(N/A) The two conditions under which heat becomes independent of path are:
$(i)$ When volume remains constant $(q_V)$
$(ii)$ When pressure remains constant $(q_p)$
Explanation:
$(i)$ At constant volume: By the first law of thermodynamics,$\Delta U = q + W$. Since $W = -P_{ext} \Delta V$ and $\Delta V = 0$,the work done is $0$. Therefore,$q_V = \Delta U$. Since internal energy $(\Delta U)$ is a state function,$q_V$ is also a state function.
$(ii)$ At constant pressure: The heat absorbed is given by $q_p = \Delta U + P \Delta V$. By definition,the change in enthalpy is $\Delta H = \Delta U + P \Delta V$. Therefore,$q_p = \Delta H$. Since enthalpy $(\Delta H)$ is a state function,$q_p$ is also a state function.