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:

  1. 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.



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