हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

P Find the Number of Molecules in 1 Cm3 of an Ideal Gas at 0°C and at a Pressure of 10−5mm of Mercury. - Physics

Advertisements
Advertisements

प्रश्न

Find the number of molecules in 1 cm3 of an ideal gas at 0°C and at a pressure of 10−5mm of mercury.

Use R = 8.31 J K-1 mol-1

योग
Advertisements

उत्तर

Given:
Volume of ideal gas, V = 1 cm3 = 10-6 m​3
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10−8 m of Hg
Density of ideal gas, ρ = 13600 kgm-3
Pressure \[\left( P \right)\] is given by 

P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,
Using the ideal gas equation, we get

\[n = \frac{PV}{RT}\] 

\[ \Rightarrow n = \frac{\rho gh \times V}{RT}\] 

\[ \Rightarrow n = \frac{13600 \times 9 . 8 \times {10}^{- 8} \times {10}^{- 6}}{8 . 31 \times 273}\] 

\[ \Rightarrow n = 5 . 87 \times  {10}^{- 13} \]

Number of molecules = N × n
                                    = 6.023 × 1023×5.874 × 10−13
                                    = 35.384 × 1010
                                    = 3.538 × 1011

shaalaa.com
Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Kinetic Theory of Gases - Exercises [पृष्ठ ३४]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 2 Kinetic Theory of Gases
Exercises | Q 3 | पृष्ठ ३४

संबंधित प्रश्न

The energy of a given sample of an ideal gas depends only on its


Which of the following quantities is zero on an average for the molecules of an ideal gas in equilibrium?


Keeping the number of moles, volume and temperature the same, which of the following are the same for all ideal gases?


Consider the quantity \[\frac{MkT}{pV}\] of an ideal gas where M is the mass of the gas. It depends on the


Calculate the volume of 1 mole of an ideal gas at STP.


Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.


A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol−1) is moving on a floor at a speed of 50 m s−1. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.


An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by `"C" ="C"_"v" +"R"/2.`


Two ideal gases have the same value of Cp / Cv = γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?


An ideal gas (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bcand (d) the change in the internal energy of the gas in the process.


Two vessels A and B of equal volume V0 are connected by a narrow tube that can be closed by a valve. The vessels are fitted with pistons that can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cp/Cv = γ) at atmospheric pressure p0 and atmospheric temperature T0. The walls of vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and pressure.


An ideal gas of density 1.7 × 10−3 g cm−3 at a pressure of 1.5 × 105 Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.


1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (Figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time ______.


Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory `pV = 2/3` E, E is ______.

  1. the total energy per unit volume.
  2. only the translational part of energy because rotational energy is very small compared to the translational energy.
  3. only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.
  4. the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero.

In a diatomic molecule, the rotational energy at a given temperature ______.

  1. obeys Maxwell’s distribution.
  2. have the same value for all molecules.
  3. equals the translational kinetic energy for each molecule.
  4. is (2/3)rd the translational kinetic energy for each molecule.

We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state?

(Hydrogen molecules can be consider as spheres of radius 1 Å).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×