NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139240
If the rms velocity of a gas is $v$, then
1 $\mathrm{v}^{2} \mathrm{~T}=$ constant
2 $\mathrm{v}^{2} / \mathrm{T}=$ constant
3 $\mathrm{vT}^{2}=$ constant
4 $\mathrm{v}$ is independent of $\mathrm{T}$
Explanation:
B The root mean square velocity of gas is, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $(\mathrm{v})^{2}=\left(\sqrt{\frac{3 R T}{M}}\right)^{2}$ $\mathrm{v}^{2}=\frac{3 R T}{M}$ $\frac{\mathrm{v}^{2}}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}}$ For a gas, $\mathrm{R}$ and $\mathrm{M}$ are constant $\therefore \quad \frac{\mathrm{v}^{2}}{\mathrm{~T}}=\mathrm{C}$
B Monoatomic gas has three translational degree of freedom. For monoatomic gas, $\mathrm{f}=3$ Specific heat at constant pressure $\left(C_{P}\right)$, $C_{\mathrm{P}} =\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}$ $=\left(\frac{3}{2}+1\right) \mathrm{R}$ $=\frac{3+2}{2}$ $\mathrm{C}_{\mathrm{P}} =\frac{5}{2} \mathrm{R}$
JCECE-2006
Kinetic Theory of Gases
139242
The average energy of molecules in a sample of oxygen gas at $300 \mathrm{~K}$ are $6.21 \times 10^{-21} \mathrm{~J}$. The corresponding values at $600 \mathrm{~K}$ are
1 $12.12 \times 10^{-21} \mathrm{~J}$
2 $8.78 \times 10^{-21} \mathrm{~J}$
3 $6.21 \times 10^{-21} \mathrm{~J}$
4 $12.42 \times 10^{-21} \mathrm{~J}$
Explanation:
D Given, Energy at $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{E}_{1}=6.21 \times 10^{-21} \mathrm{~J}$ Energy at $\mathrm{T}_{2}=600 \mathrm{~K}, \mathrm{E}_{2}=$ ? We know that, Energy of molecules $=\frac{3}{2} \mathrm{kT}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{2} \mathrm{kT}_{1}}{\frac{3}{2} \mathrm{kT}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{6.21 \times 10^{-21}}{\mathrm{E}_{2}}=\frac{300}{600}$ Energy at $\mathrm{T}_{2}$, $\mathrm{E}_{2}=12.42 \times 10^{-21} \mathrm{~J}$
COMEDK 2015
Kinetic Theory of Gases
139243
Velocities of three molecules are 2,3 and $4 \mathrm{~m} / \mathrm{s}$. Ratio of mean velocity to R.M.S. velocity is
1 $ \lt 1$
2 $=1$
3 $>1$
4 $>2$
Explanation:
A Given, $\mathrm{v}_{1}=2 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{2}=3 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{3}=4 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}_{\text {arg }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}+\mathrm{v}_{3}}{3}$ $\mathrm{v}_{\text {avg }}=\frac{2+3+4}{3}=\frac{9}{3}$ $\mathrm{v}_{\text {avg }}=3 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}}{3}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{(2)^{2}+(3)^{2}+(4)^{2}}{3}}=\sqrt{\frac{29}{3}}$ Ratio of mean velocity to rms velocity $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}}=\frac{3}{\sqrt{\frac{29}{3}}} \quad\left(\because \sqrt{\frac{29}{3}}=3.11\right)$ $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}} \lt 1$
B The root mean square velocity of gas is, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $(\mathrm{v})^{2}=\left(\sqrt{\frac{3 R T}{M}}\right)^{2}$ $\mathrm{v}^{2}=\frac{3 R T}{M}$ $\frac{\mathrm{v}^{2}}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}}$ For a gas, $\mathrm{R}$ and $\mathrm{M}$ are constant $\therefore \quad \frac{\mathrm{v}^{2}}{\mathrm{~T}}=\mathrm{C}$
B Monoatomic gas has three translational degree of freedom. For monoatomic gas, $\mathrm{f}=3$ Specific heat at constant pressure $\left(C_{P}\right)$, $C_{\mathrm{P}} =\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}$ $=\left(\frac{3}{2}+1\right) \mathrm{R}$ $=\frac{3+2}{2}$ $\mathrm{C}_{\mathrm{P}} =\frac{5}{2} \mathrm{R}$
JCECE-2006
Kinetic Theory of Gases
139242
