139213
Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :
1 decreases
2 increases
3 remains same
4 increase of decreases depending on the nature of gas
Explanation:
C We know that, K.E $=\frac{3}{2} \mathrm{KT}$ And temperature $\mathrm{T}=$ Constant But K.E only depends on temperature So, Kinetic energy remains same.
Karnataka CET-2018
Kinetic Theory of Gases
139229
Mean free path of a gas molecule is
1 inversely proportional to number of molecules per unit volume
2 inversely proportional to diameter of the molecule
3 directly proportional to the square root of the absolute temperature
4 directly proportional to the molecular mass
5 independent of temperature
Explanation:
A We know that, mean free path $(\lambda)=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{nd}^{2}}$ Where, $\mathrm{n}=$ number of molecules per unit volume $\mathrm{d}=$ diameter of the molecule. Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.
Kerala CEE - 2009
Kinetic Theory of Gases
139230
The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 \mathrm{~K}$ is
1 $600 \mathrm{~K}$
2 $2400 \mathrm{~K}$
3 $1200 \mathrm{~K}$
4 $300 \mathrm{~K}$
5 $4800 \mathrm{~K}$
Explanation:
E Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Then, $\quad \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R}(300)}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Or, $\quad \mathrm{T}=300 \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}=300 \times \frac{32}{2}=300 \times 16=4800 \mathrm{~K}$
Kerala CEE - 2009
Kinetic Theory of Gases
139231
The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{-14} \mathrm{~J}$ is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
1 $2 \times 10^{9} \mathrm{~K}$
2 $10^{9} \mathrm{~K}$
3 $6 \times 10^{9} \mathrm{~K}$
4 $3 \times 10^{9} \mathrm{~K}$
5 $4.5 \times 10^{9} \mathrm{~K}$
Explanation:
A Given that, Boltzmann constant $(\mathrm{k})=1.38 \times$ $10^{-23} \mathrm{JK}^{-1}$, Energy of proton gas $=4.14 \times 10^{-14} \mathrm{~J}$ We know, Energy of proton gas $(\mathrm{E})=\frac{3}{2} \mathrm{k} \cdot \mathrm{T}$ Putting these value, we get- $4.14 \times 10^{-14}=\frac{3}{2} \times 1.38 \times 10^{-23} \times \mathrm{T}$ Or $\quad \mathrm{T}=\frac{2 \times 4.14 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=\frac{8.28 \times 10^{-14}}{4.14 \times 10^{-23}}=2 \times 10^{9} \mathrm{~K}$
139213
Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :
1 decreases
2 increases
3 remains same
4 increase of decreases depending on the nature of gas
Explanation:
C We know that, K.E $=\frac{3}{2} \mathrm{KT}$ And temperature $\mathrm{T}=$ Constant But K.E only depends on temperature So, Kinetic energy remains same.
Karnataka CET-2018
Kinetic Theory of Gases
139229
Mean free path of a gas molecule is
1 inversely proportional to number of molecules per unit volume
2 inversely proportional to diameter of the molecule
3 directly proportional to the square root of the absolute temperature
4 directly proportional to the molecular mass
5 independent of temperature
Explanation:
A We know that, mean free path $(\lambda)=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{nd}^{2}}$ Where, $\mathrm{n}=$ number of molecules per unit volume $\mathrm{d}=$ diameter of the molecule. Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.
Kerala CEE - 2009
Kinetic Theory of Gases
139230
The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 \mathrm{~K}$ is
1 $600 \mathrm{~K}$
2 $2400 \mathrm{~K}$
3 $1200 \mathrm{~K}$
4 $300 \mathrm{~K}$
5 $4800 \mathrm{~K}$
Explanation:
E Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Then, $\quad \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R}(300)}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Or, $\quad \mathrm{T}=300 \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}=300 \times \frac{32}{2}=300 \times 16=4800 \mathrm{~K}$
Kerala CEE - 2009
Kinetic Theory of Gases
139231
The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{-14} \mathrm{~J}$ is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
1 $2 \times 10^{9} \mathrm{~K}$
2 $10^{9} \mathrm{~K}$
3 $6 \times 10^{9} \mathrm{~K}$
4 $3 \times 10^{9} \mathrm{~K}$
5 $4.5 \times 10^{9} \mathrm{~K}$
Explanation:
A Given that, Boltzmann constant $(\mathrm{k})=1.38 \times$ $10^{-23} \mathrm{JK}^{-1}$, Energy of proton gas $=4.14 \times 10^{-14} \mathrm{~J}$ We know, Energy of proton gas $(\mathrm{E})=\frac{3}{2} \mathrm{k} \cdot \mathrm{T}$ Putting these value, we get- $4.14 \times 10^{-14}=\frac{3}{2} \times 1.38 \times 10^{-23} \times \mathrm{T}$ Or $\quad \mathrm{T}=\frac{2 \times 4.14 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=\frac{8.28 \times 10^{-14}}{4.14 \times 10^{-23}}=2 \times 10^{9} \mathrm{~K}$
139213
Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :
1 decreases
2 increases
3 remains same
4 increase of decreases depending on the nature of gas
Explanation:
C We know that, K.E $=\frac{3}{2} \mathrm{KT}$ And temperature $\mathrm{T}=$ Constant But K.E only depends on temperature So, Kinetic energy remains same.
