Thermodynamics and Thermochemistry - Result Question 82
####7. Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
(2019 Main, 8 April I) (a) Cyclic process : $q=-W$
(b) Adiabatic process : $\Delta U=-W$
(c) Isochoric process : $\Delta U=q$
(d) Isothermal process : $q=-W$
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Solution:
- From the 1st law of thermodynamics,
$$ \Delta U=q+W $$
where, $\Delta U=$ change in internal energy
$$ \begin{gathered} q=\text { heat } \ W=\text { work done } \end{gathered} $$
The above equation can be represented for the given processes involving ideal gas as follows:
(a) Cyclic process For cyclic process, $\Delta U=0$
$$ \therefore \quad q=-W $$
Thus, option (a) is correct.
(b) Adiabatic process For adiabatic process,
$$ q=0 $$
$$ \therefore \quad \Delta U=W $$
Thus, option (b) is incorrect.
(c) Isochoric process For isochoric process, $\Delta V=0$.
$$ \begin{array}{lrl} \text { Thus, } & W & =0 \ \therefore & \Delta V & =q \end{array} \quad(\because W=p \Delta V) . $$
Thus, option (c) is correct.
(d) Isothermal process For isothermal process, $\quad \Delta U=0$ $\therefore \quad q=-W$
Thus, option (d) is correct.
