## NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

**QUESTIONS FROM TEXTBOOK** ( Kinetic Theory Notes )

**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 3 A.****Answer: **Diameter of an oxygen molecule, d = 3 A = 3 x 10^{-10} m. Consider one mole of oxygen gas at STP, which contain total NA = 6.023 x 10^{23} molecules

**Question 13. 3. Following figure shows plot of PV/T versus P for 1.00 x 10 ^{-3} kg of oxygen gas at two different temperatures.**

**(a) What does the dotted plot signify?**

**(b) Which is true : T**

_{1}> T_{2}or T_{1}< T_{2}?**(c) What is the value of PV/T where the curves meet on the y-axis?**

**(d) If we obtained similar plots for 1.00 x 10**

^{-3}kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for the low-pressure high-temperature region of the plot) ? (Molecular mass of H_{2}= 2.02 u, of O_{2}= 32.0 u, R = 8.31 J mol^{-1}K^{-1}.)**Question 13. 4. An oxygen cylinder of volume 30 Hire has an initial gauge pressure of 15 atmosphere and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atmosphere and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder. (R = 8.31 J mol ^{-1} K^{-1}, molecular mass of O_{2} = 32 u.)**

**Question 13. 11. A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?****Answer: ** When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.

.’. P_{1 }= 76 cm and V_{1 }= 15 cm^{3}

**Question 13. 13. A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n _{2} = n_{1} exp [ – mg (h_{2} – h_{1})/k_{B}T]**

**where n**

_{2}, n_{1}refer to number density at heights h_{2}and h_{1}respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :**n**

_{2}= n_{1}exp [ – mg N_{A}(ρ – P) (h_{2}– h_{1})/(ρ RT)] where ρ is the density of the suspended particle, and ρ that of surrounding medium. [N_{A}is Avogadro’s number, and R the universal gas constant.]**[Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]**

**Answer:**Considering the particles and molecules to be spherical, the weight of the particle is

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