Question

A gas is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these two paths are \( Q_1 \) and \( W_1 \), and \( Q_2 \) and \( W_2 \), respectively. Then

• A. \( Q_1 - W_1 = Q_2 - W_2 \)
• B. \( Q_1 = Q_2 \)
• C. \( W_1 = W_2 \)
• D. \( W_1 - W_2 = Q_1 - Q_2 \)

Hint:

The first law of thermodynamics implies that the change in internal energy is path-independent, so the difference between heat absorbed and work done is the same for any two paths between the same initial and final states.

Answer 1

The Correct Option is A

In thermodynamics, the first law of thermodynamics is given by: \[ \Delta U = Q - W \] where: - \( \Delta U \) is the change in internal energy of the system, - \( Q \) is the heat absorbed by the system, - \( W \) is the work done by the system. Since the gas is taken from state A to state B along two different paths, the change in internal energy, \( \Delta U \), is the same for both paths, as internal energy depends only on the initial and final states of the system, not on the path taken. Therefore, we have: \[ \Delta U = Q_1 - W_1 = Q_2 - W_2 \] This shows that the difference between the heat absorbed and the work done is the same for both paths. Thus, the correct relationship is: \[ Q_1 - W_1 = Q_2 - W_2 \]