
The absolute temperature of a gas has increased $3$ times. What will be the increase in root mean square velocity of the gas molecule?
Answer
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Hint: Gas molecules are continuously moving with different speeds in random directions. The root mean square velocity (rms) is a way to find a single velocity value for these molecules. The average velocity of gas molecules is found using the root mean square velocity formula.
Complete answer:
In the question, it is given that the absolute temperature of gas is increased 3 times. We have to find the increase in the Root mean square velocity (rms)
The formula of root mean square velocity is;
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, ${v_{rms}}$ = Root mean square velocity
$R$ = Gas constant
$T$ = Absolute temperature
$M$ = Molar mass of gas
Root mean square velocity varies with temperature.
Let ${v_{rms(1)}}$ be Root mean square velocity of gas molecules at temp ${T_1}$
And, ${v_{rms(2)}}$ be Root mean square velocity of gas molecule at temp ${T_2}$
It is given that absolute temperature is increased $3$ times. Thus;
${T_2} = 3{T_1}$
Thus, the Root mean square velocity at these two temperature will be;
${v_{rms(1)}} = \sqrt {\dfrac{{3R{T_1}}}{M}} $ ${v_{rms(2)}} = \sqrt {\dfrac{{3R{T_2}}}{M}} $ = $\sqrt {\dfrac{{3R(3{T_1})}}{M}} $
Now, we’ll divide the root mean square velocities at temp ${T_1}$ and temp ${T_2}$
$\dfrac{{{v_{rms(1)}}}}{{{v_{rms(2)}}}} = \dfrac{{\sqrt {\dfrac{{3R{T_1}}}{M}} }}{{\sqrt {\dfrac{{3R(3{T_1})}}{M}} }}$ = $\dfrac{1}{{\sqrt 3 }}$
Thus, ${v_{rms(2)}} = \sqrt 3 {v_{rms(1)}}$
The root mean square velocity at temp ${T_2}$ will increase $\sqrt 3 $ times the root mean square velocity at temp ${T_1}$ .
Note:
We have seen that when the temperature of the gas is increased $3$ times, the root mean square velocity will increase $\sqrt 3 $ times. It should be noted that Root mean square velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass of gas. While solving the numerical, temperature should always be taken in kelvin.
Complete answer:
In the question, it is given that the absolute temperature of gas is increased 3 times. We have to find the increase in the Root mean square velocity (rms)
The formula of root mean square velocity is;
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, ${v_{rms}}$ = Root mean square velocity
$R$ = Gas constant
$T$ = Absolute temperature
$M$ = Molar mass of gas
Root mean square velocity varies with temperature.
Let ${v_{rms(1)}}$ be Root mean square velocity of gas molecules at temp ${T_1}$
And, ${v_{rms(2)}}$ be Root mean square velocity of gas molecule at temp ${T_2}$
It is given that absolute temperature is increased $3$ times. Thus;
${T_2} = 3{T_1}$
Thus, the Root mean square velocity at these two temperature will be;
${v_{rms(1)}} = \sqrt {\dfrac{{3R{T_1}}}{M}} $ ${v_{rms(2)}} = \sqrt {\dfrac{{3R{T_2}}}{M}} $ = $\sqrt {\dfrac{{3R(3{T_1})}}{M}} $
Now, we’ll divide the root mean square velocities at temp ${T_1}$ and temp ${T_2}$
$\dfrac{{{v_{rms(1)}}}}{{{v_{rms(2)}}}} = \dfrac{{\sqrt {\dfrac{{3R{T_1}}}{M}} }}{{\sqrt {\dfrac{{3R(3{T_1})}}{M}} }}$ = $\dfrac{1}{{\sqrt 3 }}$
Thus, ${v_{rms(2)}} = \sqrt 3 {v_{rms(1)}}$
The root mean square velocity at temp ${T_2}$ will increase $\sqrt 3 $ times the root mean square velocity at temp ${T_1}$ .
Note:
We have seen that when the temperature of the gas is increased $3$ times, the root mean square velocity will increase $\sqrt 3 $ times. It should be noted that Root mean square velocity is directly proportional to the square root of temperature and inversely proportional to the square root of molar mass of gas. While solving the numerical, temperature should always be taken in kelvin.
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