Question

At what temperature will oxygen molecules have the same root mean square speed as hydrogen molecules at $200{\text{ K}}$.
$(i){\text{ 527 }}{}^ \circ {\text{C}}$
$(ii){\text{ 1027 }}{}^ \circ {\text{C}}$
$(iii){\text{ 1527 }}{}^ \circ {\text{C}}$
$(iv){\text{ 2927 }}{}^ \circ {\text{C}}$

Answer 1

Hint: The root means square speed measures the speed of gas particles at ideal conditions. Root means square speed depends on the molecular weight of the gas and the temperature of the gas at that point of time. Thus it measures the speed of all particles moving in random directions.

Formula Used:
${V_{r.m.s}}{\text{ = }}\sqrt {\dfrac{{3RT}}{M}} $
Where,
 $T{\text{ = }}$Temperature of gas
${\text{R = }}$Universal gas constant
${\text{M = }}$ Molecular weight of gas

Complete answer:
We know that root mean square speed is the measure of speed of gas particles moving in random direction at particular temperature. It is denoted by${V_{r.m.s}}$. We have a formula for the root means square speed as:
${V_{r.m.s}}{\text{ = }}\sqrt {\dfrac{{3RT}}{M}} $
From the above formula we observe that ${V_{r.m.s}}$only depends on the temperature of gas and the molecular weight of gas. Other quantities are constant in the above formula. Thus it can be deduced as:
${V_{r.m.s}}{\text{ }} \propto {\text{ }}\sqrt {\dfrac{T}{M}} $ ________________$(1)$
Now according to given question,
For oxygen molecule we have,
Molecular weight $ = {\text{ 16 }} \times {\text{ 2 g}}$
Molecular weight $ = {\text{ 32 g}}$
Temperature of gas $ = {\text{ T }}$
Now putting above value in equation $(1)$ we have,
${V_{r.m.s}}{\text{ }} \propto {\text{ }}\sqrt {\dfrac{T}{{32}}} $
Similarly for hydrogen molecule we have,
Molecular weight $ = {\text{ 2 g}}$
Temperature of gas $ = {\text{ 200 K}}$
Substituting values in equation $(1)$ we have,
${V_{r.m.s}}{\text{ }} \propto {\text{ }}\sqrt {\dfrac{{200}}{2}} $
Now according to question ${V_{r.m.s}}$ for both gases are equal. Therefore we can write as,
$\sqrt {\dfrac{T}{{32}}} {\text{ = }}\sqrt {\dfrac{{200}}{2}} $
Taking square of both sides and solving the equation we get,
$T{\text{ = 3200 K}}$
For converting it into degree celcius, subtract $273{\text{ K}}$from this value.
$T{\text{ = 3200 - 273 }}{}^ \circ {\text{C}}$
$T{\text{ = 2927 }}{}^ \circ {\text{C}}$
Hence at ${\text{2927 }}{}^ \circ {\text{C}}$ oxygen molecules have the same root mean square speed as hydrogen molecules at $200{\text{ K}}$.

Note:
Universal gas constant is constant for a given standard of units. Its value changes only when units are changed. The temperature should be taken in kelvin. After finding the result we may convert it into degree celcius. The weight which is taken must be the weight of a molecule of gas. It should not be atomic weight.