NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Kinetic Theory of Gases
139108
The temperature of an ideal gas is increased from 200 K to 800 K . If r.m.s. speed of gas at 200 K is $v_0$ Then, r.m.s. speed of the gas at 800 K will be:
1 $\mathrm{v}_0$
2 $4 \mathrm{v}_0$
3 $\frac{v_0}{4}$
4 $2 \mathrm{v}_0
Explanation:
D Given, temperature of ideal gas- $\begin{aligned} & \text { } \\ & \mathrm{T}_1=200 \mathrm{~K} \\ & \mathrm{~T}_2=800 \mathrm{~K} \\ & \text { RMS speed at } 200 \mathrm{~K} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_1=\mathrm{v}_{\mathrm{o}} \\ & \text { We know that, } \\ & \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ & \mathrm{v}_{\text {rms }} \propto \sqrt{\mathrm{T}} \\ & \text { Ratio of RMS speed at } 800 \mathrm{~K} \text { and } 200 \mathrm{~K} . \\ & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \\ & \hline \end{aligned}$ We know that, Ratio of RMS speed at 800 K and 200 K . $\begin{aligned} & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \end{aligned}$
Shift-II
Kinetic Theory of Gases
139110
The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given, $k=1.38$ $\times 10^{-23} \mathrm{JK}^{-1}$ l
1 $60 \mathrm{~mm} \mathrm{~s}^{-1}$
2 $12 \mathrm{~mm} \mathrm{~s}^{-1}$
3 $15 \mathrm{~mm} \mathrm{~s}^{-1}$
4 $36 \mathrm{~mm} \mathrm{~s}^{-1}$
Explanation:
C Given, $\mathrm{m}=5 \times 10^{-17} \mathrm{~kg}, \mathrm{~T}=273 \mathrm{~K}, \mathrm{k}=$ $1.38 \times 10^{-23} \mathrm{JK}^{-1}$ According to kinetic theory of gases an average K.E. of a gas molecule - $\text { K.E. }=\frac{3}{2} \mathrm{kT}$ $\frac{1}{2} \mathrm{mV}_{\mathrm{rms}}^{2}=\frac{3}{2} \mathrm{kT}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$ $\mathrm{~V}_{\mathrm{rms}}=1.5 \times 10^{-2} \mathrm{~m} / \mathrm{s}=15 \mathrm{~mm} / \mathrm{s}$
Shift-II
Kinetic Theory of Gases
139111
A vessel $A$ contains hydrogen and another vessel $B$ whose volume is twice of $A$ contains same mass of oxygen at the same temperature. What will be the ratio of r.m.s speed of the molecules?
1 $1: 1$
2 $2: 1$
3 $4: 1$
4 $8: 1$
Explanation:
C For hydrogen, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{m}_{1}=\mathrm{m}, \mathrm{M}_{1}=2$ For oxygen, $\mathrm{V}_{2}=2 \mathrm{~V}$ $\mathrm{~m}_{2}=\mathrm{m}$ $\mathrm{M}_{1}=32$ We know that, $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} \quad \mathrm{V}_{\mathrm{rms}} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{32}{2}}=4: 1$
Assam CEE-31.07.2022
Kinetic Theory of Gases
139112
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure $A B$, $C D, E F$ are all isotherms. If $v_{P}$ is the most probable speed of the molecules, then
1 $v_{\mathrm{P}}$ at $3-\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
2 $v_{P}$ at $3-v_{P}$ at $1>v_{P}$ at $2>v_{P}$ at 4
3 $v_{P}$ at $3>v_{P}$ at $2>v_{P}$ at $4>v_{P}$ at 1
