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Correct relation between T_C & T_B is (T_C= critical temperature, T_B=Boyle’s temperature)

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$large {T_c}, = ,0.3,{T_B}$

${T_c}, = ,frac{{27{T_B}}}{8}$

${T_B}, = ,0.3,{T_C}$

${T_B}, = ,frac{{8{T_C}}}{3}$

answer is A.

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Detailed Solution

${T_C} = frac{{8a}}{{27Rb}};{T_B} = frac{a}{{Rb}}$

${T_C} = frac{{8{T_B}}}{{27}}$

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