If the root mean square velocity of hydrogen molecule at a given temperature and pressure is 2 km/s, the root mean square velocity of oxygen at the same condition in km/s is :
- 2.0
- 0.5
- 1.5
- 1.0
The Correct Option is B
Approach Solution - 1
To find the root mean square velocity of oxygen at the same temperature and pressure, we use the formula for root mean square velocity, which is given by:
\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\)
where \(k\) is the Boltzmann's constant, \(T\) is the absolute temperature, and \(m\) is the mass of the molecule.
We know that the root mean square velocity for hydrogen is given as 2 km/s. Let this be \(v_{\text{rms, H}_2}\) and for oxygen it will be \(v_{\text{rms, O}_2}\).
We use the relation between the root mean square velocities and the molar masses:
\(\frac{v_{\text{rms, H}_2}}{v_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}}\)
Where:
- \(M_{\text{H}_2} = 2 \text{ g/mol}\) (molar mass of hydrogen)
- \(M_{\text{O}_2} = 32 \text{ g/mol}\) (molar mass of oxygen)
Substituting the values, we get:
\(\frac{2}{v_{\text{rms, O}_2}} = \sqrt{\frac{32}{2}}\)
\(\frac{2}{v_{\text{rms, O}_2}} = \sqrt{16}\)
\(\frac{2}{v_{\text{rms, O}_2}} = 4\)
Solving for \(v_{\text{rms, O}_2}\):
\(v_{\text{rms, O}_2} = \frac{2}{4} = 0.5 \text{ km/s}\)
Hence, the root mean square velocity of oxygen at the same condition is 0.5 km/s.
Approach Solution -2
Given: - Root mean square (rms) velocity of hydrogen (\(v_{H_2}\)) = 2 km/s - Molecular mass of hydrogen (\(M_{H_2}\)) = 2 g/mol - Molecular mass of oxygen (\(M_{O_2}\)) = 32 g/mol
Step 1: Relationship for Root Mean Square Velocity
The root mean square velocity of a gas is given by:
\[ v_{\text{rms}} \propto \frac{1}{\sqrt{M}} \]
where \(M\) is the molar mass of the gas.
Step 2: Calculating the Ratio of Velocities
Using the inverse square root relationship for hydrogen and oxygen:
\[ \frac{v_{O_2}}{v_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]
Substituting the given values:
\[ \frac{v_{O_2}}{2} = \sqrt{\frac{2}{32}} \] \[ \frac{v_{O_2}}{2} = \sqrt{\frac{1}{16}} \] \[ \frac{v_{O_2}}{2} = \frac{1}{4} \]
Multiplying both sides by 2:
\[ v_{O_2} = \frac{1}{4} \times 2 = 0.5 \, \text{km/s} \]
Conclusion:
The root mean square velocity of oxygen at the same conditions is 0.5 km/s.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main The Kinetic Theory of Gases Questions
- The temperature of a gas having \( 2.0 \times 10^{25} \) molecules per cubic meter at 1.38 atm (Given, \( k = 1.38 \times 10^{-23} \, \text{JK}^{-1} \)) is:
- The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use \( k_B = 1.38 \times 10^{-23} \, \text{J/mol-K} \))
- The parameter that remains the same for molecules of all gases at a given temperature is :
- The temperature of a gas is \(-78^\circ\text{C}\) and the average translational kinetic energy of its molecules is \( K \). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \( 2K \) is:
- The speed of sound in oxygen at S.T.P. will be approximately: Given, \( R = 8.3 \, \text{J K}^{-1} \), \( \gamma = 1.4 \))
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.