139292
If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by
1 $1+\frac{2}{\mathrm{f}}$
2 $1-\frac{2}{\mathrm{f}}$
3 $1+\frac{1}{\mathrm{f}}$
4 $1-\frac{1}{\mathrm{f}}$
Explanation:
A We know that, Ratio of two specific heat of gas is:- $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ We know that, $\gamma=1+\frac{2}{\mathrm{f}}$ Where, $f=$ degree of Freedom of gas
AMU-2002
Kinetic Theory of Gases
139293
At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} \mathrm{C}$ ?
1 $313^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $393^{\circ} \mathrm{C}$
4 $586^{\circ} \mathrm{C}$
Explanation:
A Given that, $\mathrm{T}_{1}=20^{\circ} \mathrm{C}=273+20$ $\mathrm{T}_{2}=273+\mathrm{T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=2 \mathrm{E}$ We know that, Kinetic energy $(\mathrm{E})=\frac{3}{2} \mathrm{RT}$ $\mathrm{E} \propto \mathrm{T}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}}{2 \mathrm{E}}=\frac{273+20}{273+\mathrm{T}}$ $586=273+\mathrm{T}$ $\mathrm{T}=313^{\circ} \mathrm{C}$
AMU-2002
Kinetic Theory of Gases
139295
The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then
1 $\mathrm{E}_{1}=\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{1} \lt \mathrm{E}_{2}$
4 $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ cannot be compared
Explanation:
A We know that, $\mathrm{KE} \propto \frac{3}{2} \mathrm{kT}$ Here, Hence, $\mathrm{CO}$ and $\mathrm{N}_{2}$ both are diatomic gas $\left(\mathrm{KE}_{1}\right)_{\mathrm{Co}}=\frac{5}{2} \mathrm{kT}$ $\left(\mathrm{KE}_{2}\right)_{\mathrm{N} 2}=\frac{5}{2} \mathrm{kT}$ Hence, For same Temperature, $\mathrm{E}_{1}=\mathrm{E}_{2}$.
MHT-CET 2005
Kinetic Theory of Gases
139296
The degrees of freedom of a molecule of a triatomic gas are
1 2
2 4
3 6
4 8
Explanation:
C Degree of Freedom (f) $=3 \mathrm{~N}-\mathrm{n}$ Where, $\mathrm{N}=$ total no of particle $\mathrm{n}=$ holonomic constraints For triatomic molecule- no of particle $(\mathrm{N})=3$ and Separation between 3 atom is Fixed i.e. no of holonomic constraints $(\mathrm{n})=3$ Hence, $\mathrm{f}= 3 \mathrm{~N}-\mathrm{n}$ $=3 \times 3-3$ $=9-3$ $\mathrm{f} =6$ Degree of Freedom of diatomic molecule is 6.
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Kinetic Theory of Gases
139292
If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by
1 $1+\frac{2}{\mathrm{f}}$
2 $1-\frac{2}{\mathrm{f}}$
3 $1+\frac{1}{\mathrm{f}}$
4 $1-\frac{1}{\mathrm{f}}$
Explanation:
A We know that, Ratio of two specific heat of gas is:- $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ We know that, $\gamma=1+\frac{2}{\mathrm{f}}$ Where, $f=$ degree of Freedom of gas
AMU-2002
Kinetic Theory of Gases
139293
At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} \mathrm{C}$ ?
