Question

The figure shown below represents two paths that are taken by an ideal gas to go from the state A to state C. In process $\text{AB},400\text{J}$ of heat is added to the system and in process $\text{BC},100\text{J}$ of heat is added to system. The heat absorbed by system in process $\text{AC}$ will be

• A. $460\text{J}$
• B. $300\text{J}$
• C. $380\text{J}$
• D. $500\text{J}$

Answer 1

The correct option is A $460\text{J}$

For process or path $\text{AB}$, (isochoric process)
$W_{AB}=0$
$\Rightarrow Q_{AB}=W_{AB}+\Delta U_{AB}$
$\Rightarrow Q_{AB}=\Delta U_{AB}$
$\therefore \Delta U_{AB}=+400J$
During process $\text{BC}$ (isobaric process)
$W_{BC}=P\Delta V$
$=6\times {10}^4\times (4-2)\times {10}^{-3}=+120\text{J}$
$Q_{BC}=+100J$
$\Rightarrow Q_{BC}=W_{BC}+\Delta U_{BC}$
or, $100=120+\Delta U_{BC}$
$\therefore \Delta U_{BC}=-20J$
The $\Delta U$ for path $A\to B\to C$ and $A\to C$ will be same because initial and final states are same for both paths.
$\Rightarrow \Delta U_{ABC}=\Delta U_{AC}$
or $\Delta U_{AB}+\Delta U_{BC}=(Q_{AC}-W_{AC})$
$[\because$ from $1^{st}$ law of thermodynamics, $Q_{AC}=W_{AC}+\Delta U_{AC}]$
work done by gas in AC,
$W_{AC}$ =Area under P-V graph
$\Rightarrow W_{AC}=\frac{1}{2}\times (8\times {10}^4)\times (4-2)\times {10}^{-3}$
$\therefore W_{AC}=80\text{J}$
Substituting in Eq. ($i$),
$\Rightarrow 400+(-20)=Q_{AC}-80$
$\Rightarrow \Delta Q_{AC}=+460J$
Since $Q_{AC}$ is + ve, heat absorbed by system is $460J$.