Question

The $KE$ of $N$ molecule of $O_{2}$ is x Joules at $-123^{\circ}C$. Another sample of $O_{2}$ at $27^{\circ}C$ has a $KE$ of $2x$ Joules. The latter sample contains

• A. $N$ molecules of $O_{2}$
• B. $2N$ molecules of $O_{2}$
• C. $N/2$ molecules of $O_{2}$
• D. $N/4$ molecule of $O_{2}$

Answer 1

Answer: (A)

$KE=\frac{3}{2}RT; T=-12+273=+150\,L$
$\frac{3}{2} \times R \times 150=3 \times 8.314 \times75 =x\,J$
$=225 \times 8.314=x$
At $27^{\circ}C=27+223=300\,K$
KE for $=2x\,J =\frac{3}{2}\times 8.314\times 300$
N molecules
$\therefore $ $x$ Joule $=3 \times 8.314 \times 75$
In both the cases x J correspond to N molecules