature increases and when gas expands, temperature decreases.
The mathematical form of the first law of thermodynamics is;
Q = U + W ---(eq 1)
Where,
Q = the total heat transferred,
U = the internal energy and
W = the work done.
When we differentiate the equation above, we will get;
Δ Q = Δ U + Δ W ---(eq 2)
We are told in the question, that the gas is well insulated. This means the walls of the container of the container carrying the gas is well insulated. We can as a result of the insulation, say that there is no heat transfer between the system and the environment.
Thus;
Δ Q = 0
∴ the work done is given as;
Δ W = P Δ V
Putting P Δ V for Δ W in eq 2 gives;
Δ Q = Δ U + P Δ V
Since Δ Q = 0, then;
Δ U + P Δ V = 0
Δ U = -P Δ V
- When gas is compressed, the final volume will be less than the initial volume and as such Δ V will be negative. This means that Δ U will be positive which denotes that it increases which is a sign of increase in temperature.
- In contrast, when the gas expands, the final volume will be more than the initial volume and as such Δ V will be positive. This means that Δ U will be negative which denotes that it decreases which is a sign of decrease in temperature.
Read more at; https://brainly.com/question/19863474
