English
Karnataka Board PUCPUC Science Class 11

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. - Physics

Advertisements
Advertisements

Question

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).

Short/Brief Note
Advertisements

Solution

Assuming hydrogen molecules as spheres of radius 1 Å.

So, r = 1 Å = radius

The volume of hydrogen molecules = `4/3 pir^3`

= `4/3 (3.14)(10^-10)^3`

= `4 xx 10^-30  m^3`

Number of moles of H2 = `"Mass"/"Molecular mass"`

= `0.5/2`

= 0.25

Molecules of H2 present = Number of moles of H2 present × 6.023 × 1023

= 0.25 × 6.023 × 1023

∴ Volume of molecules present = Molecules number × Volume of each molecule

= 0.25 × 6.023 × 1023 × 4 × 10–30

= 6.023 × 1023 × 10–30

= 6 × 10–7 m3   ......(i)

Now, if the ideal gas law is considered to be constant,

`p_iV_i = p_fV_f`

`V_f = (p_i/p_f)`

`V_i = (1/100)(3 xx 10^-2)^3`

= `(27 xx 10^-6)/10^2`

= 2.7 × 10–7 m3  ......(ii)

Hence, on compression, the volume of the gas is of the order of the molecular volume [form equation (i) and equation (ii)]. The intermolecular forces will play a role and the gas will deviate from ideal gas behaviour.

shaalaa.com
Interpretation of Temperature in Kinetic Theory - Introduction of Kinetic Theory of an Ideal Gas
  Is there an error in this question or solution?
Chapter 13: Kinetic Theory - Exercises [Page 95]

APPEARS IN

NCERT Exemplar Physics [English] Class 11
Chapter 13 Kinetic Theory
Exercises | Q 13.22 | Page 95

RELATED QUESTIONS

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?


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


A sample of 0.177 g of an ideal gas occupies 1000 cm3 at STP. Calculate the rms speed of the gas molecules.


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


Let Q and W denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.
(a) Q = 0
(b) W = 0
(c) Q = W
(d) Q ≠ W


A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1° C if 3.0 cal of heat is added to it. The gas may be
(a) helium
(b) argon
(c) oxygen
(d) carbon dioxide


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.


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?


Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take  `"R" = 25/3"J""K"^-1 "mol"^-1 `. (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.


Consider a given sample of an ideal gas (Cp/Cv = γ) having initial pressure p0 and volume V0. (a) The gas is  isothermally taken to a pressure p0/2 and from there, adiabatically to a pressure p0/4. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure p0/2 and from there, isothermally to a pressure p0/4. Find the final volume.


Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.


Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.


A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of 500 ms–1 in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground ______.


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.

When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases ______.

  1. because of collisions with moving parts of the wall only.
  2. because of collisions with the entire wall.
  3. because the molecules gets accelerated in their motion inside the volume.
  4. because of redistribution of energy amongst the molecules.

The container shown in figure has two chambers, separated by a partition, of volumes V1 = 2.0 litre and V2 = 3.0 litre. The chambers contain µ1 = 4.0 and µ2 = 5.0 moles of a gas at pressures p1 = 1.00 atm and p2 = 2.00 atm. Calculate the pressure after the partition is removed and the mixture attains equilibrium.

V1 V2
µ1, p1 µ2
  p2

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×