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11.Thermodynamics
medium
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
${V_b}{V_c}$
Solution
(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 $(ii)$ $\frac{{{V_b}}}{{{V_c}}} = \frac{{{V_a}}}{{{V_d}}}$
Standard 11
Physics
