
The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is
A. $899\,^\circ {\text{C}}$
B.$1492\,^\circ {\text{C}}$
C.$273\;{\text{K}}$
D. $819\,^\circ {\text{C}}$
Answer
585k+ views
Hint: Root mean square (RMS) velocity of the gas is directly proportional to the square root of the absolute temperature.
Complete Step by step answer:
From kinetic theory of gases, the pressure exerted by one mole of an ideal gas is $P = \dfrac{1}{3}\rho v_{rms}^2$
We know density $\rho = \dfrac{M}{V}$ therefore the above equation become $P = \dfrac{1}{3}\dfrac{M}{V}v_{rms}^2$
Therefore equation becomes $PV = \dfrac{1}{3}Mv_{rms}^2$
But according to ideal gas equation,$PV = RT$
Thus RMS velocity of the gas is given by,
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Here $R$ is the universal gas constant, $T$ is the absolute temperature and $M$ is the molar mass of the gas.
Let the velocity of hydrogen is ${V_2}$ at NTP.
At NTP, the temperature of the hydrogen is ${T_2} = 20^\circ {\text{C = 293}}\;{\text{K}}$ .
Molar mass of hydrogen is ${M_2} = 2\;{\text{g}}$.
Hence the velocity of hydrogen is, ${V_1} = 0.5{V_2}$.
Let the temperature of the hydrogen is ${T_1}$.
Molar mass of oxygen is ${M_1} = 32\;{\text{g}}$.
Since ${V_{rms}}$ is directly proportional to $\sqrt {\dfrac{T}{M}} $.Hence,
$\dfrac{{{V_1}}}{{{V_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}} \times \dfrac{{{M_2}}}{{{M_1}}}} $
Hence, substituting the values of ${V_1} = 0.5{V_2}$, ${T_2} = 293\;{\text{K}}$, ${M_1} = 32\;{\text{g}}$ and ${M_2} = 2\;{\text{g}}$, it comes out to be.
$
\dfrac{{0.5\,{V_2}}}{{{V_2}}} = \sqrt {\dfrac{{2{T_1}}}{{32 \times 293}}} \\
{T_1} = 1172\;{\text{K}} \\
{T_1} = 899^\circ {\text{C}} \\
$
Therefore, the correct option is (A).
Note:
- The square root of the sum of the squares of the velocities of all the gases divided by the total number of values is the root mean square velocity. Rates of effusion and diffusion are determined by the RMS velocity.
- The RMS velocity is taken into account to overcome the directional component of velocity.
Complete Step by step answer:
From kinetic theory of gases, the pressure exerted by one mole of an ideal gas is $P = \dfrac{1}{3}\rho v_{rms}^2$
We know density $\rho = \dfrac{M}{V}$ therefore the above equation become $P = \dfrac{1}{3}\dfrac{M}{V}v_{rms}^2$
Therefore equation becomes $PV = \dfrac{1}{3}Mv_{rms}^2$
But according to ideal gas equation,$PV = RT$
Thus RMS velocity of the gas is given by,
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Here $R$ is the universal gas constant, $T$ is the absolute temperature and $M$ is the molar mass of the gas.
Let the velocity of hydrogen is ${V_2}$ at NTP.
At NTP, the temperature of the hydrogen is ${T_2} = 20^\circ {\text{C = 293}}\;{\text{K}}$ .
Molar mass of hydrogen is ${M_2} = 2\;{\text{g}}$.
Hence the velocity of hydrogen is, ${V_1} = 0.5{V_2}$.
Let the temperature of the hydrogen is ${T_1}$.
Molar mass of oxygen is ${M_1} = 32\;{\text{g}}$.
Since ${V_{rms}}$ is directly proportional to $\sqrt {\dfrac{T}{M}} $.Hence,
$\dfrac{{{V_1}}}{{{V_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}} \times \dfrac{{{M_2}}}{{{M_1}}}} $
Hence, substituting the values of ${V_1} = 0.5{V_2}$, ${T_2} = 293\;{\text{K}}$, ${M_1} = 32\;{\text{g}}$ and ${M_2} = 2\;{\text{g}}$, it comes out to be.
$
\dfrac{{0.5\,{V_2}}}{{{V_2}}} = \sqrt {\dfrac{{2{T_1}}}{{32 \times 293}}} \\
{T_1} = 1172\;{\text{K}} \\
{T_1} = 899^\circ {\text{C}} \\
$
Therefore, the correct option is (A).
Note:
- The square root of the sum of the squares of the velocities of all the gases divided by the total number of values is the root mean square velocity. Rates of effusion and diffusion are determined by the RMS velocity.
- The RMS velocity is taken into account to overcome the directional component of velocity.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

