Question

Total kinetic energy of sample of a gas which contains $1\times {10}^{22}$ molecules is $24\times {10}^2$ $J$ at ${127}^{o}C$. Another sample of gas at ${27}^{o}C$ has a total kinetic energy of $48\times {10}^2$ $J$. The number of molecules in the second sample are:

• A. $2.67\times {10}^{22}$
• B. $1\times {10}^{22}$
• C. $1\times {10}^{24}$
• D. $5.5\times {10}^{21}$

Answer 1

Answer: (A)

$2.67\times {10}^{22}$

$\frac{K{E}_1}{K{E}_2}=\frac{n_1{T}_1}{n_2{T}_2}$
$\frac{24\times {10}^2}{48\times {10}^2}=\frac{n_1\times 400}{n_2\times 300}$
As number of molecules, say $x=n\times N_A$

$n_1=1\times {10}^{22}\times 6.023\times {10}^{23}$

$\therefore n_2=4/90$

$x_2=n_2\times N_A$
Thus, number of molecules of gas in the second sample are $2.67\times {10}^{22}$.