Q. In the following $P$-$V$ diagram two adiabatics cut two isothermals at temperatures $T_{1}$ and $T_{2}$ (fig.). The value of $\frac{V_{a}}{V_{d}}$ will be
A) $\frac{V_{b}}{V_{c}}$
B) $\frac{V_{c}}{V_{b}}$
C) $\frac{V_{d}}{V_{a}}$
D) $V_{b}V_{c}$
Explanation
(a) For adiabatic process $T_{1}V_{b}^{\gamma - 1} =$ Constant
For $bc$ curve $T_{1}V_{b}^{\gamma - 1} = T_{2}V_{c}^{\gamma - 1}$ or $\frac{T_{2}}{T_{1}} = \left( \frac{V_{b}}{V_{c}} \right)^{\gamma - 1}$ …(i)
For $ad$ curve $T_{1}V_{a}^{\gamma - 1} = T_{2}V_{d}^{\gamma - 1}$ or $\frac{T_{2}}{T_{1}} = \left( \frac{V_{a}}{V_{d}} \right)^{\gamma - 1}$ …(ii)
From equation (i) and (ii) $\frac{V_{b}}{V_{c}} = \frac{V_{a}}{V_{d}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.