Answer
(a) $T=-205C^{\circ}$
(b) $T=817C^{\circ}$
Work Step by Step
(a) According to given condition
$(v_{rms})_T=\frac{1}{2}(v_{rms})_{STP}$
$\implies \sqrt{\frac{3K_BT}{m}}=\frac{1}{2} \sqrt{\frac{3K_B(273K)}{m}}$
This simplifies to:
$\sqrt{T}=\frac{1}{2}\sqrt {273K}$
Squaring both sides, we obtain:
$T=\frac{1}{4}(273K)$
$\implies T=68K=-205C^{\circ}$
(b) As given that
$(v_{rms})_T=2(v_{rms})_{STP}$
$\implies \sqrt{\frac{3K_BT}{m}}=2\sqrt{\frac{3K_B(273K)}{m}}$
This simplifies to:
$\sqrt{T}=2\sqrt{273K}$
Squaring both sides, we obtain:
$T=4(273K)$
$\implies T=1090K=817C^{\circ}$
