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**SHORT ANSWER TYPE QUESTIONS**

**Q.1.The container shown in Figure has two chambers, separated by a partition, of volumes V _{1} = 2.0 litre and V_{2}= 3.0 litre. The chambers contain µ_{1} = 4.0 and µ_{2} = 5.0 moles of a gas at pressures p_{1} = 1.00 atm and p_{2} = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.**

µ

P

P

P

µ = µ

V = V

For 1 mole PV = 2/3 E

For µ

For µ

Total energy is

This form of ideal gas law represented by Equation marked becomes very useful for adiabatic changes.

Ans.

v

∵ m

v

(Hydrogen molecules can be consider as spheres of radius 1 Å).

Ans.

Molecular volume = 2.5 × 10

Supposing Ideal gas law is valid.

Final volume

which is about the molecular volume. Hence, intermolecular forces cannot be neglected. Therfore the ideal gas situation does not hold.

Ans.

(a) the number of atoms of helium in the balloon,

(b) the total internal energy of the system.

Ans.

T = 7 + 273 = 280 K

(a) Number of atoms of he is 5 mole = 5 x 6.023 x 10

= 30.115 x 10

= 3.0115 x 10

(b) He atom is mono atomic so degree of freedom is 3 So average kinetic energy

= 3/2 K

= 3/2 K

= 3/2 x 1.38 x 10

Total E of 15 mole of He = 1.74 x 10

Ans.

∴ Number of molecules in 1 cc of hydrogen

As each diatomic molecule has 5 degrees of freedom, hydrogen being diatomic also has 5 degrees of freedom

∴ Total no of degrees of freedom = 5 × 2.688 × 10

Ans.

where n = no: of moles.

If its temperature changes by ∆T , then

= n(3/2)RΔT

= 1/2mn v

∴ ΔT =

**LONG ANSWER TYPE QUESTIONS**

**Q.1. Explain why****(a) there is no atmosphere on moon. ****(b) there is fall in temperature with altitude.****Ans. **The moon has small gravitational force and hence the escape velocity is small. As the moon is in the proximity of the Earth as seen from the Sun, the moon has the same amount of heat per unit area as that of the Earth. The air molecules have large range of speeds. Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of molecules have speed greater than escape velocity and they escape. Now rest of the molecules arrange the speed distribution for the equilibrium temperature. Again a significant number of molecules escape as their speeds exceed escape speed. Hence, over a long time the moon has lost most of its atmosphere.

At 300 K

V_{esc} for moon = 4.6 km/s

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces.

At greater height more volume is available and gas expands and hence some cooling takes place.**Q.2. Consider an ideal gas with following distribution of speeds.**

Speed (m/s) | % of molecules |

200 | 10 |

400 | 20 |

600 | 40 |

800 | 20 |

1000 | 10 |

**(i) Calculate V _{rms} and hence T. (m = 3.0× 10^{−26} kg)**(This problem is designed to give an idea about cooling by evaporation)

(ii) If all the molecules with speed 1000 m/s escape from the system, calculate new V_{rms} and hence T.

Ans.

= 1000 × (4 + 32 + 144 +128 + 100) = 408 ×1000 m

∴ v

= 2.96 ×10

v

Ans.

N = number of particles per unit volume V = N/Volume

d = 2 x 10 m = 20 m = 20 x 10

∴

Ans.

Initial pressure = 1.5 atm = P

Final pressure = 1.5 – 0.1 = 1.4 atm =P’

Air pressure out side box = P

Initial temperature T

Final temperature T

a = area of hole = 0.01 mm

initial pressure difference between tyre and atmosphere ΔP = (1.5 - 1)atm

mass of a N

K

Let P

Number of molecule colliding in time Δt on a wall of cube

1/2 is multiplied as other 1/2 molecule will strike to opposite wall

Then

KE of gas molecule = 3/2 K

Number of N

Temperature inside the box and air are equal to T

The number of air molecule striking to hole in Δt

inward = a inward

ρ

Net number of molecule (going outward)

Net number of molecules going out from hole in Δt time

Gas equation

As for box (μ = No. of moles of gas in box)

= P

Let after time T pressure reduced by 0.1 and becomes (1.5 - 1) = 1.4 atm P'

Then final new density of N

Net number of molecules going out from volume V

(iv) (from II, III)

P

From (I) total number of molecule going out in time τ from hole

(P

Net number of molecule going out in τ time from above

From (V) and IV

τ = 1.34 x 10

Ans.

v

When block is moving with speed v

Coming head on, momentum transferred to block per collission

= 2m (v+v

No. of collission in time where A = area of cross section of block and factor of 1/2 appears due to particles moving towards block.

∴ Momentum transferred in time Δt = m(v + v

Similarly momentum transferred in time Δt = m(v + v

∴ Net force (drag force) = mnA[(v + v

= mnA (4vv

= (4ρAv)v

We also have 1/2 mv

Therefore,

Thus drag

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