Thermodynamics and Thermochemistry - Result Question 7
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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Answer:
Correct Answer: 7. (b)
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, $\Delta V=0$.
$\text { Thus, } W = 0 $$\quad(\because W = p \Delta V)$
$ \therefore \quad \Delta V =q$
Thus, option (c) is correct.
(d) Isothermal process For isothermal process, $ \Delta U=0$
$\therefore \quad q=-W$
Thus, option (d) is correct.
BETA
