- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- Chapter 3- Motion in a Straight Line
- Chapter 4- Motion in a plane
- Chapter 5- Laws of motion
- Chapter 6- Work Energy and power
- Chapter 7- System of particles and Rotational Motion
- Chapter 8- Gravitation
- Chapter 9- Mechanical Properties Of Solids
- Chapter 10- Mechanical Properties Of Fluids
- Chapter 11- Thermal Properties of matter
- Chapter 12- Thermodynamics
- Chapter 14- Oscillations
- Chapter 15- Waves

Chapter 1- Physical World |
Chapter 2- Units and Measurements |
Chapter 3- Motion in a Straight Line |
Chapter 4- Motion in a plane |
Chapter 5- Laws of motion |
Chapter 6- Work Energy and power |
Chapter 7- System of particles and Rotational Motion |
Chapter 8- Gravitation |
Chapter 9- Mechanical Properties Of Solids |
Chapter 10- Mechanical Properties Of Fluids |
Chapter 11- Thermal Properties of matter |
Chapter 12- Thermodynamics |
Chapter 14- Oscillations |
Chapter 15- Waves |

**Answer
1** :

Diameter of an oxygen molecule, d = 3 Å

Radius, r = d / 2

r = 3 / 2 = 1.5 Å = 1.5 x 10^{ -8 }cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm^{3}

Molecular volume of oxygen gas, V = 4 / 3 πr^{3}. N

Where, N is Avogadro’s number = 6.023 x 10^{23} molecules/mole

Hence,

V = 4 / 3 x 3.14 x (1.5 x 10^{-8})^{3} x 6.023 x 10^{23}

We get,

V = 8.51 cm^{3}

Therefore, ratio of the molecular volume to the actual volume ofoxygen = 8.51/ 22400 = 3. 8 x 10^{-4}

**Answer
2** :

The ideal gas equation relating pressure (P), volume (V), andabsolute temperature (T) is given as:

PV = nRT

Where, R is the universal gas constant = 8.314 J mol^{-1}K^{-1}

n = Number of moles = 1

T = Standard temperature = 273 K

P = Standard pressure = 1 atm = 1.013 x 10^{5}Nm^{-2}

Hence,

V = nRT / P

= 1 x 8.314 x 273 / 1.013 x 10^{5}

= 0.0224 m^{3}

= 22.4 litres

Therefore, the molar volume of a gas at STP is 22.4 litres

**(a) What does thedotted plot signify?**

**(b) Which is true: T _{1} > T_{2} or T_{1} < T_{2}?**

**(c) What is the value of PV/Twhere the curves meet on the y-axis?**

**(d) If we obtained similar plotsfor 1.00×10 ^{-3} kg of hydrogen,would we get the same value of PV/T at the point where the curves meet on they-axis? If not, what mass of hydrogen yields the same value of PV/T (for lowpressure high temperature region of the plot)? (Molecular mass of H_{2} = 2.02 u, of O_{2} = 32.0 u, R = 8.31J mo1^{–1} K^{–1}.)**

**Answer
3** :

(a) dotted plot is parallel to X-axis, signifying that nR [PV/T= nR] is independent of P. Thus it is representing ideal gas behaviour

(b) the graph at temperature T_{1} is closer to ideal behaviour (because closer todotted line) hence, T_{1} >T_{2} (higher thetemperature, ideal behaviour is the higher)

(c) use PV = nRT

PV/ T = nR

Mass of the gas = 1 x 10^{-3} kg= 1 g

Molecular mass of O_{2} = 32g/ mol

Hence,

Number of mole = given weight / molecular weight

= 1/ 32

So, nR = 1/ 32 x 8.314 = 0.26 J/ K

Hence,

Value of PV / T = 0.26 J/ K

(d) 1 g of H_{2} doesn’trepresent the same number of mole

Eg. molecular mass of H_{2} = 2 g/mol

Hence, number of moles of H_{2} require is 1/32 (as per the question)

Therefore,

Mass of H_{2} required= no. of mole of H_{2} xmolecular mass of H_{2}

= 1/ 32 x 2

= 1 / 16 g

= 0.0625 g

= 6.3 x 10^{-5} kg

Hence, 6.3 x 10^{-5} kgof H_{2} wouldyield the same value

**Answer
4** :

Volume of gas, V_{1} =30 litres = 30 x 10^{-3 }m^{3}

Gauge pressure, P_{1} =15 atm = 15 x 1.013 x 10^{5} P a

Temperature, T_{1} =27^{0} C = 300 K

Universal gas constant, R = 8.314 J mol^{-1} K^{-1}

Let the initial number of moles of oxygen gas in the cylinder ben_{1}

The gas equation is given as follows:

P_{1}V_{1} = n_{1}RT_{1}

Hence,

n_{1} =P_{1}V_{1} / RT_{1}

= (15.195 x 10^{5} x30 x 10^{-3}) / (8.314 x 300)

= 18.276

But n_{1} =m_{1} / M

Where,

m_{1} =Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

Thus,

m_{1} =N_{1}M = 18.276 x 32 =584.84 g

After some oxygen is withdrawn from the cylinder, the pressureand temperature reduce.

