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Which thermodynamic parameter is not a state function?
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Verified Answer
The correct answer is:
$W$ at isothermal
$H$ and $U$ are state functions but $W$ and $q$ are not state functions.
From the equation, $\Delta H=\Delta U+\Delta P V$
At constant pressure, $\Delta H=\Delta U+P \Delta V$
At constant volume, $\Delta H=\Delta U+V \Delta P$
At constant pressure, $\Delta P=0, \Delta H=q_p$
so, it is a state function.
At constant volume, $\Delta V=0, \Delta U=q_v$ so, it is a state function.
Work done in any adiabatic process is state function.
$\Delta U=q-W \quad(\because q=0)$
$\Delta U=-W$
Work done in isothermal process is not a state function.
$W=-q \quad(\because \Delta T=0, q \neq 0)$
From the equation, $\Delta H=\Delta U+\Delta P V$
At constant pressure, $\Delta H=\Delta U+P \Delta V$
At constant volume, $\Delta H=\Delta U+V \Delta P$
At constant pressure, $\Delta P=0, \Delta H=q_p$
so, it is a state function.
At constant volume, $\Delta V=0, \Delta U=q_v$ so, it is a state function.
Work done in any adiabatic process is state function.
$\Delta U=q-W \quad(\because q=0)$
$\Delta U=-W$
Work done in isothermal process is not a state function.
$W=-q \quad(\because \Delta T=0, q \neq 0)$
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