139261
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes
1 $\mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $2 \mathrm{v}$
4 $4 \mathrm{v}$
Explanation:
C rms speed $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ When temperature is doubled and molecular dissociates into atoms, $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T} / \mathrm{M}}{2 \mathrm{~T} / \mathrm{M} / 2}}=\sqrt{\frac{1}{4}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2}$ $\frac{\mathrm{v}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2} \Rightarrow\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}=2 \mathrm{v}$
WB JEE 2015
Kinetic Theory of Gases
139262
The r.m.s speed of the molecules of a gas at $100^{\circ} \mathrm{C}$ is $\mathrm{v}$. The temperature at which the r.m.s speed will be $\sqrt{3} \mathrm{v}$ is
139266
The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E=\mathrm{as}^{2}$ where, $\mathrm{a}$ is constant. The force acting on the particle is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
139261
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes
1 $\mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $2 \mathrm{v}$
4 $4 \mathrm{v}$
Explanation:
C rms speed $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ When temperature is doubled and molecular dissociates into atoms, $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T} / \mathrm{M}}{2 \mathrm{~T} / \mathrm{M} / 2}}=\sqrt{\frac{1}{4}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2}$ $\frac{\mathrm{v}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2} \Rightarrow\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}=2 \mathrm{v}$
WB JEE 2015
Kinetic Theory of Gases
139262
The r.m.s speed of the molecules of a gas at $100^{\circ} \mathrm{C}$ is $\mathrm{v}$. The temperature at which the r.m.s speed will be $\sqrt{3} \mathrm{v}$ is
139266
The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E=\mathrm{as}^{2}$ where, $\mathrm{a}$ is constant. The force acting on the particle is
139261
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes
1 $\mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $2 \mathrm{v}$
4 $4 \mathrm{v}$
Explanation:
C rms speed $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ When temperature is doubled and molecular dissociates into atoms, $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T} / \mathrm{M}}{2 \mathrm{~T} / \mathrm{M} / 2}}=\sqrt{\frac{1}{4}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2}$ $\frac{\mathrm{v}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2} \Rightarrow\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}=2 \mathrm{v}$
WB JEE 2015
Kinetic Theory of Gases
139262
The r.m.s speed of the molecules of a gas at $100^{\circ} \mathrm{C}$ is $\mathrm{v}$. The temperature at which the r.m.s speed will be $\sqrt{3} \mathrm{v}$ is
139266
The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E=\mathrm{as}^{2}$ where, $\mathrm{a}$ is constant. The force acting on the particle is
139261
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes
1 $\mathrm{v}$
2 $\sqrt{2} \mathrm{v}$
3 $2 \mathrm{v}$
4 $4 \mathrm{v}$
Explanation:
C rms speed $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$ When temperature is doubled and molecular dissociates into atoms, $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\sqrt{\frac{\mathrm{T} / \mathrm{M}}{2 \mathrm{~T} / \mathrm{M} / 2}}=\sqrt{\frac{1}{4}}$ $\frac{\left(\mathrm{v}_{1}\right)_{\mathrm{rms}}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2}$ $\frac{\mathrm{v}}{\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}}=\frac{1}{2} \Rightarrow\left(\mathrm{v}_{2}\right)_{\mathrm{rms}}=2 \mathrm{v}$
WB JEE 2015
Kinetic Theory of Gases
139262
The r.m.s speed of the molecules of a gas at $100^{\circ} \mathrm{C}$ is $\mathrm{v}$. The temperature at which the r.m.s speed will be $\sqrt{3} \mathrm{v}$ is
139266
The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends on the distance covered. It is given as $K E=\mathrm{as}^{2}$ where, $\mathrm{a}$ is constant. The force acting on the particle is