Question

The temperature at which the rms speed of gas molecules becomes double its value at 0${}^o$C is :

• A. 819${}^o$C
• B. 760${}^o$C
• C. 273${}^o$C
• D. 224${}^o$C

Answer 1

The correct option is A 819${}^o$C

According to the kinetic theory of gases

$v_{rms}=\sqrt{3\frac{RT}{M}}$

so $v_{rms}\alpha \sqrt{T}$

$\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}$ eq(1)

$v_2=2v_1$

$T_1=0^{o}C=273K$

substitute values back to eq(1)

$2=\sqrt{\frac{T_2}{273}}$

$T_2=4\times 273=1092K={819}^{o}C$

$T_2={819}^{o}C$

answer is A.