Question

The temperature at which the average speed of perfect gas molecule is double than at ${17}^{0}C$ is

• A. ${346}^{0}C$
• B. ${68}^{0}C$
• C. ${162}^{0}C$
• D. ${887}^{0}C$

Answer 1

The correct option is D ${887}^{0}C$

Average speed $V_{av}\propto (T{)}^{\frac{1}{2}}$
$\therefore \frac{(V_{av}{)}_2}{(V_{av}{)}_1}=\sqrt{\frac{T_2}{T_1}}(Given\frac{(V_{av}{)}_2}{(V_{av}{)}_t}=2)$
$(2{)}^2=\frac{T_2}{T_1}\Rightarrow T_2=4T_1$
$T_1$ = 273 + 17 = 290 K
$\therefore T_2=4\times 290=1160K=1160-273={887}^{\circ }C$