The average energy of molecules in a sample of oxygen gas at $300 \mathrm{~K}$ are $6.21 \times 10^{-21} \mathrm{~J}$. The corresponding values at $600 \mathrm{~K}$ are
1 $12.12 \times 10^{-21} \mathrm{~J}$
2 $8.78 \times 10^{-21} \mathrm{~J}$
3 $6.21 \times 10^{-21} \mathrm{~J}$
4 $12.42 \times 10^{-21} \mathrm{~J}$
Explanation:
D Given, Energy at $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{E}_{1}=6.21 \times 10^{-21} \mathrm{~J}$ Energy at $\mathrm{T}_{2}=600 \mathrm{~K}, \mathrm{E}_{2}=$ ? We know that, Energy of molecules $=\frac{3}{2} \mathrm{kT}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{2} \mathrm{kT}_{1}}{\frac{3}{2} \mathrm{kT}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{6.21 \times 10^{-21}}{\mathrm{E}_{2}}=\frac{300}{600}$ Energy at $\mathrm{T}_{2}$, $\mathrm{E}_{2}=12.42 \times 10^{-21} \mathrm{~J}$
COMEDK 2015
Kinetic Theory of Gases
139243
Velocities of three molecules are 2,3 and $4 \mathrm{~m} / \mathrm{s}$. Ratio of mean velocity to R.M.S. velocity is
1 $ \lt 1$
2 $=1$
3 $>1$
4 $>2$
Explanation:
A Given, $\mathrm{v}_{1}=2 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{2}=3 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{3}=4 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}_{\text {arg }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}+\mathrm{v}_{3}}{3}$ $\mathrm{v}_{\text {avg }}=\frac{2+3+4}{3}=\frac{9}{3}$ $\mathrm{v}_{\text {avg }}=3 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}}{3}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{(2)^{2}+(3)^{2}+(4)^{2}}{3}}=\sqrt{\frac{29}{3}}$ Ratio of mean velocity to rms velocity $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}}=\frac{3}{\sqrt{\frac{29}{3}}} \quad\left(\because \sqrt{\frac{29}{3}}=3.11\right)$ $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}} \lt 1$
B The root mean square velocity of gas is, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $(\mathrm{v})^{2}=\left(\sqrt{\frac{3 R T}{M}}\right)^{2}$ $\mathrm{v}^{2}=\frac{3 R T}{M}$ $\frac{\mathrm{v}^{2}}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}}$ For a gas, $\mathrm{R}$ and $\mathrm{M}$ are constant $\therefore \quad \frac{\mathrm{v}^{2}}{\mathrm{~T}}=\mathrm{C}$
B Monoatomic gas has three translational degree of freedom. For monoatomic gas, $\mathrm{f}=3$ Specific heat at constant pressure $\left(C_{P}\right)$, $C_{\mathrm{P}} =\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}$ $=\left(\frac{3}{2}+1\right) \mathrm{R}$ $=\frac{3+2}{2}$ $\mathrm{C}_{\mathrm{P}} =\frac{5}{2} \mathrm{R}$
JCECE-2006
Kinetic Theory of Gases
139242
The average energy of molecules in a sample of oxygen gas at $300 \mathrm{~K}$ are $6.21 \times 10^{-21} \mathrm{~J}$. The corresponding values at $600 \mathrm{~K}$ are
1 $12.12 \times 10^{-21} \mathrm{~J}$
2 $8.78 \times 10^{-21} \mathrm{~J}$
3 $6.21 \times 10^{-21} \mathrm{~J}$
4 $12.42 \times 10^{-21} \mathrm{~J}$
Explanation:
D Given, Energy at $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{E}_{1}=6.21 \times 10^{-21} \mathrm{~J}$ Energy at $\mathrm{T}_{2}=600 \mathrm{~K}, \mathrm{E}_{2}=$ ? We know that, Energy of molecules $=\frac{3}{2} \mathrm{kT}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{2} \mathrm{kT}_{1}}{\frac{3}{2} \mathrm{kT}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{6.21 \times 10^{-21}}{\mathrm{E}_{2}}=\frac{300}{600}$ Energy at $\mathrm{T}_{2}$, $\mathrm{E}_{2}=12.42 \times 10^{-21} \mathrm{~J}$
COMEDK 2015
Kinetic Theory of Gases
139243
Velocities of three molecules are 2,3 and $4 \mathrm{~m} / \mathrm{s}$. Ratio of mean velocity to R.M.S. velocity is
1 $ \lt 1$
2 $=1$
3 $>1$
4 $>2$
Explanation:
