Question

550 kJ / cycle work is done on 1 mole of an ideal gas in a cyclic process. The amount of heat absorbed by the system in one cycle is:
(a)- 550 kJ
(b)- 590 kJ
(c)- 1100 kJ
(d)- Zero

Answer 1

Hint: This question can be solved by the formula $\Delta U=q+W$, where $\Delta U$ is the change in the internal energy, q is the heat absorbed or evolved by the system, and W is the done by or on the system.

Complete step by step answer:
There are two types of functions in thermodynamics i.e., state function and path function. State functions are those properties that depend only on the initial and final state of the system but path functions are those properties that depend on the path of the system. So some examples of state functions are changes in internal energy, pressure, temperature, etc, and some examples of path functions are work done, heat, etc. So in a cyclic process, the initial point and the final point are the same, which means that those properties that are state functions will have zero value in the cyclic process because of the same initial and final point. As we know the first law of thermodynamics states that:
$\Delta U=q+W$, where $\Delta U$ is the change in the internal energy, q is the heat absorbed or evolved by the system, and W is the done by or on the system.
So for a cyclic process, the $\Delta U$ will be zero. (it is a state function)
The equation will be:
$q=W$
So the amount of work done is equal to the heat absorbed by the system.
$q=550\text{ kJ / mol}$

Therefore, the correct answer is an option (a)- 550 kJ / mol.

Note: There are some sign conventions used, i.e., when work is done by the system (work of expansion) then it is negative and when the work is done on the system (work of compression) then it is positive.