Volume, V_{2} =30 litres = 30 x 10^{-3} m^{3}

Gauge pressure, P_{2} =11 atm

= 11 x 1.013 x 10^{5 }Pa

Temperature, T_{2} =17^{0} C = 290 K

Let n_{2} bethe number of moles of oxygen left in the cylinder

The gas equation is given as:

P_{2}V_{2} = n_{2}RT_{2}

Hence,

n_{2} =P_{2}V_{2} / RT_{2}

= (11.143 x 10^{5} x30 x 10^{-30}) / (8.314 x 290)

= 13.86

But

n_{2} =m_{2} / M

Where,

m_{2 }isthe mass of oxygen remaining in the cylinder

Therefore,

m_{2} =n_{2} x M = 13.86 x32 = 453.1 g

The mass of oxygen taken out of the cylinder is given by therelation:

Initial mass of oxygen in the cylinder – Final mass of oxygen inthe cylinder

= m_{1} –m_{2}

= 584.84 g – 453.1 g

We get,

= 131.74 g

= 0.131 kg

Hence, 0.131 kg of oxygen is taken out of the cylinder

**Answer
5** :

Volume of the air bubble, V_{1} = 1.0 cm^{3}

= 1.0 x 10^{-6 }m^{3}

Air bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T_{1} = 12^{0} C= 285 K

Temperature at the surface of the lake, T_{2 }= 35^{0} C= 308 K

The pressure on the surface of the lake:

P_{2} =1 atm = 1 x 1.013 x 10^{5} Pa

The pressure at the depth of 40 m:

P_{1}= 1atm + dρg

Where,

ρ is the density of water = 10^{3} kg/ m^{3}

g is the acceleration due to gravity = 9.8 m/s^{2}

Hence,

P_{1} =1.013 x 10^{5} + 40 x 10^{3} x 9.8

We get,

= 493300 Pa

We have

P_{1}V_{1} / T_{1} = P_{2}V_{2 }/ T_{2}

Where, V_{2} isthe volume of the air bubble when it reaches the surface

V_{2} =P_{1}V_{1}T_{2} / T_{1}P_{2}

= 493300 x 1 x 10^{-6} x308 / (285 x 1.013 x 10^{5})

We get,

= 5.263 x 10^{-6} m^{3} or 5.263 cm^{3}

Hence, when the air bubble reaches the surface, its volumebecomes 5.263 cm^{3}

**Answer
6** :

Volume of the room, V = 25.0 m^{3}

Temperature of the room, T = 27^{0} C= 300 K

Pressure in the room, P = 1 atm = 1 x 1.013 x 10^{5} Pa

The ideal gas equation relating pressure (P), Volume (V), andabsolute temperature (T) can be written as:

PV = (k_{B}NT)

Where,

K_{B} isBoltzmann constant = (1.38 x 10^{-23}) m^{2} kg s^{-2} K^{-1}

N is the number of air molecules in the room

Therefore,

N = (PV / k_{B}T)

= (1.013 x 10^{5} x25) / (1.38 x 10^{-23} x 300)

We get,

= 6.11 x 10^{26} molecules

Hence, the total number of air molecules in the given room is6.11 x 10^{26}

Estimate the average thermal energy of a helium atom at

(i) room temperature (27 °C),

(ii) the temperature on the surface of the Sun (6000 K),

(iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

**Answer
7** :

(i) At room temperature, T = 27^{0} C= 300 K

Average thermal energy = (3 / 2) kT

Where,

k is the Boltzmann constant = 1.38 x 10^{-23} m^{2} kg s^{-2} K^{-1}

Hence,

(3 / 2) kT = (3 / 2) x 1.38 x 10^{-23} x300

On calculation, we get,

= 6.21 x 10^{-21} J

Therefore, the average thermal energy of a helium atom at roomtemperature of 27^{0} C is 6.21 x 10^{-21} J

