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Difference between Internal Energy and Enthalpy

A thermodynam...

Question

A thermodynamical system is changed from state $(P_{1}, V_{1})$ to $(P_{2}, V_{2})$ by two different processes, the quantity which will remain same will be

A

Δ Q

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B

Δ W

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C

Δ Q + Δ W

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D

Δ Q - Δ W

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Solution

The correct option is D $Δ Q - Δ W$
For all processes, change in internal energy $Δ U = (Δ Q - Δ W)$ does not change. It depends only on initial and final states.

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Q. 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.

Q. Match the thermodynamic processes taking place in a system with the correct conditions.
In the table,

Δ Q

is the heat supplied,

Δ W

is the work done and

Δ U

is the change in internal energy of the system.

Process	Condition
(i) Adiabatic	(A) $Δ W = 0$
(ii) Isothermal	(B) $Δ Q = 0$
(iii) Isochoric	(C) $Δ U \neq 0$ , $Δ W \neq 0$ , $Δ Q \neq 0$
(iv) Isobaric	(D) $Δ U = 0$

Δ U_{1}

Δ U_{2}

be the change in internal energy in process A and B respectively,

Δ Q

be the net heat given to the system in process (A+B) and

Δ W

be the net work done by the system in the process (A+ B).
i)

Δ U_{1} + Δ U_{2} = 0

Δ U_{1} - Δ U_{2} = 0

Δ Q + Δ W = 0

Δ Q + Δ W = 0

Q. Refer to figure. Let ∆U₁ and ∆U₂ be the change in internal energy in processes A and B respectively, ∆Q be the net heat given to the system in process A + B and ∆W be the net work done by the system in the process A + B.

(a) ∆U₁ + ∆U₂ = 0.
(b) ∆U₁ − ∆U₂ = 0.
(c) ∆Q − ∆W = 0.
(d) ∆Q + ∆W = 0

Q. A thermodynamic system is changed from state

(P_{1}, V_{1})

(P_{2}, V_{2})

by two different processes, the quantity which will remain same will be :

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