A gas undergoes a thermodynamic process along the path AB as depicted in the figure. Determine the change in the internal energy of the system given that 70 cal of heat is extracted from the gas during this process.

Step-by-step solution

From the given graph, the initial and final states of the gas are: P_A = 200 kPa , V_A = 250 cc P_B = 500 kPa , V_B = 100 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 = P , dV = 1 2 (P_A + P_B)(V_B - V_A) W = 1 2 (200 + 500) 10^3 (100 - 250) 10⁻⁶ J W = 1 2 (700 10^3) (-150 10⁻⁶) J = -52.5 J Since 70 cal of heat is extracted from the gas, the heat added to the system is negative. Using the conversion 1 cal 4.2 J : Q = -70 cal -70 4.2 J = -294 J Accordi