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**Multiple Choice Questions - I**

**Q.1. 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 500m s ^{–1} in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground.**

(a) Remains the same because 500m s

(b) Remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.

(d) Will be different on the top wall and bottom wall of the vessel.

**Q. Choose the correct option.Ans. **(b)

- According to the ideal gas law, P=nRT/V, here temperature of the vessel remain unchanged hence, the pressure remains same from that point of view.
- Now, let us discuss the phenomenon inside the vessel. The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic.
- The number of collisions per unit volume in a gas remains constant. So, the pressure of the gas inside the vessel remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls.

**Q.2. 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,**

**(a) The pressure on EFGH would be zero.(b) The pressure on all the faces will be equal.(c) The pressure of EFGH would be double the pressure on ABCD.(d) The pressure on EFGH would be half that on ABCD.Ans.** (d)

So, the momentum transferred to the face ABCD = 2mv And the gas molecule is absorbed by the face EFGH. Hence it does not rebound. So, momentum transferred to the face EFGH = mv.

And the pressure on the faces is due to the total momentum to the faces. So, pressure on EFGH would be half that on ABCD.**Q.3. Boyle’s law is applicable for an(a) Adiabatic process(b) Isothermal process(c) Isobaric process(d) Isochoric processAns.** (b)

So we can say that when temperature is constant, Boyle’s law is applicable.

i.e., PV= nRT= constant

⇒ PV = constant (at constant temperature)

i.e. p ∝ 1/V - [where, P = pressure. V= volume]

So, this law is applicable for an isothermal process, in which temperature remain constant.**Q.4. A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction in Figure. If the temperature is increased:**

**(a) Both p and V of the gas will change.(b) Only p will increase according to Charle’s law.(c) V will change but not p.(d) p will change but not V.Ans. **(c)

According to the problem, piston can move up and down without friction. So, the only force present is weight of the piston.

where P

A = Area of cross-section of the piston

Mg = Weight of piston

Weight of piston and atmospheric pressure are constant, hence pressure remains constant.

__According to ideal gas law:__PV = nRT

As the pressure remains constant, so if the temperature is increased, only the volume increases as the piston moves up without friction.

**(a) P _{1} > P_{2} **. (a)

(b) P_{1} = P_{2}

(c) P_{1} < P_{2}

(d) Data is insufficient

Ans

So, if the slope is greater, pressure will be smaller and vice-versa.

Slope of the V-T graph, m = dV/dT = nR/P

⇒ m∝1/P [∴ nR = constant]

⇒ p ∝ 1/m

Hence, P

So, P

where, m

Hence the correct option is (a), i.e. P

**Q.6. 1 mole of H _{2} gas is contained in a box of volume V = 1.00 m^{3} at T = 300K. The gas is heated to a temperature of T = 3000K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)**(d)

(a) Same as the pressure initially

(b) 2 times the pressure initially

(c) 10 times the pressure initially

(d) 20 times the pressure initially

Ans.

P = Pressure of gas, n = Number of moles

R = Gas constant, T = Temperature PV = nRT

As volume (V) of the container is constant. Hence, when temperature (T) becomes 10 times, (from 300 K to 3000 K) pressure (P) also becomes 10 times, as P∝ T.

Pressure is due to the bombardment of particles and as gases break, the number of moles becomes twice of initial, so n_{2} = 2n_{1}

So P ∝ nT

⇒

⇒ P_{2} = 20P_{1}

Hence, final pressure of the gas would be 20 times the pressure initially.**Q.7. A vessel of volume V contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let f _{1} (v)dv, denote the fraction of molecules with speed between v and (v + dv) with f_{2} (v)dv, similarly for oxygen. Then** (c)

(a) f_{1}(v) + f_{2} (v)= f (v) obeys the Maxwell’s distribution law.

(b) f_{1}( v), f_{2} (v) will obey Maxwell’s distribution law separately.

(c) Neither f_{1} (v) nor f_{2} (v) will obey Maxwell’s distribution law.

(d) f_{2} (v) and f_{1} (v) will be the same.

Ans.

where dN = Number of molecules with speeds between v and v + dv

The masses of hydrogen and oxygen molecules are different.

For a function f(v), the number of molecules dn = f[v), which are having speeds between v and v + dv.

The Maxwell-Boltzmann speed distribution function ( N_{v} = dn/dv depends on the mass of the gas molecules.