4 $\mathrm{v}_{\mathrm{P}}$ at $2=\mathrm{v}_{\mathrm{P}}$ at $3>\mathrm{v}_{\mathrm{P}}$ at $1>\mathrm{v}_{\mathrm{P}}$ at 4
Explanation:
A Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ For isothermal process $\mathrm{PV}=\text { constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Line $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are all isotherm. $\therefore \quad \mathrm{T}_{\mathrm{AB}}=\mathrm{T}_{1}$ $\mathrm{~T}_{\mathrm{CD}}=\mathrm{T}_{2}$ $\mathrm{~T}_{\mathrm{EF}}=\mathrm{T}_{3}=\mathrm{T}_{4}$ Therefore, $\mathrm{T}_{\mathrm{EF}}>\mathrm{T}_{\mathrm{CD}}>\mathrm{T}_{\mathrm{AB}}$ $\therefore \mathrm{T}_{3}=\mathrm{T}_{4}>\mathrm{T}_{2}>\mathrm{T}_{1}$ Now, Root square mean velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $\therefore \mathrm{v}_{\mathrm{P}}=\mathrm{v}_{\mathrm{rms}}$ $\therefore \mathrm{v}_{\mathrm{P}} \propto \sqrt{\mathrm{T}}$ $\therefore \mathrm{v}_{\mathrm{P}}$ at $3=\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
139108
The temperature of an ideal gas is increased from 200 K to 800 K . If r.m.s. speed of gas at 200 K is $v_0$ Then, r.m.s. speed of the gas at 800 K will be:
1 $\mathrm{v}_0$
2 $4 \mathrm{v}_0$
3 $\frac{v_0}{4}$
4 $2 \mathrm{v}_0
Explanation:
D Given, temperature of ideal gas- $\begin{aligned} & \text { } \\ & \mathrm{T}_1=200 \mathrm{~K} \\ & \mathrm{~T}_2=800 \mathrm{~K} \\ & \text { RMS speed at } 200 \mathrm{~K} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_1=\mathrm{v}_{\mathrm{o}} \\ & \text { We know that, } \\ & \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ & \mathrm{v}_{\text {rms }} \propto \sqrt{\mathrm{T}} \\ & \text { Ratio of RMS speed at } 800 \mathrm{~K} \text { and } 200 \mathrm{~K} . \\ & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \\ & \hline \end{aligned}$ We know that, Ratio of RMS speed at 800 K and 200 K . $\begin{aligned} & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \end{aligned}$
Shift-II
Kinetic Theory of Gases
139110
The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given, $k=1.38$ $\times 10^{-23} \mathrm{JK}^{-1}$ l
1 $60 \mathrm{~mm} \mathrm{~s}^{-1}$
2 $12 \mathrm{~mm} \mathrm{~s}^{-1}$
3 $15 \mathrm{~mm} \mathrm{~s}^{-1}$
4 $36 \mathrm{~mm} \mathrm{~s}^{-1}$
Explanation:
C Given, $\mathrm{m}=5 \times 10^{-17} \mathrm{~kg}, \mathrm{~T}=273 \mathrm{~K}, \mathrm{k}=$ $1.38 \times 10^{-23} \mathrm{JK}^{-1}$ According to kinetic theory of gases an average K.E. of a gas molecule - $\text { K.E. }=\frac{3}{2} \mathrm{kT}$ $\frac{1}{2} \mathrm{mV}_{\mathrm{rms}}^{2}=\frac{3}{2} \mathrm{kT}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$ $\mathrm{~V}_{\mathrm{rms}}=1.5 \times 10^{-2} \mathrm{~m} / \mathrm{s}=15 \mathrm{~mm} / \mathrm{s}$
Shift-II
Kinetic Theory of Gases
139111
A vessel $A$ contains hydrogen and another vessel $B$ whose volume is twice of $A$ contains same mass of oxygen at the same temperature. What will be the ratio of r.m.s speed of the molecules?
