Kinetic Theory

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New answer posted•a year ago

Kinetic Theory Class 11th

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).

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.9 Temperature of the helium atom, $T_{H e}$ = – 20 °C = 253 K and temperature of argon atom be = $T_{A r}$

Atomic mass of helium, $M_{H e}$ = 4.0 u

Atomic mass of Argon, $M_{A r}$ = 39.9 u

Let ${V_{r m s}}_{A r}$ be the rms speed of Argon and ${V_{r m s}}_{H e}$ be the rms speed of Helium

From the relation of

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New answer posted•a year ago

Kinetic Theory Class 11th

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?

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.8 (a) 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 $* 10^{23}$

(b) The root mean square speed ( $V_{r m s})$ of a gas of mass m and temperature T is given by the relation $V_{r m s} = \sqrt{\frac{3 k T}{m}}$ . Where k

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New answer posted•a year ago

Kinetic Theory Class 11th

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).

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.7 (i) At room temperature, T = 27 $?$ = 300 K

$k_{B}$ is Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

Average thermal energy = $\frac{3}{2} k_{B} T$ = $\frac{3 \times 1.38 \times 10^{- 23} \times 300}{2}$ = 6.21 $\times 10^{- 21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21 $\times 10^{- 21}$ J

(ii) On the surface of the Sun, T = 6000

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New answer posted•a year ago

Kinetic Theory Class 11th

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 $m^{3}$ at a temperature of 27 °C and 1 atm pressure.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Temperature of the room, T = 27 $?$ = 300 K

Pressure of the room, P = 1 atm = 1 $\times$ 1.013 $\times 10^{5}$ Pa

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

PV = $k_{B}$ NT, where $k_{B}$ &nb

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New answer posted•a year ago

Kinetic Theory Class 11th

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?

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.5 Volume of the air bubble $,$ $V_{1}$ = 1.0 ${c m}^{3}$ = 1 $\times 10^{- 6}$ $m^{3}$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_{1}$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_{2}$ = 35 $?$ = 308 K

The pressure at the surface of the lake,&

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New answer posted•a year ago

Kinetic Theory Class 11th

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 (R = 8.31 J mol^–1 K^–1, molecular mass of O₂ = 32 u).

18 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.4 Volume of oxygen, $V_{1}$ = 30 litres = 30 $\times 10^{- 3} m^{3}$

Gauge pressure, $P_{1}$ = 15 atm = 15 $\times 1.013 \times 10^{5}$ Pa

Temperature, $T_{1}$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_{1}$

From the gas equation, we get&nbs

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New answer posted•a year ago

Kinetic Theory Class 11th

**13.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

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

(d) If we obtained similar plots for 1.00×10^–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 H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

25 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{P V}{T}$ = $?$ R, (where $?$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperatu

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New answer posted•a year ago

Kinetic Theory Class 11th