Karnataka CET-2018
Kinetic Theory of Gases
139229
Mean free path of a gas molecule is
1 inversely proportional to number of molecules per unit volume
2 inversely proportional to diameter of the molecule
3 directly proportional to the square root of the absolute temperature
4 directly proportional to the molecular mass
5 independent of temperature
Explanation:
A We know that, mean free path $(\lambda)=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{nd}^{2}}$ Where, $\mathrm{n}=$ number of molecules per unit volume $\mathrm{d}=$ diameter of the molecule. Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.
Kerala CEE - 2009
Kinetic Theory of Gases
139230
The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 \mathrm{~K}$ is
1 $600 \mathrm{~K}$
2 $2400 \mathrm{~K}$
3 $1200 \mathrm{~K}$
4 $300 \mathrm{~K}$
5 $4800 \mathrm{~K}$
Explanation:
E Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Then, $\quad \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R}(300)}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Or, $\quad \mathrm{T}=300 \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}=300 \times \frac{32}{2}=300 \times 16=4800 \mathrm{~K}$
Kerala CEE - 2009
Kinetic Theory of Gases
139231
The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{-14} \mathrm{~J}$ is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
1 $2 \times 10^{9} \mathrm{~K}$
2 $10^{9} \mathrm{~K}$
3 $6 \times 10^{9} \mathrm{~K}$
4 $3 \times 10^{9} \mathrm{~K}$
5 $4.5 \times 10^{9} \mathrm{~K}$
Explanation:
A Given that, Boltzmann constant $(\mathrm{k})=1.38 \times$ $10^{-23} \mathrm{JK}^{-1}$, Energy of proton gas $=4.14 \times 10^{-14} \mathrm{~J}$ We know, Energy of proton gas $(\mathrm{E})=\frac{3}{2} \mathrm{k} \cdot \mathrm{T}$ Putting these value, we get- $4.14 \times 10^{-14}=\frac{3}{2} \times 1.38 \times 10^{-23} \times \mathrm{T}$ Or $\quad \mathrm{T}=\frac{2 \times 4.14 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=\frac{8.28 \times 10^{-14}}{4.14 \times 10^{-23}}=2 \times 10^{9} \mathrm{~K}$
139213
Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules :
1 decreases
2 increases
3 remains same
4 increase of decreases depending on the nature of gas
Explanation:
C We know that, K.E $=\frac{3}{2} \mathrm{KT}$ And temperature $\mathrm{T}=$ Constant But K.E only depends on temperature So, Kinetic energy remains same.
Karnataka CET-2018
Kinetic Theory of Gases
139229
Mean free path of a gas molecule is
1 inversely proportional to number of molecules per unit volume
2 inversely proportional to diameter of the molecule
3 directly proportional to the square root of the absolute temperature
4 directly proportional to the molecular mass
5 independent of temperature
Explanation:
A We know that, mean free path $(\lambda)=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{nd}^{2}}$ Where, $\mathrm{n}=$ number of molecules per unit volume $\mathrm{d}=$ diameter of the molecule. Mean free path of a gas molecule is inversely proportional to number of molecules per unit volume.
Kerala CEE - 2009
Kinetic Theory of Gases
139230
The temperature at which oxygen molecules have the same root mean square speed as that of hydrogen molecules at $300 \mathrm{~K}$ is
1 $600 \mathrm{~K}$
2 $2400 \mathrm{~K}$
3 $1200 \mathrm{~K}$
4 $300 \mathrm{~K}$
5 $4800 \mathrm{~K}$
Explanation:
E Given that, $\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{O}_{2}}=\left(\mathrm{v}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}$ We know that, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$ Then, $\quad \sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{O}_{2}}}}=\sqrt{\frac{3 \mathrm{R}(300)}{\mathrm{M}_{\mathrm{H}_{2}}}}$ Or, $\quad \mathrm{T}=300 \times \frac{\mathrm{M}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}=300 \times \frac{32}{2}=300 \times 16=4800 \mathrm{~K}$
Kerala CEE - 2009
Kinetic Theory of Gases
139231
The temperature at which protons in proton gas would have enough energy to overcome Coulomb barrier of $4.14 \times 10^{-14} \mathrm{~J}$ is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
1 $2 \times 10^{9} \mathrm{~K}$
2 $10^{9} \mathrm{~K}$
3 $6 \times 10^{9} \mathrm{~K}$
4 $3 \times 10^{9} \mathrm{~K}$
5 $4.5 \times 10^{9} \mathrm{~K}$
Explanation:
A Given that, Boltzmann constant $(\mathrm{k})=1.38 \times$ $10^{-23} \mathrm{JK}^{-1}$, Energy of proton gas $=4.14 \times 10^{-14} \mathrm{~J}$ We know, Energy of proton gas $(\mathrm{E})=\frac{3}{2} \mathrm{k} \cdot \mathrm{T}$ Putting these value, we get- $4.14 \times 10^{-14}=\frac{3}{2} \times 1.38 \times 10^{-23} \times \mathrm{T}$ Or $\quad \mathrm{T}=\frac{2 \times 4.14 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=\frac{8.28 \times 10^{-14}}{4.14 \times 10^{-23}}=2 \times 10^{9} \mathrm{~K}$
