Question

A gas is held within a closed chamber. It is undergoing a cyclic process as shown in the figure. The energy removed as heat $Q_{CA}$ during iso-choric process $CA$ is $20\text{J}$. No energy is transferred as heat during adiabatic process $BC$ and net work done during the cyclic process is $15\text{J}$. Determine the energy transferred by the system as heat during iso-baric process $AB$.

• A. $10\text{J}$ of heat is absorbed by the gas.
• B. $10\text{J}$ of heat is released by the gas.
• C. $5\text{J}$ of heat is absorbed by the gas.
• D. $5\text{J}$ of heat is released by the gas.

Answer 1

The correct option is C $5\text{J}$ of heat is absorbed by the gas.

We know that , as the initial and final states of the gas are same . The change in internal energy in a cyclic process is zero $(\Delta U=0)$ .
So, Net Heat absorbed/rejected by the gas $=$ Net work done during cyclic process.
From first law of thermodynamics, we can say that,
$Q=\Delta U+W$
$\Rightarrow Q=W$
$\therefore$ we can write that ,
$Q_{AB}+Q_{BC}+Q_{CA}=W_{AB}+W_{BC}+W_{CA}$
$\Rightarrow Q_{AB}=W_{net}-Q_{BC}-Q_{CA}(\because W_{net}=W_{AB}+W_{BC}+W_{CA})$
From the data given in the question,
$Q_{AB}=-15-0-(-20)$
$\Rightarrow Q_{AB}=+5\text{J}$
Positive sign shows that, heat has been absorbed by the gas during the iso-baric process
Hence, option $\left(c\right)$ is the correct answer