An ideal monoatomic gas is carried along the cycle $A B C D A$ as shown in the figure. The total heat…
An ideal monoatomic gas is carried along the cycle $A B C D A$ as shown in the figure. The total heat absorbed during this process is
- $10.5 p_0 V_0$
- $7.5 p_0 V_0$
- $2.5 p_0 V_0$
- $1.5 p_0 V_0$
Solution
Heat absorbed means heat is extracted from source, i.e.Q must be positive.
This occurs along path
$
\begin{aligned}
\quad A \rightarrow B \rightarrow C . & \\
\therefore & \text { Heat added }=\Delta Q_{A B C}=\Delta U_{A B C}+\Delta W_{A B C} \\
= & n C_V\left(T_C-T_A\right)+3 p_0\left(2 V_0-V_0\right) \\
= & n \frac{3}{2} R\left(T_C-T_A\right)+3 p_0 V_0=\frac{3}{2}\left(n R T_C-n R T_A\right)+3 p_0 V_0 \\
= & \frac{3}{2}\left(p_C V_C-p_A V_A\right)+3 p_0 V_0 \\
= & \frac{3}{2}\left(3 p_0 2 V_0-p_0 V_0\right)+3 p_0 V_0 \\
= & \frac{3}{2} \times 5 p_0 V_0+3 p_0 V_0=\frac{21}{2} p_0 V_0=10.5 p_0 V_0
\end{aligned}
$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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