Physics M.C.Q (1 Marks)

1. Obeys Maxwell’s distribution.

4. Is (2/3)rd the translational kinetic energy for each molecule.

Explanation:

Consider a diatomic molecule along z-axis so its rotational energy about z-axis is zero. So energy of diatomic molecule,

$\text{E}=\frac{1}{2}\text{mv}_\text{x}^2+\frac{1}{2}\text{mv}_\text{y}^2+\frac{1}{2}\text{mv}_\text{z}^2+\frac{1}{2}\text{I}_\text{x}\omega_\text{x}^2+\frac{1}{2}\text{I}_\text{y}\omega_\text{y}^2$ (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.

2. Isothermal process.

Explanation:

Boyle’s law is applicable at constant temperature, and temperature remains constant in isothermal process,

PV = nRT (n, R and T are constant)

$\therefore$ PV = constant

$\text{P}\alpha\frac{1}{\text{V}}$ (where constant = nRT)

1. P₁ > P₂

Explanation:

$\text{V}\propto\text{T}$ as n, R and P are constant

$\frac{\text{V}_1}{\text{T}_1}$ = constant or slope of graph is constant

$\text{V}=\frac{\text{nRT}}{\text{P}}$

$\frac{\text{dv}}{\text{dt}}=\frac{\text{nR}}{\text{P}}\text{ so }\frac{\text{dV}}{\text{dT}}$ increase when P decreases

$\frac{\text{dV}}{\text{dT}}\alpha\frac{1}{\text{P}}$ as slope of P₁ is smaller than P₂.

Hence, P₁ > P₂ verifies option (a).

1. 20 times the pressure initially.

Explanation:

The situation is shown in the diagram, H₂ gas is contained in a box is heated and gets converted to a gas of hydrogen atoms. Then the number of moles would become twice.

According to gas equation,

PV = nRT

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 300K to 3000K) 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 n₂ = 2n₁

So P ∝ nT

$\Rightarrow\frac{\text{P}_2}{\text{P}_1}=\frac{\text{n}_2\text{T}_2}{\text{n}_1\text{T}_1}=\frac{(2\text{n}_1)(3000)}{\text{n}_1(300)}=20$

$\Rightarrow\text{P}_2=20\text{P}_1$

Hence, final pressure of the gas would be 20 times the pressure initially.

4. Between P and 1.1

Explanation:

PV = nRT n and R are constant for the system here

$\frac{\text{PV}}{\text{T}}$ = constant

or

$\frac{\text{P}_1\text{V}_1}{\text{T}_1}=\frac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\text{P}_2=\frac{\text{P}_1\text{V}_1}{\text{T}_1}\times\frac{\text{T}_2}{\text{V}_2}$

$=\frac{\text{p}\text{V1.1T}}{\text{T1.05V}}=\frac{1.1}{1.05}\text{p}$

$=(1.0476)\text{p}$

i.e. P2 is between P and 1.1p.

So option (d) verifies.

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

Explanation:

According to kinetic theory equation, $\text{PV} = \frac{2}{3}$ E [where P= Pressure V = volume]

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.

Explanation:

For ideal gas behaviour,

PV = nRT

1. When pressure, P = constant.

From (i) Volume V $\propto$ Temperature T

Graph of V versus T will be straight line.

2. When T = constant.

So, graph of P versus V will be a rectangular hyperbola.

Hence this graph is wrong.

The correct graph is shown below:

3. When V = constant.

From (i) $\text{P}\propto\text{T}$

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

4. From (i) $\text{PV}\propto\text{T}$

$\Rightarrow\frac{\text{PV}}{\text{T}}$ = 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.