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INTRODUCTION - Chapter 13 Kinetic Theory class 11 ncert solutions Physics - SaraNextGen [2024-2025]


Question 13.1:

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

Answer:

Diameter of an oxygen molecule, d = 3Å

Radius, r

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4677/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m1d9e59c0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4677/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m143c6df7.gif  = 1.5 Å = 1.5 × 10–8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm3

Molecular volume of oxygen gas,https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4677/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_1bb3aa9e.gif

Where, N is Avogadro’s number = 6.023 × 1023 molecules/mole

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4677/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m68737ab.gif

Ratio of the molecular volume to the actual volume of oxygen https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4677/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m1694740b.gif

= 3.8 × 10–4

Question 13.2:

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

Answer:

The ideal gas equation relating pressure (P), volume (V), and absolute 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 × 105 Nm–2

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4678/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_4b1ba0cc.gif

= 0.0224 m3

= 22.4 litres

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

Question 13.3:

Figure 13.8 shows plot of PV/versus Pfor 1.00×10–3 kg of oxygen gas at two different temperatures.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_7cf05733.jpg

(a) What does the dotted plot signify?

(b) Which is true: TT2 or T1 < T2?

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

(d) If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get the same value of PV/at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/(for low pressure high temperature region of the plot)? (Molecular mass of H= 2.02 u, of O2 = 32.0 u, = 8.31 J mo1–1 K–1.)

Answer:

(a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratiohttps://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m378b85ae.gif  is equal. μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is not dependent on the pressure of the gas.

(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T1 is closer to the dotted plot than the curve of the gas at temperature T2. A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore, T1 > T2 is true for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas equation is given as:

PV = μRT

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m3993fe78.gif

Where,

P is the pressure

T is the temperature

is the volume

μ is the number of moles

R is the universal constant

Molecular mass of oxygen = 32.0 g

Mass of oxygen = 1 × 10–3 kg = 1 g

R = 8.314 J mole–1 K–1

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m3012e7ae.gif https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_60539240.gif

= 0.26 J K–1

Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is

0.26 J K–1.

(d) If we obtain similar plots for 1.00 × 10–3 kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).

We have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m10d145b3.gif

R = 8.314 J mole–1 K–1

Molecular mass (M) of H2 = 2.02 u

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m63571402.gif

m = Mass of H2

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m50c0cea7.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/191/4679/NS_15-10-08_Sravana_11_Physics_13_14_NRJ_LVN_html_m1d4089ab.gif

= 6.3 × 10–2 g = 6.3 × 10–5 kg

Hence, 6.3 × 10–5 kg of H2 will yield the same value of PV/T.

Also Read : Page-No-334:-Chapter-13-Kinetic-Theory-class-11-ncert-solutions-Physics

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