CBSE 11th Standard Physics Subject Kinetic Theory Ncert Exemplar 1 Mark Questions 2021
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CBSE 11th Standard Physics Subject Kinetic Theory Ncert Exemplar 1 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Physics
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Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mol-1 .
(a)Molar mass of gold is 197 g mol-1 , the number of atoms
= 6 x 1023
\(\therefore\) Number of atoms in 39.4 g
= \(\frac { 6.0\times { 10 }^{ 23 }\times 39.4 }{ 197 }\)
= 1.2 x 1023 -
When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased." What about Boyle's law in this case?
(a)When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
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The volume of a given mass of a gas at 270C, 1 atm is 100 cc. What will be its volume at 3270C?
(a)Keeping p constant, we have
\({ V }_{ 2 }\) = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } }\)
= \(\frac { 100\times 600 }{ 300 } \)
= 200cc -
Two molecules of a gas have speeds of 9 \(\times\)106 ms-1 and 1 \(\times\) 106 ms-1, respectively. What is the root mean square speed of these molecules ?
(a)\({ v }_{ rms }\) = \(\sqrt { \frac { { v }_{ 1\quad }^{ 2 }+{ { v }_{ 2 }^{ 2 } } }{ 2 } } \)
= \(\sqrt { \frac { (9\times { 10 }^{ 6 })^{ 2 }+(1\times { 10 }^{ 6 })^{ 2 } }{ 2 } }\)= \(\sqrt { \frac { (81+1)\times { 10 }^{ 12 } }{ 2 } }\)
= \(\sqrt { 41 } \times { 10 }^{ 6 }ms^{ -1 }\) -
Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters \(1\overset { 0 }{ A } \) and \(2\overset { 0 }{ A } \). The gases may be considered under identical conditions of temperature, pressure and volume.
(a)As, we know, mean free path,
\(\lambda \propto \frac { 1 }{ { d }^{ 2 } } \)
\(Given,\ { d }_{ 1 }=1\overset { 0 }{ A } \ and\ { d }_{ 2 }=2\overset { 0 }{ A } \Rightarrow { \lambda }_{ 1 }:{ \lambda }_{ 2 }=4:1\)
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CBSE 11th Standard Physics Subject Kinetic Theory Ncert Exemplar 1 Mark Questions 2021 Answer Keys
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Molar mass of gold is 197 g mol-1 , the number of atoms
= 6 x 1023
∴ Number of atoms in 39.4 g
= \(\frac{6.0×10^{23} ×39.4}{197}\)
= 1.2 x 1023Molar mass of gold is 197 g mol-1 , the number of atoms
= 6 x 1023
∴ Number of atoms in 39.4 g
= \(\frac{6.0×10^{23} ×39.4}{197}\)
= 1.2 x 1023 -
When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.
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Keeping p constant, we have
V2/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200ccKeeping p constant, we have
V2/2 = \(\frac{V_1T_2}{T_1}\)
= \(\frac{100×600}{300}\)
= 200cc -
\(v_{r m s}=\sqrt{\frac{v_1^2+v_2^2}{2}} \)
\(=\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^{\circ}\right)^2}{2}}=\sqrt{\frac{(81+1) \times 10^{12}}{2}}\)
\(=\sqrt{41} \times 10^6 \mathrm{~ms}^{-1}\)
\(v_{r m s}=\sqrt{\frac{v_1^2+v_2^2}{2}} \)
\(=\sqrt{\frac{\left(9 \times 10^6\right)^2+\left(1 \times 10^{\circ}\right)^2}{2}}=\sqrt{\frac{(81+1) \times 10^{12}}{2}}\)
\(=\sqrt{41} \times 10^6 \mathrm{~ms}^{-1}\)
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As, we know, mean free path,
λ ∝ 1/d2
Given, d1= 1Ao and d2 = 2Ao⇒ λ1 : λ2=4:1As, we know, mean free path,
λ ∝ 1/d2
Given, d1= 1Ao and d2 = 2Ao⇒ λ1 : λ2=4:1
