Question

Figure below shows two paths that may be taken by a gas to go from a state A to state C.
In process AB, $400J$ of heat is added to the system and in process BC, $100J$ of heat is added to the system. The heat absorbed by the system in the process AC will be

• A. $460J$
• B. $300J$
• C. $380J$
• D. $500J$

Answer 1

The correct option is A $460J$

In process A to B change in internal energy is
$\Delta U_{AB}=\Delta Q_{AB}-\Delta W_{AB}$
process AB is Isochoric, therefore heat added is equal to change in internal energy.
$\Delta U_{AB}=\Delta Q_{AB}=400J$
process BC change in internal energy is
$\Delta U_{BC}=\Delta Q_{BC}-\Delta W_{BC}$
$\Delta U_{BC}=100-P\Delta V$
$\Delta U_{BC}=100-6\times {10}^4(4\times {10}^{-3}-2\times {10}^{-3})$
$\Delta U_{BC}=100-6\times {10}^4\times 2\times {10}^{-3}$
$\Delta U_{BC}=100-120J=-20J$
$\Delta U_{BC}=100-6\times {10}^4\times 2\times {10}^{-3}$
total change in internal energy will be :
$\Delta U_{AC}=400-20=380J$
heat added in process AC will be:
$\Delta Q_{AC}=\Delta U_{AC}+\Delta W_{AC}$
$\Delta Q_{AC}=380+areaofP-Vdiagram$
$\Delta Q_{AC}=380+\frac{1}{2}(6\times {10}^4+2\times {10}^4)2\times {10}^{-3}$
$\Delta Q_{AC}=380+80=460J$