1 $1: 1$
2 $2: 1$
3 $4: 1$
4 $8: 1$
Explanation:
C For hydrogen, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{m}_{1}=\mathrm{m}, \mathrm{M}_{1}=2$ For oxygen, $\mathrm{V}_{2}=2 \mathrm{~V}$ $\mathrm{~m}_{2}=\mathrm{m}$ $\mathrm{M}_{1}=32$ We know that, $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} \quad \mathrm{V}_{\mathrm{rms}} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{32}{2}}=4: 1$
Assam CEE-31.07.2022
Kinetic Theory of Gases
139112
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure $A B$, $C D, E F$ are all isotherms. If $v_{P}$ is the most probable speed of the molecules, then
1 $v_{\mathrm{P}}$ at $3-\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
2 $v_{P}$ at $3-v_{P}$ at $1>v_{P}$ at $2>v_{P}$ at 4
3 $v_{P}$ at $3>v_{P}$ at $2>v_{P}$ at $4>v_{P}$ at 1
4 $\mathrm{v}_{\mathrm{P}}$ at $2=\mathrm{v}_{\mathrm{P}}$ at $3>\mathrm{v}_{\mathrm{P}}$ at $1>\mathrm{v}_{\mathrm{P}}$ at 4
Explanation:
A Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ For isothermal process $\mathrm{PV}=\text { constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Line $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are all isotherm. $\therefore \quad \mathrm{T}_{\mathrm{AB}}=\mathrm{T}_{1}$ $\mathrm{~T}_{\mathrm{CD}}=\mathrm{T}_{2}$ $\mathrm{~T}_{\mathrm{EF}}=\mathrm{T}_{3}=\mathrm{T}_{4}$ Therefore, $\mathrm{T}_{\mathrm{EF}}>\mathrm{T}_{\mathrm{CD}}>\mathrm{T}_{\mathrm{AB}}$ $\therefore \mathrm{T}_{3}=\mathrm{T}_{4}>\mathrm{T}_{2}>\mathrm{T}_{1}$ Now, Root square mean velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $\therefore \mathrm{v}_{\mathrm{P}}=\mathrm{v}_{\mathrm{rms}}$ $\therefore \mathrm{v}_{\mathrm{P}} \propto \sqrt{\mathrm{T}}$ $\therefore \mathrm{v}_{\mathrm{P}}$ at $3=\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
139108
The temperature of an ideal gas is increased from 200 K to 800 K . If r.m.s. speed of gas at 200 K is $v_0$ Then, r.m.s. speed of the gas at 800 K will be:
1 $\mathrm{v}_0$
2 $4 \mathrm{v}_0$
3 $\frac{v_0}{4}$
4 $2 \mathrm{v}_0
Explanation:
D Given, temperature of ideal gas- $\begin{aligned} & \text { } \\ & \mathrm{T}_1=200 \mathrm{~K} \\ & \mathrm{~T}_2=800 \mathrm{~K} \\ & \text { RMS speed at } 200 \mathrm{~K} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_1=\mathrm{v}_{\mathrm{o}} \\ & \text { We know that, } \\ & \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ & \mathrm{v}_{\text {rms }} \propto \sqrt{\mathrm{T}} \\ & \text { Ratio of RMS speed at } 800 \mathrm{~K} \text { and } 200 \mathrm{~K} . \\ & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \\ & \hline \end{aligned}$ We know that, Ratio of RMS speed at 800 K and 200 K . $\begin{aligned} & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \end{aligned}$
Shift-II
Kinetic Theory of Gases
139110
The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given, $k=1.38$ $\times 10^{-23} \mathrm{JK}^{-1}$ l
1 $60 \mathrm{~mm} \mathrm{~s}^{-1}$
2 $12 \mathrm{~mm} \mathrm{~s}^{-1}$
3 $15 \mathrm{~mm} \mathrm{~s}^{-1}$
4 $36 \mathrm{~mm} \mathrm{~s}^{-1}$
Explanation:
C Given, $\mathrm{m}=5 \times 10^{-17} \mathrm{~kg}, \mathrm{~T}=273 \mathrm{~K}, \mathrm{k}=$ $1.38 \times 10^{-23} \mathrm{JK}^{-1}$ According to kinetic theory of gases an average K.E. of a gas molecule - $\text { K.E. }=\frac{3}{2} \mathrm{kT}$ $\frac{1}{2} \mathrm{mV}_{\mathrm{rms}}^{2}=\frac{3}{2} \mathrm{kT}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$ $\mathrm{~V}_{\mathrm{rms}}=1.5 \times 10^{-2} \mathrm{~m} / \mathrm{s}=15 \mathrm{~mm} / \mathrm{s}$
Shift-II
Kinetic Theory of Gases
139111
A vessel $A$ contains hydrogen and another vessel $B$ whose volume is twice of $A$ contains same mass of oxygen at the same temperature. What will be the ratio of r.m.s speed of the molecules?
