Figure shows the ?pV?-diagram for an isothermal expansion of 1.50 mol of an ideal gas, at a temperature of 15.0°C. (a) What is the change in internal energy of the gas? Explain. (b) Calculate the work done by (or on) the gas and the heat absorbed (or released) by the gas during the expansion. Figure:

Solution 38P Step 1: (a).Internal energy is due to motion of particles in a system. As internal energy depends on temperature. As we know temperature in isothermal process is constant so the internal energy will also be constant thus the change in internal energy will be zero. U = 0 Step 2: (b).The ideal gas equation PV = nRT V2 Work done by the gas is given by W = nRT/V dV V 1 W = nRT ln(V /V2) d1 W = 1.50 × (8.314)(288 K)ln(0.02/0.01) W = 1.50 × (8.314)(288 K) × 0.693 W = 2489.01 J