A gas can be taken from A to B via two different processes $A C B$ and $A D B$. When path $A C B$ is used 60 J of heat flows into the system and 30J of work is done by the system. If path ADB is used work done by the system is 10 J . The heat Flow into the system in path ADB is :

• Correct answer
  A

  40 J
• B

  80 J
• C

  100 J
• D

  20 J

Solution

$\triangle \mathrm{U}$ remains same for both paths ACB and ADB $\Delta Q_{A C B}=\Delta W_{A C B}+\Delta U_{A C B}$ $\Rightarrow 60 \mathrm{~J}=30 \mathrm{~J}+\Delta \mathrm{U}_{\mathrm{ACB}} \Rightarrow \mathrm{U}_{\mathrm{ACB}}=30 \mathrm{~J}$ As change in internal energy depends only on initial and final point $\therefore \Delta \mathrm{U}_{\mathrm{ADB}}=\Delta \mathrm{U}_{\mathrm{ACB}}=30 \mathrm{~J}$ $\Delta \mathrm{Q}_{\mathrm{ADB}}=\Delta \mathrm{U}_{\mathrm{ADB}}+\Delta \mathrm{W}_{\mathrm{ADB}}=10 \mathrm{~J}+30 \mathrm{~J}=40 \mathrm{~J}$

Hence, the correct answer is option: 'A'.