Two different adiabatic paths for the same ga | vedclass

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(C) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.For the adiabatic path

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$\frac{V_a}{V_d}=\frac{V_b}{V_c}$

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(C) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.For the adiabatic path connecting points $a$ and $d$,we have:$T_a V_a^{\gamma-1} = T_d V_d^{\gamma-1}$$\left(\frac{V_a}{V_d}ight)^{\gamma-1} = \frac{T_d}{T_a}$For the adiabatic path connecting points $b$ and $c$,we have:$T_b V_b^{\gamma-1} = T_c V_c^{\gamma-1}$$\left(\frac{V_b}{V_c}ight)^{\gamma-1} = \frac{T_c}{T_b}$Since points $a$ and $b$ lie on the same isothermal curve,$T_a = T_b$. Similarly,points $d$ and $c$ lie on the same isothermal curve,so $T_d = T_c$.Substituting these into the equations,we get:$\frac{T_d}{T_a} = \frac{T_c}{T_b}$Therefore,$\left(\frac{V_a}{V_d}ight)^{\gamma-1} = \left(\frac{V_b}{V_c}ight)^{\gamma-1}$This implies $\frac{V_a}{V_d} = \frac{V_b}{V_c}$.

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in the $P-V$ diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

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