1 $1: 1$
2 $2: 1$
3 $4: 1$
4 $8: 1$
Explanation:
C For hydrogen, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{m}_{1}=\mathrm{m}, \mathrm{M}_{1}=2$ For oxygen, $\mathrm{V}_{2}=2 \mathrm{~V}$ $\mathrm{~m}_{2}=\mathrm{m}$ $\mathrm{M}_{1}=32$ We know that, $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} \quad \mathrm{V}_{\mathrm{rms}} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{32}{2}}=4: 1$
Assam CEE-31.07.2022
Kinetic Theory of Gases
139112
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure $A B$, $C D, E F$ are all isotherms. If $v_{P}$ is the most probable speed of the molecules, then
1 $v_{\mathrm{P}}$ at $3-\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
2 $v_{P}$ at $3-v_{P}$ at $1>v_{P}$ at $2>v_{P}$ at 4
3 $v_{P}$ at $3>v_{P}$ at $2>v_{P}$ at $4>v_{P}$ at 1
4 $\mathrm{v}_{\mathrm{P}}$ at $2=\mathrm{v}_{\mathrm{P}}$ at $3>\mathrm{v}_{\mathrm{P}}$ at $1>\mathrm{v}_{\mathrm{P}}$ at 4
Explanation:
A Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ For isothermal process $\mathrm{PV}=\text { constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Line $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are all isotherm. $\therefore \quad \mathrm{T}_{\mathrm{AB}}=\mathrm{T}_{1}$ $\mathrm{~T}_{\mathrm{CD}}=\mathrm{T}_{2}$ $\mathrm{~T}_{\mathrm{EF}}=\mathrm{T}_{3}=\mathrm{T}_{4}$ Therefore, $\mathrm{T}_{\mathrm{EF}}>\mathrm{T}_{\mathrm{CD}}>\mathrm{T}_{\mathrm{AB}}$ $\therefore \mathrm{T}_{3}=\mathrm{T}_{4}>\mathrm{T}_{2}>\mathrm{T}_{1}$ Now, Root square mean velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $\therefore \mathrm{v}_{\mathrm{P}}=\mathrm{v}_{\mathrm{rms}}$ $\therefore \mathrm{v}_{\mathrm{P}} \propto \sqrt{\mathrm{T}}$ $\therefore \mathrm{v}_{\mathrm{P}}$ at $3=\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
139108
The temperature of an ideal gas is increased from 200 K to 800 K . If r.m.s. speed of gas at 200 K is $v_0$ Then, r.m.s. speed of the gas at 800 K will be:
1 $\mathrm{v}_0$
2 $4 \mathrm{v}_0$
3 $\frac{v_0}{4}$
4 $2 \mathrm{v}_0
Explanation:
D Given, temperature of ideal gas- $\begin{aligned} & \text { } \\ & \mathrm{T}_1=200 \mathrm{~K} \\ & \mathrm{~T}_2=800 \mathrm{~K} \\ & \text { RMS speed at } 200 \mathrm{~K} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_1=\mathrm{v}_{\mathrm{o}} \\ & \text { We know that, } \\ & \mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \\ & \mathrm{v}_{\text {rms }} \propto \sqrt{\mathrm{T}} \\ & \text { Ratio of RMS speed at } 800 \mathrm{~K} \text { and } 200 \mathrm{~K} . \\ & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \\ & \hline \end{aligned}$ We know that, Ratio of RMS speed at 800 K and 200 K . $\begin{aligned} & \frac{\left(\mathrm{v}_{\mathrm{rms}}\right)_2}{\left(\mathrm{v}_{\mathrm{rms}}\right)_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \times\left(\mathrm{v}_{\mathrm{rms}}\right)_1 \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=\sqrt{\frac{800}{200}} \times \mathrm{v}_{\mathrm{o}} \\ & \left(\mathrm{v}_{\mathrm{rms}}\right)_2=2 \mathrm{v}_{\mathrm{o}} \end{aligned}$
Shift-II
Kinetic Theory of Gases
139110
The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given, $k=1.38$ $\times 10^{-23} \mathrm{JK}^{-1}$ l
1 $60 \mathrm{~mm} \mathrm{~s}^{-1}$
2 $12 \mathrm{~mm} \mathrm{~s}^{-1}$
3 $15 \mathrm{~mm} \mathrm{~s}^{-1}$
4 $36 \mathrm{~mm} \mathrm{~s}^{-1}$
Explanation:
C Given, $\mathrm{m}=5 \times 10^{-17} \mathrm{~kg}, \mathrm{~T}=273 \mathrm{~K}, \mathrm{k}=$ $1.38 \times 10^{-23} \mathrm{JK}^{-1}$ According to kinetic theory of gases an average K.E. of a gas molecule - $\text { K.E. }=\frac{3}{2} \mathrm{kT}$ $\frac{1}{2} \mathrm{mV}_{\mathrm{rms}}^{2}=\frac{3}{2} \mathrm{kT}$ $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$ $\mathrm{~V}_{\mathrm{rms}}=1.5 \times 10^{-2} \mathrm{~m} / \mathrm{s}=15 \mathrm{~mm} / \mathrm{s}$
Shift-II
Kinetic Theory of Gases
139111
A vessel $A$ contains hydrogen and another vessel $B$ whose volume is twice of $A$ contains same mass of oxygen at the same temperature. What will be the ratio of r.m.s speed of the molecules?
