The heat absorbed by a system in going through the given cyclic process is :

Question

Given $C_p = a + bT$
$a = 19.5$, $b = 0.042$, $n = 1$ mole
Temperature changes from $27^\circ C$ to $327^\circ C$.
Find $\Delta H$.

Answer 1

In a cyclic process, the internal energy change ($\Delta U$) is nil. Consequently, the absorbed heat ($\Delta Q$) equates to the work performed ($\Delta W$), represented by the area under the P-V curve. This area is computed as:

\[ \Delta Q = \Delta W = \pi \times (140 \times 10^3) \, \text{Pa} \times (140 \times 10^{-6} \, \text{m}^3) \]

\[ \Delta Q = 61.6 \, \text{J} \]

The heat absorbed by a system in going through the given cyclic process is :

The Correct Option is A

Solution and Explanation

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