Question

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

• A. ${34}^{\circ }C$
• B. ${68}^{\circ }C$
• C. ${162}^{\circ }C$
• D. ${887}^{\circ }C$

Answer 1

Answer: (D)

${887}^{\circ }C$

As we know,
Average speed $V_{av}\propto (T{)}^{\frac{1}{2}}$
$\therefore \frac{(V_{av}{)}_2}{(V_{av}{)}_1}=\sqrt{\frac{T_2}{T_1}}$

$(Given\frac{(V_{av}{)}_2}{(V_{av}{)}_t}=2)$
$(2{)}^2=\frac{T_2}{T_1}\Rightarrow T_2=4T_1$
$T_1=273+17=290K$

$\therefore T_2=4\times 290=1160K=1160-273={887}^{\circ }C$

Hence, the correct answer is option $\text{D}$.