SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

Page No 334: - Chapter 13 Kinetic Theory class 11 ncert solutions Physics - SaraNextGen [2024]


Question 13.4:

An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (= 8.31 J mol–1 K–1, molecular mass of O2 = 32 u).

Answer:

Volume of oxygen, V1 = 30 litres = 30 × 10–3 m3

Gauge pressure, P1 = 15 atm = 15 × 1.013 × 105 Pa

Temperature, T1 = 27°C = 300 K

Universal gas constant, R = 8.314 J mole–1 K–1

Let the initial number of moles of oxygen gas in the cylinder be n1.

The gas equation is given as:

P1V1 = n1RT1

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m7c9a0adb.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_77e505ac.gif  = 18.276

But,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m4aaf189c.gif

Where,

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g

m1 = n1= 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.

Volume, V2 = 30 litres = 30 × 10–3 m3

Gauge pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa

Temperature, T2 = 17°C = 290 K

Let n2 be the number of moles of oxygen left in the cylinder.

The gas equation is given as:

P2V2 = n2RT2

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_6d26588d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_77acbb80.gif  = 13.86

But,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4680/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_4efe7a20.gif

Where,

m2 is the mass of oxygen remaining in the cylinder

m2 = n2= 13.86 × 32 = 453.1 g

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

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

m1 – m2

= 584.84 g – 453.1 g

= 131.74 g

= 0.131 kg

Therefore, 0.131 kg of oxygen is taken out of the cylinder.

Question 13.5:

An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Answer:

Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 12°C = 285 K

Temperature at the surface of the lake, T2 = 35°C = 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 ×1.013 × 105 Pa

The pressure at the depth of 40 m:

P1 = 1 atm + g

Where,

ρ is the density of water = 103 kg/m3

g is the acceleration due to gravity = 9.8 m/s2

P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have: https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4681/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_6a9e52b4.gif

Where, V2 is the volume of the air bubble when it reaches the surface

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4681/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_5f41a074.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4681/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m438397f5.gif

= 5.263 × 10–6 m3 or 5.263 cm3

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3.

Question 13.6:

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.

Answer:

Volume of the room, V = 25.0 m3

Temperature of the room, T = 27°C = 300 K

Pressure in the room, P = 1 atm = 1 × 1.013 × 105 Pa

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

PV kBNT

Where,

KB is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

N is the number of air molecules in the room

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4682/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m6e9ce875.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4682/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m1091c4bb.gif  = 6.11 × 1026 molecules

Therefore, the total number of air molecules in the given room is 6.11 × 1026.

Question 13.7:

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:

(i) At room temperature, T = 27°C = 300 K

Average thermal energy https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m59843966.gif

Where is Boltzmann constant = 1.38 × 10–23 m2 kg s–2 K–1

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_157bc9d9.gif

= 6.21 × 10–21J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10–21 J.

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

Average thermal energy https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m59843966.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m17f7dfa2.gif

= 1.241 × 10–19 J

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10–19 J.

(iii) At temperature, T = 107 K

Average thermal energy https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m59843966.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4683/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_f685f20.gif

= 2.07 × 10–16 J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10–16 J.

Question 13.8:

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?

Answer:

Yes. All contain the same number of the respective molecules.

No. The root mean square speed of neon is the largest.

Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature.

According to Avogadro’s law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro’s number, N = 6.023 × 1023.

The root mean square speed (vrms) of a gas of mass m, and temperature T, is given by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4684/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_2413e828.gif

Where, k is Boltzmann constant

For the given gases, k and T are constants.

Hence vrms depends only on the mass of the atoms, i.e.,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4684/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_6dfac72.gif

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

Question 13.9:

At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

Answer:

Temperature of the helium atom, THe = –20°C= 253 K

Atomic mass of argon, MAr = 39.9 u

Atomic mass of helium, MHe = 4.0 u

Let, (vrms)Ar be the rms speed of argon.

Let (vrms)He be the rms speed of helium.

The rms speed of argon is given by:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4685/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_3f7f6294.gif  … (i)

Where,

R is the universal gas constant

TAr is temperature of argon gas

The rms speed of helium is given by:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4685/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_2cd05ace.gif  … (ii)

It is given that:

(vrms)Ar = (vrms)He

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4685/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m58fc9de5.gif

= 2523.675 = 2.52 × 103 K

Therefore, the temperature of the argon atom is 2.52 × 103 K.

Question 13.10:

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

Answer:

Mean free path = 1.11 × 10–7 m

Collision frequency = 4.58 × 109 s–1

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa

Temperature inside the cylinder, T = 17°C =290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m

Diameter, d = 2 × 1 × 1010 = 2 × 1010 m

Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_270606da.gif

Where,

R is the universal gas constant = 8.314 J mole–1 K–1

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m272ad9da.gif  = 508.26 m/s

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m25323c06.gif

Where,

k is the Boltzmann constant = 1.38 × 10–23 kg m2 s–2K–1

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m51f7ff19.gif

= 1.11 × 10–7 m

Collision frequency https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_619b94be.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m2c65d165.gif = 4.58 × 109 s–1

Collision time is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_59517ab9.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_770c02bc.gif  = 3.93 × 10–13 s

Time taken between successive collisions:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_5839b1bf.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_768373c1.gif  = 2.18 × 10–10 s

∴ https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4686/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_6f224122.gif

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

Also Read : Page-No-335:-Chapter-13-Kinetic-Theory-class-11-ncert-solutions-Physics

SaraNextGen