An ideal gas taken through a process ABCA as shown in figure. If the net heat supplied to gas in the cycle is 5 J , then the work done by the gas in process from C to A is

For a cyclic process, the net work done equals the net heat transfer, $W_{\text{net}} = Q_{\text{net}}$, by the first law of thermodynamics since the change in internal energy over a cycle is zero. Given $Q_{\text{net}} = 5\text{ J}$, it follows that $W_{\text{net}} = 5\text{ J}$.

Process $A \to B$ is isobaric with $P = 10\text{ N/m}^2$, and the volume increases from $1\text{ m}^3$ to $2\text{ m}^3$, so the work done is $W_{AB} = 10(2 - 1) = 10\text{ J}$.

Process $B \to C$ is isochoric ($\Delta V = 0$), so $W_{BC} = 0$.

Process $C \to A$ contributes $W_{CA}$ to the cycle, and since $W_{\text{net}} = W_{AB} + W_{BC} + W_{CA}$, substitution yields $5 = 10 + 0 + W_{CA}$, giving $W_{CA} = -5\text{ J}$.

The work done by the gas during process $C \to A$ is therefore $\boxed{-5\text{ J}}$.