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Physics Class 12th 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

3 months ago

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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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:

Vishal Baghel

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.

A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be:

Take R = 8.32 J mole1 k1.

Vishal Baghel

$m_{N_{2}} = 14 g, T = 27 ° C = 300 k$

To double the Vrms Tf = 4Ti = 4 × 300 = 1200 k.

$∴ Q = n C_{v} Δ T$

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= 9360 J.

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecules per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).

Choose the most appropriate answer form the options given below

Payal Gupta

$\bar{x}$ (mean free path) = $\frac{1}{\sqrt[]{2} π d^{2} n_{v}}$

$n_{v} = \frac{n N_{A}}{v},$ n → No. of moles in volume v N_A ® Avogadro’s Number

$\frac{n}{v} = \frac{P}{R_{T}}$

$\bar{x} \propto v \propto \frac{1}{ρ (d e n s i t y} & \bar{x} α T a t c o n s t a n t P)$

^{$ρ ↑ \Leftrightarrow \bar{x} ↓ | \bar{x} ↑ \Leftrightarrow T ↑$}

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is $= \frac{1}{2} k_{B} T$ (for Mono atomic gases)

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