An ideal gas is taken from state 1 to state 2 through a process...

The work done by the gas is the area under the P-V graph from state 1 to state 2. Since the process is represented by a straight line, the area under the curve forms a trapezium.

Work done, $\Delta W = \dfrac{1}{2} \times (P_1 + P_2) \times (V_2 - V_1)$

Given $P_1 = 20 \text{ kPa}$, $P_2 = 60 \text{ kPa}$, $V_1 = 10 \text{ L}$, and $V_2 = 30 \text{ L}$.

$\Delta W = \dfrac{1}{2} \times (20 + 60) \times 10^3 \text{ Pa} \times (30 - 10) \times 10^{-3} \text{ m}^3$

$\Delta W = \dfrac{1}{2} \times 80 \times 20 \text{ J} = 800 \text{ J}$

According to the first law of thermodynamics:

$\Delta Q = \Delta U + \Delta W$

Given the heat supplied to the system, $\Delta Q = 2000 \text{ J}$.

$2000 \text{ J} = \Delta U + 800 \text{ J}$

$\Delta U = 1200 \text{ J}$

Answer: $1200 \text{ J}$