Question

On an isothermal process, there are two POINTs $A$ and $B$ at which pressures and volumes are $(2P_o,V_o)$ and $(P_o,2V_o)$ respectively. If $A$ and $B$ are connected by a straight line, the temperature of the isothermal process is how much lower the maximum temperature

• A. $\frac{P_o{V}_o}{nR}$
• B. $\frac{P_o{V}_o}{2nR}$
• C. $\frac{3P_o{V}_o}{2nR}$
• D. $\frac{P_o{V}_o}{4nR}$

Answer 1

The correct option is D $\frac{P_o{V}_o}{4nR}$

Using symmetry, the temperature will be maximum at a point where the pressure and the volume attain the same value.
Therefore, on a straight line, joining the 2 points A and B as given in the question, the temperature will be maximum at their midpoint.
Hence, the pressure will be $\frac{3P_o}{2}$ and the volume will be $\frac{3V_o}{2}$
Thus, for n moles of a gas, the maximum temperature is,
$T=\frac{9P_o{V}_o}{4nR}$
However, the isothermal temperature is,
$T=\frac{2P_o{V}_o}{nR}$
Thus, the difference in 2 temperatures will be,
$\frac{P_o{V}_o}{4nR}$