Question

If the temperature is raised from ${27^ \circ }C$ to ${327^ \circ }C$ , by what factor will the rms velocity change?
A. $\sqrt 2 $ times
B. $2$ times
C. $2\sqrt 2 $ times
D. $4$ times

Answer 1

Hint:Root mean square velocity of individual gas molecules is given by ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ , where T is the temperature of gas molecules in Kelvin, M is the molar mass and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . To solve the above question, use this formula of root mean square velocity.



Formula used:
 RMS velocity of gas molecules is given by:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Here, T is the temperature of gas molecules in Kelvin,
M is the molar mass and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$.


Complete answer:
RMS velocity of gas molecules is given by:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ … (1)
It is given that the temperature of a gas molecule is raised from ${27^ \circ }C$ to ${327^ \circ }C$ .
Now, for a given gas molecule,
Let the initial temperature, ${T_1} = {27^ \circ }C = 300{\text{ K}}$.
Using formula (1) and calculating the RMS velocity at ${27^ \circ }C$ ,
${\left( {{v_{rms}}} \right)_1} = \sqrt {\dfrac{{3(300)R}}{M}} = \sqrt {\dfrac{{900R}}{M}} $ … (2)
Let the final temperature, ${T_2} = {327^ \circ }C = 600{\text{ K}}$ .
Using formula (1) and calculating the RMS velocity at ${52^ \circ }C$ ,
${\left( {{v_{rms}}} \right)_2} = \sqrt {\dfrac{{3(600)R}}{M}} = \sqrt {\dfrac{{1800R}}{M}} $ … (3)
Dividing (3) by (2),

$\dfrac{{{{\left( {{v_{rms}}} \right)}_2}}}{{{{\left( {{v_{rms}}} \right)}_1}}} = \sqrt {\dfrac{{1800}}{{900}}} = \sqrt 2 $
Therefore, ${\left( {{v_{rms}}} \right)_2} = \sqrt 2 {\left( {{v_{rms}}} \right)_1}$ , that is, the root mean square velocity will change by $\sqrt 2 $ times.
Thus, the correct option is A.


Note: For individual gas molecules, It is calculated as ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $ . Root Mean Square velocity is directly proportional to the absolute temperature of the gas and inversely proportional to the molar mass. The above question can also be solved using $\dfrac{{{{\left( {{v_{rms}}} \right)}_2}}}{{{{\left( {{v_{rms}}} \right)}_1}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}} $ .