1 $313^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $393^{\circ} \mathrm{C}$
4 $586^{\circ} \mathrm{C}$
Explanation:
A Given that, $\mathrm{T}_{1}=20^{\circ} \mathrm{C}=273+20$ $\mathrm{T}_{2}=273+\mathrm{T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=2 \mathrm{E}$ We know that, Kinetic energy $(\mathrm{E})=\frac{3}{2} \mathrm{RT}$ $\mathrm{E} \propto \mathrm{T}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}}{2 \mathrm{E}}=\frac{273+20}{273+\mathrm{T}}$ $586=273+\mathrm{T}$ $\mathrm{T}=313^{\circ} \mathrm{C}$
AMU-2002
Kinetic Theory of Gases
139295
The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then
1 $\mathrm{E}_{1}=\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{1} \lt \mathrm{E}_{2}$
4 $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ cannot be compared
Explanation:
A We know that, $\mathrm{KE} \propto \frac{3}{2} \mathrm{kT}$ Here, Hence, $\mathrm{CO}$ and $\mathrm{N}_{2}$ both are diatomic gas $\left(\mathrm{KE}_{1}\right)_{\mathrm{Co}}=\frac{5}{2} \mathrm{kT}$ $\left(\mathrm{KE}_{2}\right)_{\mathrm{N} 2}=\frac{5}{2} \mathrm{kT}$ Hence, For same Temperature, $\mathrm{E}_{1}=\mathrm{E}_{2}$.
MHT-CET 2005
Kinetic Theory of Gases
139296
The degrees of freedom of a molecule of a triatomic gas are
1 2
2 4
3 6
4 8
Explanation:
C Degree of Freedom (f) $=3 \mathrm{~N}-\mathrm{n}$ Where, $\mathrm{N}=$ total no of particle $\mathrm{n}=$ holonomic constraints For triatomic molecule- no of particle $(\mathrm{N})=3$ and Separation between 3 atom is Fixed i.e. no of holonomic constraints $(\mathrm{n})=3$ Hence, $\mathrm{f}= 3 \mathrm{~N}-\mathrm{n}$ $=3 \times 3-3$ $=9-3$ $\mathrm{f} =6$ Degree of Freedom of diatomic molecule is 6.
139292
If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by
1 $1+\frac{2}{\mathrm{f}}$
2 $1-\frac{2}{\mathrm{f}}$
3 $1+\frac{1}{\mathrm{f}}$
4 $1-\frac{1}{\mathrm{f}}$
Explanation:
A We know that, Ratio of two specific heat of gas is:- $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ We know that, $\gamma=1+\frac{2}{\mathrm{f}}$ Where, $f=$ degree of Freedom of gas
AMU-2002
Kinetic Theory of Gases
139293
At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} \mathrm{C}$ ?
1 $313^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $393^{\circ} \mathrm{C}$
4 $586^{\circ} \mathrm{C}$
Explanation:
A Given that, $\mathrm{T}_{1}=20^{\circ} \mathrm{C}=273+20$ $\mathrm{T}_{2}=273+\mathrm{T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=2 \mathrm{E}$ We know that, Kinetic energy $(\mathrm{E})=\frac{3}{2} \mathrm{RT}$ $\mathrm{E} \propto \mathrm{T}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}}{2 \mathrm{E}}=\frac{273+20}{273+\mathrm{T}}$ $586=273+\mathrm{T}$ $\mathrm{T}=313^{\circ} \mathrm{C}$
AMU-2002
Kinetic Theory of Gases
139295
The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then
1 $\mathrm{E}_{1}=\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{1} \lt \mathrm{E}_{2}$
4 $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ cannot be compared
Explanation:
A We know that, $\mathrm{KE} \propto \frac{3}{2} \mathrm{kT}$ Here, Hence, $\mathrm{CO}$ and $\mathrm{N}_{2}$ both are diatomic gas $\left(\mathrm{KE}_{1}\right)_{\mathrm{Co}}=\frac{5}{2} \mathrm{kT}$ $\left(\mathrm{KE}_{2}\right)_{\mathrm{N} 2}=\frac{5}{2} \mathrm{kT}$ Hence, For same Temperature, $\mathrm{E}_{1}=\mathrm{E}_{2}$.
