139175
According to the assumption made in the kinetic theory of gases, when two molecules of a gas collide with each other then
1 neither K.E. nor momentum is conserved.
2 both K.E. and momentum are conserved.
3 momentum is conserved but K.E. is not conserved.
4 K.E. is conserved but momentum is not conserved.
Explanation:
B There are two types of collisions (i) Elastic collisions (ii) Inelastic collision Though in elastic collision both energy and velocity does not get changed. Therefore momentum and kinetic energy both are conserved. That means energy is preserved and velocity also does not change. But in inelastic collision both energy and velocity get changed due to number of factors during the collision or after the collision.
MHT-CET 2020
Kinetic Theory of Gases
139177
Mean free path of molecules in a polyatomic gas is independent of
1 number density of the molecules
2 volume of the molecule
3 temperature of the gas
4 gas constant $\mathrm{R}$
Explanation:
D The mean free path of a gas molecule is the average distance between two successive collisions mean path of molecule is represent $\lambda$. $\lambda=\frac{1}{\sqrt{2} \cdot \pi \mathrm{d}^{2} \mathrm{n}}$ Where, $\mathrm{d}=$ Diameter of molecule $\mathrm{n}=$ Number of molecule per unit volume Mean free path of molecule is depends on number of the molecules, volume of molecule and temperature of gas.
TS- EAMCET-14.09.2020
Kinetic Theory of Gases
139181
If the average kinetic energy of a molecule of a hydrogen gas at $300 \mathrm{~K}$ is $\mathrm{E}$, then the average kinetic energy of a molecule of a nitrogen gas at the same temperature is
1 $7 \mathrm{E}$
2 $\mathrm{E} / 14$
3 $14 \mathrm{E}$
4 $\mathrm{E} / 7$
5 $\mathrm{E}$
Explanation:
E We know that, average translational kinetic energy of gas molecule $=\frac{3}{2} K_{B} T$ $\text { K.E. } \propto \mathrm{T} \quad\left(\because \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}_{\mathrm{A}}}\right)$ Given, a molecule of a hydrogen gas (K.E) $=\mathrm{E}$ Then, the kinetic energy of a molecule of $\mathrm{N}_{2}$ gas $=\mathrm{E}$
AP EAMCET (18.09.2020) Shift-I
Kinetic Theory of Gases
139188
The ratio of R.M.S. velocities of hydrogen molecules to oxygen molecules at $273^{\circ} \mathrm{C}$ is (molecular wt. of hydrogen and oxygen is 2 and 32 respectively.
1 $1: 8$
2 $16: 1$
3 $1: 4$
4 $4: 1$
Explanation:
D We know that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O} 2}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H} 2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}=\sqrt{\frac{32}{2}}=\frac{4}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}:\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=4: 1$
139175
According to the assumption made in the kinetic theory of gases, when two molecules of a gas collide with each other then
1 neither K.E. nor momentum is conserved.
2 both K.E. and momentum are conserved.
3 momentum is conserved but K.E. is not conserved.
4 K.E. is conserved but momentum is not conserved.
Explanation:
B There are two types of collisions (i) Elastic collisions (ii) Inelastic collision Though in elastic collision both energy and velocity does not get changed. Therefore momentum and kinetic energy both are conserved. That means energy is preserved and velocity also does not change. But in inelastic collision both energy and velocity get changed due to number of factors during the collision or after the collision.
MHT-CET 2020
Kinetic Theory of Gases
139177
Mean free path of molecules in a polyatomic gas is independent of
1 number density of the molecules
2 volume of the molecule
3 temperature of the gas
4 gas constant $\mathrm{R}$
Explanation:
D The mean free path of a gas molecule is the average distance between two successive collisions mean path of molecule is represent $\lambda$. $\lambda=\frac{1}{\sqrt{2} \cdot \pi \mathrm{d}^{2} \mathrm{n}}$ Where, $\mathrm{d}=$ Diameter of molecule $\mathrm{n}=$ Number of molecule per unit volume Mean free path of molecule is depends on number of the molecules, volume of molecule and temperature of gas.
TS- EAMCET-14.09.2020
Kinetic Theory of Gases
139181
If the average kinetic energy of a molecule of a hydrogen gas at $300 \mathrm{~K}$ is $\mathrm{E}$, then the average kinetic energy of a molecule of a nitrogen gas at the same temperature is
1 $7 \mathrm{E}$
2 $\mathrm{E} / 14$
3 $14 \mathrm{E}$
4 $\mathrm{E} / 7$
5 $\mathrm{E}$
Explanation:
E We know that, average translational kinetic energy of gas molecule $=\frac{3}{2} K_{B} T$ $\text { K.E. } \propto \mathrm{T} \quad\left(\because \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}_{\mathrm{A}}}\right)$ Given, a molecule of a hydrogen gas (K.E) $=\mathrm{E}$ Then, the kinetic energy of a molecule of $\mathrm{N}_{2}$ gas $=\mathrm{E}$
AP EAMCET (18.09.2020) Shift-I
Kinetic Theory of Gases
139188
The ratio of R.M.S. velocities of hydrogen molecules to oxygen molecules at $273^{\circ} \mathrm{C}$ is (molecular wt. of hydrogen and oxygen is 2 and 32 respectively.
1 $1: 8$
2 $16: 1$
3 $1: 4$
4 $4: 1$
Explanation:
D We know that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O} 2}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H} 2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}=\sqrt{\frac{32}{2}}=\frac{4}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}:\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=4: 1$
139175
According to the assumption made in the kinetic theory of gases, when two molecules of a gas collide with each other then
1 neither K.E. nor momentum is conserved.
