Question

At what temperature pressure remaining constant will the root mean square velocity of a gas be half of its value at $0^{o}C$ ?

• A. $-{204}^{o}C$
• B. ${6825}^{o}C$
• C. ${404}^{o}C$
• D. None of these

Answer 1

Answer: (A)

$-{204}^{o}C$

We have,
$\frac{C_1^2}{C_2^2}=\frac{T_1}{T_2}$
$\frac{C_1}{\left(\frac{C_1}{2}\right)}=\frac{273}{T_2}$
$T_2=\frac{273}{4}$
$=68.25K$
$t=68.25-273$
$=-{204.25}^{o}C$