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 $B\rightarrow A\rightarrow C$ 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)
+\frac{1}{2}\left(V_{2}-V_{1}\right)\left(p_{2}-p_{1}\right)$