[QUE/SM-02010] SM-PROBLEM

Consider one mole of an ideal gas. A molecule of the gas is mono atomic and is spherically symmetric. The system of ideal gas is in equilibrium. Its temperature $T_1=300$ k and its volume $V_1=1$ litre. Since the system is in equilibrium, it can be represented by a point $A=( V_1,T_1)$, in the temperature-volume thermodynamic phase plane. We are taking Volume along the $X$ axis and Temperature along the $Y$ axis.

Now consider a process by which the system expands to a volume $V_2=2$ litres. The process is adiabatic.

• {\bf Case-1 :} The process is quasi static and reversible. Let $T_2$ be the temperature of the system at the end of the process. Let $B=(V_2,T_2)$ denote the system at the end of the process. {\bf Find} $T_2$. The process $A\to B$ can be represented by a curve joining $A$ to $B$. The curve is called an adiabat.
• {\bf Case-2 :} The process is not reversible. Hence the process can not be represented by a curve in the thermodynamic phase diagram. The system disappears from $A$ at the start of the process. At the completion of the process, if we wait long enough, the system would equilibrate and appear at a point $B^\prime=(T_2^\prime,V_2)$ in the phase diagram. There is no ready-made formula for calculating $T_2^\prime$. Nor is there a formula for calculating the increase in entropy of the system in the irreversible process. Also these quantities depend on how far away the irreversible process is from its reversible companion. However $B^\prime$ is on a line parallel to $Y$ axis and passing through $B$. Employing the Second law of thermodynamics

Answer the following questions.

1. Find if the point $B^\prime$ is vertically above or below the point $B$.
2. Calculate the increase in entropy in terms of $T_2^\prime$.