(ii) On the surface of the sun, T = 6000 K

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10^{-23} x6000

We get,

= 1.241 x 10^{-19} J

Therefore, the average thermal energy of a helium atom on thesurface of the sun is 1.241 x 10^{-19} J

(iii) At temperature, T = 10^{7} K

Average thermal energy = (3 / 2) kT

= (3 / 2) x 1.38 x 10^{-23} x10^{7}

We get,

= 2.07 x 10^{-16 }J

Therefore, the average thermal energy of a helium atom at thecore of a star is 2.07 x 10^{-16} J

**Answer
8** :

All the three vessels have the same capacity, they have the samevolume.

So, each gas has the same pressure, volume and temperature

According to Avogadro’s law, the three vessels will contain anequal number of the respective molecules.

This number is equal to Avogadro’s number, N = 6.023 x 10^{23}.

The root mean square speed (V_{rms}) of a gas of mass m and temperature T is given bythe relation:

V_{rms} = **√**3kT / m

Where,

k is Boltzmann constant

For the given gases, k and T are constants

Therefore, V_{rms} dependsonly on the mass of the atoms, i.e., V_{rms} ∝ (1/m)^{1/2}

Hence, the root mean square speed of the molecules in the threecases is not the same.

Among neon, chlorine and uranium hexafluoride, the mass of neonis the smallest.

Therefore, neon has the largest root mean square speed among thegiven gases.

**Answer
9** :

Given

Temperature of the helium atom, T_{He} = -20^{0} C= 253 K

Atomic mass of argon, M_{Ar} = 39.9 u

Atomic mass of helium, M_{He} = 4.0 u

Let (V_{rms})_{Ar }be the rms speedof argon and

Let (V_{rms})_{He }be the rms speedof helium

The rms speed of argon is given by:

(V_{rms})_{Ar }= √3RT_{Ar} / M_{Ar} ………… (i)

Where,

R is the universal gas constant

T_{Ar} istemperature of argon gas

The rms speed of helium is given by:

(V_{rms})_{He} = √3RT_{He} / M_{He} ………… (ii)

Given that,

(V_{rms})_{Ar} = (V_{rms})_{He}

√3RT_{Ar} /M_{Ar} = √3RT_{He} / M_{He}

T_{Ar} /M_{Ar} = T_{He} / M_{He}

T_{Ar} =T_{He} / M_{He} x M_{Ar}

= (253 / 4) x 39.9

We get,

= 2523.675

= 2.52 x 10^{3} K

Hence, the temperature of the argon atom is 2.52 x 10^{3} K

**Answer
10** :

Mean free path = 1.11 x 10^{-7} m

Collision frequency = 4.58 x 10^{9} s^{-1}

Successive collision time **≅**** **500 x(Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm =2.026 x 10^{5} Pa

Temperature inside the cylinder, T = 17^{0} C= 290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 x 10^{10} m

Diameter, d = 2 x 1 x 10^{10} =2 x 10^{10} m

Molecular mass of nitrogen, M = 28.0 g = 28 x 10^{-3} kg

The root mean square speed of nitrogen is given by the relation:

V_{rms}=√3RT / M

Where,

R is the universal gas constant = 8.314 J mol^{-1} K^{-1}

Hence,

V_{rms}=3 x 8.314 x 290 / 28 x 10^{-3}

On calculation, we get,

= 508.26 m/s

The mean free path (l) is given by relation:

l = KT / √2 x π x d^{2} xP

Where,

k is the Boltzmann constant = 1.38 x 10^{-23} kgm^{2} s^{-2} K^{-1}

Hence,

l = (1.38 x 10^{-23} x290) / (√2 x 3.14 x (2 x 10^{-10})^{2} x 2.026 x 10^{5}

We get,

= 1.11 x 10^{-7} m

Collision frequency = V_{rms} / l

= 508.26 / 1.11 x 10^{-7}

On calculation, we get,

= 4.58 x 10^{9} s^{-1}

Collision time is given as:

T = d / V_{rms}

= 2 x 10^{-10} /508.26

On further calculation, we get

= 3.93 x 10^{-13} s

Time taken between successive collisions:

T’ = l / V_{rms} =1.11 x 10^{-7} / 508.26

We get,

= 2.18 x 10^{-10}

Hence,

T’ / T = 2.18 x 10^{-10} /3.93 x 10^{-13}

On calculation, we get,

= 500

Therefore, the time taken between successive collisions is 500times the time taken for a collision

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