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(B) The given relation is $P = P_o(1 + \frac{V_o^2}{V^2})^{-1} = \frac{P_o V^2}{V^2 + V_o^2}$.From the ideal gas law, $PV = nRT$, so $T = \frac{PV}{nR

Vedclass · Accepted Answer

$\frac{11P_o V_o}{10R}$

Vedclass · Answer

(B) The given relation is $P = P_o(1 + \frac{V_o^2}{V^2})^{-1} = \frac{P_o V^2}{V^2 + V_o^2}$.From the ideal gas law, $PV = nRT$, so $T = \frac{PV}{nR}$.For sample $A$: $V_A = V_o$, so $P_A = \frac{P_o V_o^2}{V_o^2 + V_o^2} = \frac{P_o}{2}$.$T_A = \frac{P_A V_A}{nR} = \frac{(P_o/2) V_o}{2R} = \frac{P_o V_o}{4R}$.For sample $B$: $V_B = 3V_o$, so $P_B = \frac{P_o (3V_o)^2}{(3V_o)^2 + V_o^2} = \frac{9P_o V_o^2}{10V_o^2} = \frac{9P_o}{10}$.$T_B = \frac{P_B V_B}{nR} = \frac{(9P_o/10) (3V_o)}{2R} = \frac{27 P_o V_o}{20R}$.Now, the difference $T_B - T_A = \frac{27 P_o V_o}{20R} - \frac{P_o V_o}{4R} = \frac{27 P_o V_o}{20R} - \frac{5 P_o V_o}{20R} = \frac{22 P_o V_o}{20R} = \frac{11 P_o V_o}{10R}$.

Column-$I$	Column-$II$
$(a)$ Boyle's law	$(i)$ Pressure
$(b)$ Charles's law	$(ii)$ Volume
	$(iii)$ Temperature

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