MHT-CET 2005
Kinetic Theory of Gases
139296
The degrees of freedom of a molecule of a triatomic gas are
1 2
2 4
3 6
4 8
Explanation:
C Degree of Freedom (f) $=3 \mathrm{~N}-\mathrm{n}$ Where, $\mathrm{N}=$ total no of particle $\mathrm{n}=$ holonomic constraints For triatomic molecule- no of particle $(\mathrm{N})=3$ and Separation between 3 atom is Fixed i.e. no of holonomic constraints $(\mathrm{n})=3$ Hence, $\mathrm{f}= 3 \mathrm{~N}-\mathrm{n}$ $=3 \times 3-3$ $=9-3$ $\mathrm{f} =6$ Degree of Freedom of diatomic molecule is 6.
139292
If the degrees of freedom of a gas are $f$ then the ratio of its specific heat $\frac{C_{p}}{C_{v}}$ is given by
1 $1+\frac{2}{\mathrm{f}}$
2 $1-\frac{2}{\mathrm{f}}$
3 $1+\frac{1}{\mathrm{f}}$
4 $1-\frac{1}{\mathrm{f}}$
Explanation:
A We know that, Ratio of two specific heat of gas is:- $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma$ We know that, $\gamma=1+\frac{2}{\mathrm{f}}$ Where, $f=$ degree of Freedom of gas
AMU-2002
Kinetic Theory of Gases
139293
At which of the following temperature would the molecules of a gas have twice, the average kinetic energy they have at $20^{\circ} \mathrm{C}$ ?
1 $313^{\circ} \mathrm{C}$
2 $373^{\circ} \mathrm{C}$
3 $393^{\circ} \mathrm{C}$
4 $586^{\circ} \mathrm{C}$
Explanation:
A Given that, $\mathrm{T}_{1}=20^{\circ} \mathrm{C}=273+20$ $\mathrm{T}_{2}=273+\mathrm{T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=2 \mathrm{E}$ We know that, Kinetic energy $(\mathrm{E})=\frac{3}{2} \mathrm{RT}$ $\mathrm{E} \propto \mathrm{T}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{\mathrm{E}}{2 \mathrm{E}}=\frac{273+20}{273+\mathrm{T}}$ $586=273+\mathrm{T}$ $\mathrm{T}=313^{\circ} \mathrm{C}$
AMU-2002
Kinetic Theory of Gases
139295
The gases carbon-monoxide (CO) and nitrogen at the same temperature have kinetic energies $E_{1}$ and $E_{2}$ respectively. Then
1 $\mathrm{E}_{1}=\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{1} \lt \mathrm{E}_{2}$
4 $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$ cannot be compared
Explanation:
A We know that, $\mathrm{KE} \propto \frac{3}{2} \mathrm{kT}$ Here, Hence, $\mathrm{CO}$ and $\mathrm{N}_{2}$ both are diatomic gas $\left(\mathrm{KE}_{1}\right)_{\mathrm{Co}}=\frac{5}{2} \mathrm{kT}$ $\left(\mathrm{KE}_{2}\right)_{\mathrm{N} 2}=\frac{5}{2} \mathrm{kT}$ Hence, For same Temperature, $\mathrm{E}_{1}=\mathrm{E}_{2}$.
MHT-CET 2005
Kinetic Theory of Gases
139296
The degrees of freedom of a molecule of a triatomic gas are
1 2
2 4
3 6
4 8
Explanation:
C Degree of Freedom (f) $=3 \mathrm{~N}-\mathrm{n}$ Where, $\mathrm{N}=$ total no of particle $\mathrm{n}=$ holonomic constraints For triatomic molecule- no of particle $(\mathrm{N})=3$ and Separation between 3 atom is Fixed i.e. no of holonomic constraints $(\mathrm{n})=3$ Hence, $\mathrm{f}= 3 \mathrm{~N}-\mathrm{n}$ $=3 \times 3-3$ $=9-3$ $\mathrm{f} =6$ Degree of Freedom of diatomic molecule is 6.