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In an adiabatic expansion of ideal gas:
A: $W = - \Delta E$
B: $W = \Delta E$
C: $\Delta E = 0$
D: $W = 0$

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Last updated date: 14th Jun 2024
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Answer
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Hint: Adiabatic expansion in thermodynamics refers to a process in which the volume is increasing and there is no heat transferred into or out the system. It is not necessary that temperature will be constant.

Complete step by step answer:
We know that an adiabatic process in thermodynamics is a process in which there is no transfer of heat into or out the system. So, if we apply the first law of thermodynamics in chemistry the relation we obtain between $\Delta E,\Delta Q$and $W$is $\Delta E = \Delta Q + W$.
Where $\Delta E$is the internal energy of the system, $\Delta Q$is the change in heat and $W$is the work done. In adiabatic process $\Delta Q = 0$, so $W = \Delta E$.

So, the correct answer is Option B .

Additional Information:
In thermodynamics an adiabatic process is denoted by $P{V^\gamma } = $constant. Where $\gamma = \dfrac{{{C_P}}}{{{C_V}}}$. In adiabatic expansion the volume of the system increases. We can obtain an adiabatic process only when we can surround the system with a thermally insulated material which will not allow any heat to flow outside or inside the system. If we observe practically it is nearly impossible to build a system which is truly adiabatic. Isothermal process is a thermodynamic process in which $\Delta T = 0$.

Note:
The value of internal energy of a thermodynamic process can be calculated by $\Delta E = n{C_V}\Delta T$. The magnitude of the work done by an isothermal in expansion as well in compression is always greater than the magnitude of the work done in an adiabatic process. If an ideal gas is adiabatically expanded in vacuum, the process becomes isothermal