Question

A gas undergoes a change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A:
A ) 10 J of work will be done by the gas.
B ) 6 J of work will be done by the gas.
C ) 10 J of work will be done by the surrounding gas.
D ) 6 J of the work will be done by the surrounding gas.

Answer 1

Hint: The mathematical expression for the first law of thermodynamics gives the relationship between the change in the internal energy, the work and the heat.
\[\Delta U = q + w\]

Complete step by step answer:
In the first change from state A to state B, the heat absorbed is 5 J and the work done is 8 J.
Use the mathematical expression for the first law of thermodynamics and calculate the change in the internal energy.
\[\Delta U = q + w \\
= 5{\text{ J}} - 8{\text{ J}} \\
= - 3{\text{ J}} \\ \]
Enthalpy change is a state function. The value of the state function depends only on the initial and final states and is independent of the path followed. The magnitude of the enthalpy change for the first change from state A to state B is the same as the magnitude of the enthalpy change for the second change from state B to state A. But they have opposite signs.
Hence, for the second change from state B to state A:
\[\Delta U = q + w \\
3{\text{ J}} = - 3{\text{ J + w}} \\
w = + 6{\text{ J}} \\ \]
Thus, the surrounding does a work of 6 J on the system. Hence, the option D ) is the correct option.

Note: Heat absorbed by the system has positive signs. Heat given out by the system has negative signs. Work done on the system has positive signs. Work done by the system has negative signs.