A Given, $\mathrm{v}_{1}=2 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{2}=3 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{3}=4 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}_{\text {arg }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}+\mathrm{v}_{3}}{3}$ $\mathrm{v}_{\text {avg }}=\frac{2+3+4}{3}=\frac{9}{3}$ $\mathrm{v}_{\text {avg }}=3 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}}{3}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{(2)^{2}+(3)^{2}+(4)^{2}}{3}}=\sqrt{\frac{29}{3}}$ Ratio of mean velocity to rms velocity $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}}=\frac{3}{\sqrt{\frac{29}{3}}} \quad\left(\because \sqrt{\frac{29}{3}}=3.11\right)$ $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}} \lt 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139240
If the rms velocity of a gas is $v$, then
1 $\mathrm{v}^{2} \mathrm{~T}=$ constant
2 $\mathrm{v}^{2} / \mathrm{T}=$ constant
3 $\mathrm{vT}^{2}=$ constant
4 $\mathrm{v}$ is independent of $\mathrm{T}$
Explanation:
B The root mean square velocity of gas is, $\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ $(\mathrm{v})^{2}=\left(\sqrt{\frac{3 R T}{M}}\right)^{2}$ $\mathrm{v}^{2}=\frac{3 R T}{M}$ $\frac{\mathrm{v}^{2}}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}}$ For a gas, $\mathrm{R}$ and $\mathrm{M}$ are constant $\therefore \quad \frac{\mathrm{v}^{2}}{\mathrm{~T}}=\mathrm{C}$
B Monoatomic gas has three translational degree of freedom. For monoatomic gas, $\mathrm{f}=3$ Specific heat at constant pressure $\left(C_{P}\right)$, $C_{\mathrm{P}} =\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{R}$ $=\left(\frac{3}{2}+1\right) \mathrm{R}$ $=\frac{3+2}{2}$ $\mathrm{C}_{\mathrm{P}} =\frac{5}{2} \mathrm{R}$
JCECE-2006
Kinetic Theory of Gases
139242
The average energy of molecules in a sample of oxygen gas at $300 \mathrm{~K}$ are $6.21 \times 10^{-21} \mathrm{~J}$. The corresponding values at $600 \mathrm{~K}$ are
1 $12.12 \times 10^{-21} \mathrm{~J}$
2 $8.78 \times 10^{-21} \mathrm{~J}$
3 $6.21 \times 10^{-21} \mathrm{~J}$
4 $12.42 \times 10^{-21} \mathrm{~J}$
Explanation:
D Given, Energy at $\mathrm{T}_{1}=300 \mathrm{~K}, \mathrm{E}_{1}=6.21 \times 10^{-21} \mathrm{~J}$ Energy at $\mathrm{T}_{2}=600 \mathrm{~K}, \mathrm{E}_{2}=$ ? We know that, Energy of molecules $=\frac{3}{2} \mathrm{kT}$ $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\frac{3}{2} \mathrm{kT}_{1}}{\frac{3}{2} \mathrm{kT}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\frac{6.21 \times 10^{-21}}{\mathrm{E}_{2}}=\frac{300}{600}$ Energy at $\mathrm{T}_{2}$, $\mathrm{E}_{2}=12.42 \times 10^{-21} \mathrm{~J}$
COMEDK 2015
Kinetic Theory of Gases
139243
Velocities of three molecules are 2,3 and $4 \mathrm{~m} / \mathrm{s}$. Ratio of mean velocity to R.M.S. velocity is
1 $ \lt 1$
2 $=1$
3 $>1$
4 $>2$
Explanation:
A Given, $\mathrm{v}_{1}=2 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{2}=3 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{3}=4 \mathrm{~m} / \mathrm{s}$ We know that, $\mathrm{v}_{\text {arg }}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}+\mathrm{v}_{3}}{3}$ $\mathrm{v}_{\text {avg }}=\frac{2+3+4}{3}=\frac{9}{3}$ $\mathrm{v}_{\text {avg }}=3 \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+\mathrm{v}_{3}^{2}}{3}}$ $\mathrm{v}_{\text {rms }}=\sqrt{\frac{(2)^{2}+(3)^{2}+(4)^{2}}{3}}=\sqrt{\frac{29}{3}}$ Ratio of mean velocity to rms velocity $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}}=\frac{3}{\sqrt{\frac{29}{3}}} \quad\left(\because \sqrt{\frac{29}{3}}=3.11\right)$ $\frac{\mathrm{v}_{\text {avg }}}{\mathrm{v}_{\mathrm{rms}}} \lt 1$