In a certain process 210 105 J of heat is liberated by a sy

In a certain process, 2.10 * 10^5 J of heat is liberated by a system, and at the same time the system contracts under a constant external pressure of 9.30 * 10^5 Pa. The internal energy of the system is the same at the beginning and end of the process. Find the change in volume of the system. (The system is not an ideal gas.)

Solution

We have the 1st law of thermodynamics as: dQ = dU +PdV

dQ = Amount of heat exchange between system and surrounding = 2.10 x 10⁵ J

dU = Change in internal energy = 0;

dV = Change in volume of the system: which we need to find out

P = pressure applied to the system = 9.30 x 10⁵ Pa

Now we have,      dV = dQ/P

                                  = 2.10 / 9.30

                                  = 225.806 x 10^-3 m³ .