Question

A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used, \( 60 J \) of heat flows into the system and \( 30 J \) of work is done by the system. If path ADB is used, the work done by the system is \( 10 J \). The heat flow into the system in path ADB is: 
 

• A. \( 40 J \)
• B. \( 80 J \)
• C. \( 100 J \)
• D. \( 20 J \)

Hint:

Change in internal energy \( \Delta U \) depends only on initial and final states, not on the path.

Answer 1

The Correct Option is A

Step 1: {Using first law of thermodynamics}
\[ \Delta Q = \Delta U + \Delta W \] Step 2: {For process ACB}
\[ \Delta U = 60 - 30 = 30 J \] Step 3: {Applying same internal energy change to ADB}
\[ \Delta Q_{ADB} = 30 + 10 = 40 J \] Thus, the correct answer is \( 40 J \).

Answer 2

Step 1: Use the first law of thermodynamics:
\( Q = \Delta U + W \)
Where:
- \( Q \) is the heat supplied to the system
- \( \Delta U \) is the change in internal energy
- \( W \) is the work done by the system

Step 2: Apply the law to path ACB:
Given: \( Q_{ACB} = 60\,J \), \( W_{ACB} = 30\,J \)
So, \( \Delta U = Q_{ACB} - W_{ACB} = 60 - 30 = 30\,J \)

Step 3: Now apply the law to path ADB:
Given: \( W_{ADB} = 10\,J \)
We assume internal energy change \( \Delta U \) between states A and B remains the same (since initial and final states are identical).
So, \( Q_{ADB} = \Delta U + W_{ADB} = 30 + 10 = 40\,J \)

Final Answer: \( 40\,J \)