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2 months 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:

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alok kumar singh

Contributor-Level 10

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

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

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

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

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

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2 months ago

Kinetic Theory Class 11th

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt[]{2}$ time the speed of sound, then the value of n will be

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Payal Gupta

Contributor-Level 10

As we know that

$v = \sqrt[]{\frac{γ R T}{M}} \Rightarrow$ Speed of sound in a gas, and

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow$ Rms speed of gas molecules, so

$v_{r m s} = \sqrt[]{2} v \Rightarrow \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{2} \sqrt[]{\frac{γ R T}{M}} \Rightarrow γ = \frac{3}{2}$

According formula of specific heat ratio of mixture, we can write

$\Rightarrow$ 5n = 14 n = 2.8

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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]

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alok kumar singh

Contributor-Level 10

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

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

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

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

$= 15 m m / s$

&nb

...more

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

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

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

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

$= 15 m m / s$

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2 months ago

Kinetic Theory Class 11th

Given below are two statements:

Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature.

Statement II : The rms speed of oxygen molecules in a gas is $υ$ . If the temperature is doubles and the oxygen molecules dissociate into oxygen atoms, the rms speeds will become 2 $υ$ . In the light of the above statements, choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

$V = \sqrt[]{\frac{3 R T}{M}}$ $V' = \sqrt[]{\frac{3 R 2 T}{(\frac{M}{2})}} = 2 \frac{\sqrt[]{3 R T}}{M} = 2 v$

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2 months ago

Kinetic Theory Class 11th

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture of constant volume is $\frac{α^{2}}{4} R J / m o l K;$ then the value of a will be________.

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Vishal Baghel

Contributor-Level 10

$n_{1} = 1 \to m o n o t o m i c$

$n_{2} = 3 \to d i a t o m i c$ $C_{v} = \frac{α^{2} R}{4} J / m o l^{- k}$

As, Cv_(mix) = $\frac{1 * \frac{3}{2} R + 3 * \frac{5}{2} R}{4} = \frac{9 R}{4} = \frac{α^{2} R}{4}$

$∴ α = 3$

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2 months ago

Kinetic Theory Class 11th

0.056 kg of Nitrogen is enclose in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ___________ k cal.

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Vishal Baghel

Contributor-Level 10

$V_{r m s} α \sqrt[]{T} | T_{i} = 127 ° C = 400 k .$

$V_{r m s^{1}} = 2 V_{r m s}, m = 0.056 k g = 56 g$

T_f = 4T_i = 4 * 400 = 1600 k

$Q = n c_{v} Δ T = (\frac{m}{M}) C_{v} Δ T$

= $(\frac{56}{28}) \frac{5}{2} R Δ T$

= $2 * \frac{5}{2} * 2 * 1200$

Q = 12 * 10³ cal

= 12 k cal

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2 months ago

Kinetic Theory Class 11th

A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is

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Payal Gupta

Contributor-Level 10

According to question, we can write

Total moles of gas = n = n_Oxygen+ n_Oxygen = $\frac{16}{2} + \frac{128}{32} = 12 m o l e s$

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵cm3 $\approx 27 * 1 0^{4} c m^{3}$

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2 months ago

Kinetic Theory Class 11th

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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2 months ago

Kinetic Theory physics ncert exemplar solutions class 12th chapter two

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively will be:

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Vishal Baghel

Contributor-Level 10

The average kinetic energy (per molecule)of any (ideal) gas (be it monoatomic like Argon, diatomic like oxygen or polyatomic) is always equal to $\frac{3}{2} K_{B} T .$ If depends only on temperature, and is independent of the nature of the gas. So, Answer is 1 : 1.

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Kinetic Theory Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J at work when it expands under constant pressure. The amount of heat absorbed by the gas will be________J.

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Vishal Baghel

Contributor-Level 10

Degree of freedom, f = 8

$∴ C_{V} = \frac{8}{2} R = 4 R a n d C_{P} = 4 R + R = 5 R$

So, H = $Δ U + W$

at constant pressure, W = $P Δ v$

150 = $n R Δ T$

So, $H = n C_{V} Δ T = n 5 R Δ T = 5 (n R Δ T)$

= 5 * 150

= 750 J

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