A gas undergoes a thermodynamic process along the path AB as...

From the given graph, the initial and final states of the gas are:
$P_A = 200 \text{ kPa}$, $V_A = 250 \text{ cc}$
$P_B = 500 \text{ kPa}$, $V_B = 100 \text{ cc}$

The work done by the gas during the linear process from A to B is given by the area under the P-V curve:
$W = \int P \, dV = \dfrac{1}{2}(P_A + P_B)(V_B - V_A)$
$W = \dfrac{1}{2}(200 + 500) \times 10^3 \times (100 - 250) \times 10^{-6} \text{ J}$
$W = \dfrac{1}{2}(700 \times 10^3) \times (-150 \times 10^{-6}) \text{ J} = -52.5 \text{ J}$

Since $70 \text{ cal}$ of heat is extracted from the gas, the heat added to the system is negative. Using the conversion $1 \text{ cal} \approx 4.2 \text{ J}$:
$Q = -70 \text{ cal} \approx -70 \times 4.2 \text{ J} = -294 \text{ J}$

According to the first law of thermodynamics, the change in internal energy is:
$\Delta U = Q - W$
$\Delta U = -294 \text{ J} - (-52.5 \text{ J}) = -241.5 \text{ J}$

Rounding to the nearest integer, the change in internal energy is approximately $-241 \text{ J}$.

Answer: $-241 \text{ J}$