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 3Ãƒâ€¦.
Answer :
Diameter of an oxygen molecule, d= 3Ãƒâ€¦
Radius, r = 1.5 Ãƒâ€¦ = 1.5 × 10^{â€“8}cm
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm^{3}
Molecular volume of oxygen gas,
Where, N is Avogadro's number = 6.023 × 10^{23} molecules/mole
Ratio of the molecular volume to the actual volume of oxygen
= 3.8 × 10^{â€“4}
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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,
Ris the universal gas constant = 8.314 J mol^{â€“1}K^{â€“1}
n= Number of moles = 1
T= Standard temperature = 273 K
P= Standard pressure = 1 atm = 1.013 × 10^{5}Nm^{â€“2}
= 0.0224 m^{3}
= 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.
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Figure 13.8 shows plot of PV/T versus Pfor 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 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 mo1^{-1}K^{-1}.)
Answer :
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cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^{-1} K^{-1}, molecular mass of O_{2} = 32 u).
Answer :
An air bubble of volume 1.0 cm^{3}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 :
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^{3}at a temperature of 27 °C and 1 atm pressure.
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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 :
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 v_{rms}the largest?
Answer :
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 :
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 Ãƒ”¦. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N_{2}= 28.0 u).
Answer :
A metre 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 :
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm^{3}s^{-1}. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm^{3}s^{-1}. Identify the gas.
[Hint:Use Graham's law of diffusion: R_{1}/R_{2}= (M_{2}/M_{1})^{1/2}, where R_{1}, R_{2}are diffusion rates of gases 1 and 2, and M_{1}and M_{2}their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer :
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}(ÃÂ - P”²) (h_{2}-h_{1})/ (ÃÂRT)]
Where ÃÂ is the density of the suspended particle, and ÃÂ' that of surrounding medium. [N_{A}is Avogadro's number, and R the universal gas constant.] [Hint:Use Archimedes principle to find the apparent
Answer :
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms:
Substance |
Atomic Mass (u) |
Density (10^{3} Kg m^{-3}) |
Carbon (diamond) Gold Nitrogen (liquid) Lithium Fluorine (liquid) |
12.01 197.00 14.01 6.94 19.00 |
2.22 19.32 1.00 0.53 1.14 |
[Hint:Assume the atoms to be 'tightly packed' in a solid or liquid phase, and use the known value of Avogadro's number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Ãƒâ€¦].
Answer :
Physics Part-2 - Physics : CBSE NCERT Exercise Solutions for Class 11th for Kinetic Theory will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.