For each function f_{1}(v) and f_{2}(v), n will be different, hence each function f_{1}(v) and f_{2}(v) will obey Maxwell’s distribution law separately.**Q.8. An inflated rubber balloon contains one mole of an ideal gas, has a pressure p, volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be(a) 1.1 P(b) P(c) Less than P(d) Between P and 1.1Ans.** (d)

P = pressure

V = volume

n = number of moles of gases

R = gas constant

T = temperature

Thus we have to rewrite this equation in such a way that no. of moles is given by,

n = PV/RT

⇒

Hence, final pressure P

**Multiple Choice Questions - II**

**Q.9. ABCDEFGH is a hollow cube made of an insulator (Fig.). Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen.The usual kinetic theory expression for pressure:**

**(a) Will be valid.(b) Will not be valid since the ions would experience forces other than due to collisions with the walls.(c) Will not be valid since collisions with walls would not be elastic.(d) Will not be valid because isotropy is lost.Ans.** (b, d)

(a) The total energy per unit volume.

(b) Only the translational part of energy because rotational energy is very small compared to the translational energy.

(c) Only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum.

(d) 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.

Ans.

E is representing only translational part of energy. Internal energy contains all types of energies like translational, rotational, vibrational etc. But the molecules of an ideal gas is treated as point masses in kinetic theory, so its kinetic energy is only due to translational motion.

Point mass does not have rotational or vibrational motion. Here, we assumed that the walls only exert perpendicular forces on molecules. They do not exert any parallel force, hence there will not be any type of rotation present. The wall produces only change in translational motion.**Q.11. In a diatomic molecule, the rotational energy at a given temperature(a) Obeys Maxwell’s distribution.(b) Have the same value for all molecules.(c) Equals the translational kinetic energy for each molecule.(d) Is (2/3)rd the translational kinetic energy for each molecule.Ans. **(a, d)

(as moment of inertia along z axis is zero)

The independent terms in the above expression is 5. As we can predict velocities of molecules by Maxwell’s distribution. Hence the above expression also obeys Maxwell’s distribution. As 2 rotational and 3 translational energies are associated with each molecule. So the rotational energy at given temperature is 2/3 of its translational Kinetic energy of each molecule.**Q.12. Which of the following diagrams (Figure) depicts ideal gas behaviour?Ans.** (a, c)

PV = nRT ...(i)

From (i) Volume (V) ∝ Temperature (T)

Graph of V versus T will be straight line.

**(b)** When T = constant

From (i) PV = constant

So, graph of P versus V will be a rectangular hyperbola. Hence this graph is wrong.__The correct graph is shown below:__

From (i) P ∝ T

So, the graph is a straight line passing through the origin.

⇒ PV/T = constant

So, graph of PV versus T will be a straight line parallel to the temperature axis(x-axis).

i.e., slope of this graph will be zero.

So, (d) is not correct.

(a) Because of collisions with moving parts of the wall only.

(b) Because of collisions with the entire wall.

(c) Because the molecules gets accelerated in their motion inside the volume.

(d) Because of redistribution of energy amongst the molecules.

Ans.

**Very Short Answer Type Questions**

**Q.1. Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g mole ^{-1}.**

Molar mass = Mass of Avogadro's number of atoms (Molecules)

= 6.023 x 10

Molar mass of the gold, M = 197 g mol

Now, 197 g of gold contains 6.023 x 10

So, 1 g of gold contains

∴ 39.4 g of gold contains (6.023 x 10

Ans.

P

T

T

Units of (P

Ans.

T

v

T

M = Molar mass of gas for a gas M is constant

∴ v

v

Ans.

[v

where v

For two molecules,

According to the problem, v

∴

Ans.

Temperature of the gas = T

Since oxygen is a diatomic gas. So, the degree of freedom associated with the molecules of the oxygen is 5.

Energy (total internal) per mole of the gas = 5/2 RT

[R = universal gas constant

T = temperature]

For 2 moles of the gas total internal energy = 2 x 5/2 RT = 5RT ...(i)

Neon (Ne) is a monoatomic gas having 3 degrees of freedom.

Energy per mole = 3/2RT

Hence, Energy = 4 x 3/2 RT = 6RT ….(ii)

[Using Eqs. (i) and (ii)]

Total energy = 5RT = 6RT= 11RT

Ans.

The distance travelled by a gas molecule between two successive collisions is known as free path.

λ = Total distance travelled by a gas molecule between successive collisions / Total number of collisions during two successive collisions

A molecule of a gas moves in a straight line with constant velocity. Let λ_{1,} λ_{2 }, λ_{3} be the distance travelled by a gas molecule during n collisions respectively, then the mean free path of a gas molecule is given by:

λ =

where d = Diameter of the molecule,

n = Number of molecules per unit volume__According to the problem:__d

Since, mean free path,

λ ∝ 1/d

⇒So,

Hence, λ

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