The relation between Boyle’s Temperature T_B and critical temperature T_C is:

2 T_{B} = \frac{8}{27} T_{C}

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T_{B} = \frac{27}{8} T_{C}

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T_{B} = \frac{3}{2} T_{C}

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3 T_{B} = \frac{5}{13} T_{C}

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Solution

The correct option is B $T_{B} = \frac{27}{8} T_{C}$
Boyle’s temperature, T_B = $T_{B} = \frac{a}{R b}$
Critical temperature, T_C = $T_{C} = \frac{8 a}{27 R b}$
$T_{C} = \frac{8}{27} T_{B} T_{B} = \frac{27}{8} T_{C}$

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The relation between Boyle’s Temperature TB and critical temperature TC is:

The correct option is B TB=278TC Boyle’s temperature, TB = TB=aRb Critical temperature, TC =TC=8a27Rb TC=827TBTB=278TC

The relation between Boyle’s Temperature T_B and critical temperature T_C is:

The correct option is B $T_{B} = \frac{27}{8} T_{C}$
Boyle’s temperature, T_B = $T_{B} = \frac{a}{R b}$
Critical temperature, T_C = $T_{C} = \frac{8 a}{27 R b}$
$T_{C} = \frac{8}{27} T_{B} T_{B} = \frac{27}{8} T_{C}$