Question

An ideal gas with constant heat capacity $C_v=\frac{3}{2}nR$ made to carry out a cycle that is depicted by a triangle in the figure given below

The following statement is true about the cycle

• A. The efficiency is given by $1-\frac{p_1{V}_1}{p_2{V}_2}$
• B. The efficiency is given by $1-\frac{1}{2}\frac{p_1{V}_1}{p_2{V}_2}$
• C. Net heat absorbed in the cycle is $\left(p_2-p_1\right)\left(V_2-V_1\right)$
• D. Heat absorbed in part AC is given by$2\left(p_2{V}_2-p_1{V}_1\right)+\frac{1}{2}\left(p_1{V}_2-p_2{V}_1\right)$

Answer 1

Answer: (D)

As $B\to A\to C$ is a closed cyclicprocess, we have
$\Delta U$(complete cycle) $=0$
So, by first law of thermodynamics,we have
$\Delta Q=\Delta W$
or $\Delta Q_{AB}+\Delta Q_{BC}=\Delta W$
$=\Delta W_{AB}+\Delta W_{BC}+\Delta W_{AC}$
$=$ Area enclosed under $p-V$ graph

$\Rightarrow \Delta Q_{AC}=-\left(\Delta Q_{AB}+\Delta Q_{BC}\right)+\frac{1}{2}$
$\left(V_2-V_1\right)\left(p_2-p_1\right)$
$=\left(\frac{3}{2}nR\Delta T-p_1\left(V_2-V_1\right)+\frac{1}{2}nR\Delta T-V_2\left(p_2-p_1\right)\right)$
$+\frac{1}{2}\left(V_2-V_1\right)\left(p_2-p_1\right)$
$\therefore \Delta Q_{AC}=2\left(p_2{V}_2-p_{{}_1}{V}_1\right)<br/>+\frac{1}{2}\left(V_2-V_1\right)\left(p_2-p_1\right)$