Question

Gas molecules each of mass ${10^{ - 26}}kg$are taken in a container of volume$1d{m^3}$. The root mean square speed of gas molecules is$1km{\sec ^{ - 1}}$. What temperature of gas molecules (given:${N_A} = 6 \times {10^{23}},R = 85/mol/k$)
A.$298K$
B.$25K$
C.$250K$
D.$2500K$

Answer 1

Hint: In the question, it takes one simple formula to solve it. And it is also the question that is directly substituted and getting the final value or solution. All that has to be followed is the simple steps.
The formula used: Root Mean Square (RMS)$ = \sqrt {\dfrac{{3RT}}{M}} $$ = \sqrt {\dfrac{{3kT}}{m}} $, Where $R$- is the gas constant and the value is provided
T- Is the temperature of a gas molecule that has to be calculated, so, $T = ?$
M – Is the mass of the gas and which is also provided in the question
k – Is the Avogadro number constant, which is also given to us.

Complete step-by-step answer:Now it is clear that we have to find out the temperature of gas molecules, the formula is simple
$RMS = \sqrt {\dfrac{{3RT}}{M}} = \sqrt {\dfrac{{3kT}}{m}} $, now all the values are directly substituted but as the value of temperature of gas molecule has to be found. Hence, let us first note down the values of,
$RMS = 1km{\sec ^{ - 1}} = 1000m/s$
Gas constant; $R = 8J/mol/K$
Mass; $M = {10^{ - 26}}kg$
Avogadro number constant; ${N_A} = 6 \times {10^{23}}$
Therefore on substituting it looks like,
$1000 = \sqrt {\dfrac{{3 \times 8 \times T}}{{6 \times {{10}^{23}} \times {{10}^{ - 26}}}}} $
Now, squaring on both the side gives out,
$1000 = \dfrac{{3 \times 8 \times T}}{{{{10}^{ - 26}} \times 6 \times {{10}^{23}}}}$
$1000 \times 1000 = \dfrac{{4 \times T}}{{{{10}^{ - 3}}}}$
$\begin{gathered}
  4 \times T = 1000 \\
  T = 250K \\
\end{gathered} $

Hence, the correct option is (C)

Note: Now root mean square is calculated because gas molecules are known to have different speed in a given temperature and therefore we do not have a constant value of speed and hence we calculate root mean square value. It provides us with the average value of the speed of the gasses molecules and hence this value can be used to calculate other different values.