1 $1: 1$
2 $2: 1$
3 $4: 1$
4 $8: 1$
Explanation:
C For hydrogen, $\mathrm{V}_{1}=\mathrm{V}, \mathrm{m}_{1}=\mathrm{m}, \mathrm{M}_{1}=2$ For oxygen, $\mathrm{V}_{2}=2 \mathrm{~V}$ $\mathrm{~m}_{2}=\mathrm{m}$ $\mathrm{M}_{1}=32$ We know that, $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} \quad \mathrm{V}_{\mathrm{rms}} \propto \sqrt{\frac{1}{\mathrm{M}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}}$ $\frac{\mathrm{V}_{\mathrm{H}_{2}}}{\mathrm{~V}_{\mathrm{O}_{2}}}=\sqrt{\frac{32}{2}}=4: 1$
Assam CEE-31.07.2022
Kinetic Theory of Gases
139112
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure $A B$, $C D, E F$ are all isotherms. If $v_{P}$ is the most probable speed of the molecules, then
1 $v_{\mathrm{P}}$ at $3-\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
2 $v_{P}$ at $3-v_{P}$ at $1>v_{P}$ at $2>v_{P}$ at 4
3 $v_{P}$ at $3>v_{P}$ at $2>v_{P}$ at $4>v_{P}$ at 1
4 $\mathrm{v}_{\mathrm{P}}$ at $2=\mathrm{v}_{\mathrm{P}}$ at $3>\mathrm{v}_{\mathrm{P}}$ at $1>\mathrm{v}_{\mathrm{P}}$ at 4
Explanation:
A Ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ For isothermal process $\mathrm{PV}=\text { constant }$ $\mathrm{P} \propto \frac{1}{\mathrm{~V}}$ Line $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are all isotherm. $\therefore \quad \mathrm{T}_{\mathrm{AB}}=\mathrm{T}_{1}$ $\mathrm{~T}_{\mathrm{CD}}=\mathrm{T}_{2}$ $\mathrm{~T}_{\mathrm{EF}}=\mathrm{T}_{3}=\mathrm{T}_{4}$ Therefore, $\mathrm{T}_{\mathrm{EF}}>\mathrm{T}_{\mathrm{CD}}>\mathrm{T}_{\mathrm{AB}}$ $\therefore \mathrm{T}_{3}=\mathrm{T}_{4}>\mathrm{T}_{2}>\mathrm{T}_{1}$ Now, Root square mean velocity $\left(\mathrm{v}_{\mathrm{rms}}\right)=\sqrt{\frac{3 R T}{\mathrm{M}}}$ $\therefore \mathrm{v}_{\mathrm{P}}=\mathrm{v}_{\mathrm{rms}}$ $\therefore \mathrm{v}_{\mathrm{P}} \propto \sqrt{\mathrm{T}}$ $\therefore \mathrm{v}_{\mathrm{P}}$ at $3=\mathrm{v}_{\mathrm{P}}$ at $4>\mathrm{v}_{\mathrm{P}}$ at $2>\mathrm{v}_{\mathrm{P}}$ at 1
