Kinetic Theory

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New answer posted•a year ago

Kinetic Theory Class 11th

The average translational kinetic energy of N₂ gas molecules at_________°C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k_B = 1.38 × 10^-23 J/K) (Fill the nearest integer).

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

K.E. energy of electron = eV

Translational K.E. of N₂ = $\frac{3}{2} K T$

So eV = $\frac{3}{2} K T$

$1.6 \times 1 0^{- 19} \times 0.1 = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times T$

T = 773 – 273 = 500°C

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New answer posted•a year ago

Kinetic Theory Class 11th

Consider a mixture of gas molecule of types A, B and C having masses m_A < m_B < m_C. The ratio of their root mean square speeds at normal temperature and pressure is:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

As we know that root mean square speed is given as

$v = \sqrt[]{\frac{3 R T}{M}}, s o$

$v_{A} > v_{B} > v_{C} \Rightarrow \frac{1}{v_{A}} < \frac{1}{v_{B}} < \frac{1}{v_{C}} a s m_{A} < m_{B} < m_{C}$

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New answer posted•a year ago

Kinetic Theory Class 11th

Which of the following graphs represent the behaviour of an ideal gas? Symbol have their usual meaning.

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

According to ideal gas law we know that

PV = nRT

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New answer posted•a year ago

Kinetic Theory Class 11th

The correct relation between the degrees of freedom f and the ratio of specific heat  $γ i s :$

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$γ = 1 + \frac{2}{f} \Rightarrow \frac{2}{f} = γ - 1 \Rightarrow f = \frac{2}{γ - 1}$

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New answer posted•a year ago

Kinetic Theory Class 11th

The average translational kinetic energy of N₂ gas molecules at_________°C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given k_B = 1.38 × 10^-23 J/K) (Fill the nearest integer).

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

K.E. energy of electron = eV

Translational K.E. of N₂ = $\frac{3}{2} K T$

So eV = $\frac{3}{2} K T$

$1.6 \times 1 0^{- 19} \times 0.1 = \frac{3}{2} \times 1.38 \times 1 0^{- 23} \times T$

T = 773 – 273 = 500°C

Upvote

New answer posted•a year ago

Kinetic Theory Class 11th

If the rms speed of oxygen molecules at 0°C is 160 m/s, find the rms speed of hydrogen molecules at 0°C.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{{(v_{r m s})}_{O_{2}}}{(v_{r m s}) H_{2}} = \sqrt[]{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt[]{\frac{2}{32}}$

$\Rightarrow {(v_{r m s})}_{H_{2}} = 4 \times {(v_{r m s})}_{O_{2}} = 4 \times 160 = 640 m / s$

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New answer posted•a year ago

Kinetic Theory Class 11th

The rms speed of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are $V_{H}, V_{O} a n d V_{C}$  respectively then:

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}}$

$V_{r m s} \propto \frac{1}{\sqrt[]{M}}$ at constant T

$V_{H} : V_{O} : V_{C o_{2}} = \frac{1}{\sqrt[]{1}} : \frac{1}{\sqrt[]{16}} : \frac{1}{\sqrt[]{44}}$

$= 1 : \frac{1}{4} : \frac{1}{2 \sqrt[]{11}}$

$V_{H} > V_{O} > V_{C O_{2}}$

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New answer posted•a year ago

Kinetic Theory Class 11th

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is 1 : 4, then

(A) The r.m.s. velocity of gas molecules in two vessels will be the same.

(B) The ratio of pressure in these vessels will be 1 : 4.

(C) The ratio of pressure will be 1 : 1.

(D) The r.m.s. velocity of gas molecules in two vessels will be 1 : 4.

Choose the correct answer form the options given below:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} A s . v_{1} = v_{2}$

$V_{r m s} α \sqrt[]{T}$ $η_{2} = 4 n_{1}$

$T_{1} = T_{2} \Rightarrow V_{r m s - 1} = V_{r m s - 2}$

$P_{1} = \frac{n_{1} R T}{v} P_{z} = \frac{n_{2} R T}{v}$

$\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}} = \frac{1}{4}$

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New answer posted•a year ago

Kinetic Theory Class 11th

The relation between root mean square speed $(υ_{r m s})$  and most probable speed $(υ_{p})$  for the milar mass M of oxygen gas molecule at the temperature of 300K will be:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}}$

& $v_{P} = \sqrt[]{\frac{2 R T}{M}}$

$v_{r m s} = \sqrt[]{\frac{3}{2}} v_{P}$

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New answer posted•a year ago

Kinetic Theory Class 11th

The root mean square speed of smoke particles of mass 5 × 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 × 10^-23 JK^-1]

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

m = 5 × 10^-17kg ; k = 1.38 × 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 \times 1.38 \times 1 0^{- 23} \times 300}{5 \times 1 0^{- 17}}}$

$= \sqrt[]{18 \times 13.8 \times 1 0^{- 6}}$

$= 15.7 \times 1 0^{- 3} m / s$

$= 15 m m / s$

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