Question

A system is taken from state $a$ to state $c$ by two paths $adc$ and $abc$ as shown in the figure. The internal energy at $a$ is $U_a=10 \, J . $ Along the path $adc$ the amount of heat heat absorbed $\delta Q_1=50\, J $ and the work obtained $\delta W_1=20 \, J $ whereas along the path $abc$ the heat absorbed $\delta Q_2=36 \, J . $ The amount of work along the path $abc$ is :

• A. 10 J
• B. 12 J
• C. 36 J
• D. 6 J

Answer 1

The Correct Option is D

To solve the problem, we will use the first law of thermodynamics, which states:

\(\Delta U = \delta Q - \delta W\)

where \(\Delta U\) is the change in internal energy, \(\delta Q\) is the heat absorbed, and \(\delta W\) is the work done by the system. Let's evaluate the changes along each path.

1. Along the path \(adc\):
   • The heat absorbed, \(\delta Q_1 = 50 \, \text{J}\).
   • The work done, \(\delta W_1 = 20 \, \text{J}\).
   • According to the first law: \(\Delta U = \delta Q_1 - \delta W_1 = 50 - 20 = 30 \, \text{J}\).
2. Along the path \(abc\):
   • The heat absorbed, \(\delta Q_2 = 36 \, \text{J}\).
   • The change in internal energy, \(\Delta U\), is the same as along path \(adc\) since both paths move the system from the same initial state \(a\) to the same final state \(c\). Thus, \(\Delta U = 30 \, \text{J}\).
   • Using the first law again: \(\Delta U = \delta Q_2 - \delta W_2\), where \(\delta W_2\) is the work done along path \(abc\).
   • Substitute the known values: \(30 = 36 - \delta W_2\)
   • Solving for \(\delta W_2\), we get: \(\delta W_2 = 36 - 30 = 6 \, \text{J}\).

Therefore, the amount of work done along path \(abc\) is 6 J, which corresponds to the correct answer option.

Thus, the correct option is 6 J.