The relation between change in internal energ | vedclass

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(A) The first law of thermodynamics states that $\Delta E = q + w$. For a process at constant pressure,the heat exchanged is equal to the change in en

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$\Delta H = \Delta E + W$

Vedclass · Answer

(A) The first law of thermodynamics states that $\Delta E = q + w$. For a process at constant pressure,the heat exchanged is equal to the change in enthalpy,i.e.,$q = \Delta H$. Substituting this into the equation,we get $\Delta E = \Delta H + w$. Rearranging this gives $\Delta H = \Delta E - w$. However,in many conventions,work done by the system is represented as $W = -P\Delta V$,leading to $\Delta H = \Delta E + P\Delta V$. If $W$ is defined as the work done on the system $(W = P\Delta V)$,then the relation is $\Delta H = \Delta E + W$.

The relation between change in internal energy $(\Delta E)$,change in enthalpy $(\Delta H)$,and work done $(W)$ is represented as:

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