Question

A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let ∆Q and ∆W represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods?
(a) ∆Q (b) ∆W (c) ∆Q + ∆W (d) ∆Q − ∆W.

Answer 1

Answer:

Correct Option D.

Explanation:

The quantity ΔQ-ΔW  represents the change in internal energy of the system (ΔU) which is a state function. i.e. it only depends on the initial and final states of the system not on the path followed.

By First Law Of thermodynamics we have,

ΔQ=ΔW+ΔU

ΔU=ΔQ-ΔU

Answer 2

option d is correct

Explanation:

Two different methods take a system from an initial condition to the final state. Therefore, the work done and the heat supplied will be different in both cases because they depend on the direction followed.

On the other hand, the system's internal energy (U) is a state function, i.e. it depends only on the final and initial state of the mechanism. The two strategies above are the same.

Using the first thermodynamic law, we obtain

$\begin{aligned}&\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}\\&\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}\end{aligned}$

Here, ΔU is the internal energy shift,  ΔQ is the heat supplied to the system, and ΔW is the system's function.