A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let and 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? 1. 2. 3. + 4. -

A system can be taken from the initial state p1, V1 to the final …

Question

A system can be taken from the initial state p₁, V₁ to the final state p₂, V₂ by two different methods. Let $\Delta\text{Q}$ and $\Delta\text{W}$ 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?

$\Delta\text{Q}$
$\Delta\text{W}$
$\Delta\text{Q}+\Delta\text{W}$
$\Delta\text{Q}-\Delta\text{W}$

✓

Answer

$\Delta\text{Q}-\Delta\text{W}$

Explanation:

A system is taken from an initial state to the final state by two different methods. So, work done and heat supplied in both the cases will be different as they depend on the path followed. On the other hand, internal energy of the system (U) is a state function, i.e. it only depends on the final and initial state of the process. They are the same in the above two methods.

Using the first law of thermodynamics, we get,

$\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}$

$\Rightarrow\Delta\text{U}=\Delta\text{Q}-\Delta\text{W}$

Here, $\Delta\text{U}$ is the change in internal energy, $\Delta\text{Q}$ is the heat given to the system and $\Delta\text{W}$ is the work done by the system.

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