Explanation

(d) We have $v_{rms} = \sqrt{\frac{3RT}{M}};\ \ at\ T = T_{0}(NTP)$
$v_{rms} = \sqrt{\frac{3RT_{0}}{M}}$
But at temperature $T$,
$v_{rms} = 2 \times \sqrt{\frac{3RT_{0}}{M}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{\frac{3RT}{M}} = 2\sqrt{\frac{3RT_{0}}{M}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{T} = \sqrt{4T_{0}}$
or $T = 4T_{0}$
$T = 4 \times 273$K=1092K
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = 819℃$