2 both K.E. and momentum are conserved.
3 momentum is conserved but K.E. is not conserved.
4 K.E. is conserved but momentum is not conserved.
Explanation:
B There are two types of collisions (i) Elastic collisions (ii) Inelastic collision Though in elastic collision both energy and velocity does not get changed. Therefore momentum and kinetic energy both are conserved. That means energy is preserved and velocity also does not change. But in inelastic collision both energy and velocity get changed due to number of factors during the collision or after the collision.
MHT-CET 2020
Kinetic Theory of Gases
139177
Mean free path of molecules in a polyatomic gas is independent of
1 number density of the molecules
2 volume of the molecule
3 temperature of the gas
4 gas constant $\mathrm{R}$
Explanation:
D The mean free path of a gas molecule is the average distance between two successive collisions mean path of molecule is represent $\lambda$. $\lambda=\frac{1}{\sqrt{2} \cdot \pi \mathrm{d}^{2} \mathrm{n}}$ Where, $\mathrm{d}=$ Diameter of molecule $\mathrm{n}=$ Number of molecule per unit volume Mean free path of molecule is depends on number of the molecules, volume of molecule and temperature of gas.
TS- EAMCET-14.09.2020
Kinetic Theory of Gases
139181
If the average kinetic energy of a molecule of a hydrogen gas at $300 \mathrm{~K}$ is $\mathrm{E}$, then the average kinetic energy of a molecule of a nitrogen gas at the same temperature is
1 $7 \mathrm{E}$
2 $\mathrm{E} / 14$
3 $14 \mathrm{E}$
4 $\mathrm{E} / 7$
5 $\mathrm{E}$
Explanation:
E We know that, average translational kinetic energy of gas molecule $=\frac{3}{2} K_{B} T$ $\text { K.E. } \propto \mathrm{T} \quad\left(\because \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}_{\mathrm{A}}}\right)$ Given, a molecule of a hydrogen gas (K.E) $=\mathrm{E}$ Then, the kinetic energy of a molecule of $\mathrm{N}_{2}$ gas $=\mathrm{E}$
AP EAMCET (18.09.2020) Shift-I
Kinetic Theory of Gases
139188
The ratio of R.M.S. velocities of hydrogen molecules to oxygen molecules at $273^{\circ} \mathrm{C}$ is (molecular wt. of hydrogen and oxygen is 2 and 32 respectively.
1 $1: 8$
2 $16: 1$
3 $1: 4$
4 $4: 1$
Explanation:
D We know that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O} 2}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H} 2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}=\sqrt{\frac{32}{2}}=\frac{4}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}:\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=4: 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Kinetic Theory of Gases
139175
According to the assumption made in the kinetic theory of gases, when two molecules of a gas collide with each other then
1 neither K.E. nor momentum is conserved.
2 both K.E. and momentum are conserved.
3 momentum is conserved but K.E. is not conserved.
4 K.E. is conserved but momentum is not conserved.
Explanation:
B There are two types of collisions (i) Elastic collisions (ii) Inelastic collision Though in elastic collision both energy and velocity does not get changed. Therefore momentum and kinetic energy both are conserved. That means energy is preserved and velocity also does not change. But in inelastic collision both energy and velocity get changed due to number of factors during the collision or after the collision.
MHT-CET 2020
Kinetic Theory of Gases
139177
Mean free path of molecules in a polyatomic gas is independent of
1 number density of the molecules
2 volume of the molecule
3 temperature of the gas
4 gas constant $\mathrm{R}$
Explanation:
D The mean free path of a gas molecule is the average distance between two successive collisions mean path of molecule is represent $\lambda$. $\lambda=\frac{1}{\sqrt{2} \cdot \pi \mathrm{d}^{2} \mathrm{n}}$ Where, $\mathrm{d}=$ Diameter of molecule $\mathrm{n}=$ Number of molecule per unit volume Mean free path of molecule is depends on number of the molecules, volume of molecule and temperature of gas.
TS- EAMCET-14.09.2020
Kinetic Theory of Gases
139181
If the average kinetic energy of a molecule of a hydrogen gas at $300 \mathrm{~K}$ is $\mathrm{E}$, then the average kinetic energy of a molecule of a nitrogen gas at the same temperature is
1 $7 \mathrm{E}$
2 $\mathrm{E} / 14$
3 $14 \mathrm{E}$
4 $\mathrm{E} / 7$
5 $\mathrm{E}$
Explanation:
E We know that, average translational kinetic energy of gas molecule $=\frac{3}{2} K_{B} T$ $\text { K.E. } \propto \mathrm{T} \quad\left(\because \mathrm{K}_{\mathrm{B}}=\frac{\mathrm{R}}{\mathrm{N}_{\mathrm{A}}}\right)$ Given, a molecule of a hydrogen gas (K.E) $=\mathrm{E}$ Then, the kinetic energy of a molecule of $\mathrm{N}_{2}$ gas $=\mathrm{E}$
AP EAMCET (18.09.2020) Shift-I
Kinetic Theory of Gases
139188
The ratio of R.M.S. velocities of hydrogen molecules to oxygen molecules at $273^{\circ} \mathrm{C}$ is (molecular wt. of hydrogen and oxygen is 2 and 32 respectively.
1 $1: 8$
2 $16: 1$
3 $1: 4$
4 $4: 1$
Explanation:
D We know that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O} 2}}}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{H} 2}}}$ $\frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}{\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}}=\sqrt{\frac{32}{2}}=\frac{4}{1}$ $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}:\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=4: 1$
