Physics M.C.Q (1 Marks)

3. Helium.
4. Argon.

Explanation:

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54J of heat is added to it. So, its molar heat capacity, C_​v = 12.54J-JK^-1mol^-1, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas, ​$\text{C}_\text{v}\simeq\frac{3}{2}\text{R}=1.5\times8.314=12.54\text{JK}^{-1}\text{mol}^{-1}.$

Among the given gases, only helium and argon are inert and hence, monoatomic. Therefore, the gas may be helium or argon.

4. Number of molecules in the gas.

Explanation:

Here,

PV = nRT ...(1)

Also,

$\text{k}=\frac{\text{R}}{\text{N}}$

$\Rightarrow\text{R}=\text{kN}\ ...(2)$

Now,

PV = nkNT [From eq. (1) and eq. (2)]

$\Rightarrow\text{nN}=\frac{\text{PV}}{\text{kT}}$

nN = Number of molecules

$\frac{\text{PV}}{\text{kT}}$ = Number of molecules.

1. $\frac{9}{7}$

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).

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.

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. 2000ms^-1

Explanation:

Given,

Molecular mass of hydrogen, M_H = 2

Molecular mass of oxygen, M_O = 32

RMS speed is given by,

$\text{v}_\text{rms}=\sqrt{\frac{3\text{RT}}{\text{M}}}$

$\Rightarrow\sqrt{\frac{3\text{RT}}{\text{M}_\text{O}}}=500$

Now,

$\Rightarrow\frac{\text{v}_\text{Orms}}{\text{v}_\text{Hrms}}=\frac{\sqrt{\frac{3\text{RT}}{\text{M}_\text{O}}}}{\sqrt{\frac{3\text{RT}}{\text{M}_\text{H}}}}$

$\Rightarrow\frac{\text{v}_\text{Orms}}{\text{v}_\text{Hrms}}=\frac{\sqrt{\frac{3\text{RT}}{32}}}{\sqrt{\frac{3\text{RT}}{2}}}$

$\Rightarrow\frac{\text{v}_\text{Orms}}{\text{v}_\text{Hrms}}=\frac{1}{4}$

$\Rightarrow\frac{500}{\text{v}_\text{Hrms}}=\frac{1}{4}$

$\Rightarrow\text{v}_\text{Hrms}=4\times500=2000\text{ms}^{-1}$

1. 4.25J/ cal

Explanation:

According to first law of thermodynamics $\text{J}\triangle\text{Q}=\triangle\text{W}+\triangle\text{U},$ where J is the mechanical equivalent of heat.

J × 20 = 50 + 35

J = 4.25J/ cal

4. (C_P - C_V) << R

Explanation:

In case of solides, C_P = C_V

$\therefore$ C_P - C_V << R

4. $\text{P}=\frac{2}{3}\text{E}$

Explanation:

$\text{P}=\frac{1}{3}\rho\text{C}^2$

Mean K.E./ volume $=\text{E}=\frac{1}{2}\rho\text{C}^2$

$\therefore\text{P}=\frac{1}{3}\rho\text{C}^2=\frac{2}{3}\Big(\frac{1}{2}\rho\text{C}^2\Big)=\frac{2}{3}\text{E}$ 

2. $3\text{RT}$

Explanation:

lnternal energy, $\text{U}=\Big(\frac{3}{2}\text{k}_{\text{B}}\text{T}\Big)2\text{N}_{\text{A}}$

$=3(\text{k}_{\text{B}}\times\text{N}_{\text{A}})\text{T = 3RT}$

1. $\frac{7}{5}$

Explanation:

The value of CV for a diatomic gas is $\frac{7}{2}$ R and

CP is$​​\frac{9}{2}$ R.

Thus the value of γ is:

$​​\frac{\text{C}_\text{P}}{\text{C}_\text{V}}=​​\frac{7}{5}$

3. Pressure.

Explanation:

Pressure of an ideal gas is given by $\text{PV}=\frac{1}{3}\text{mnu}^2.$

We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change.

3. $\frac{5}{2}\text{R}$

1. 2.17×10^-20

Explanation:

Using, PV = NkT

we get: $\text{N}=\frac{\text{PV}}{\text{KT}}$

= $\frac{(3\times3\times105)}{(1.38\times10^{-23}\times300)}$

= 2.17×10^-20

4. Nature of the gas.

Explanation:

In an ideal gas, the equation of state is given by

$\text{PV}=\text{nRT}$

$\Rightarrow\text{PV}=\text{nN}_\text{A}\frac{\text{R}}{\text{N}_\text{A}}\text{T}$

$\Rightarrow\text{PV}=\text{nN}_\text{A}\text{kT}$

$\Rightarrow\frac{1}{\text{nN}_\text{A}}=\frac{\text{kT}}{\text{PV}}$

Multiplying both sides by mass of the gas M, we get

$\frac{\text{M}}{\text{nN}_\text{A}}=\frac{\text{MkT}}{\text{PV}}$

Now, nN_A gives the total number of molecules of the gas.

Also, $\frac{\text{M}}{\text{nN}_\text{A}}$ gives the mass of a single molecule.

Hence, $\frac{\text{MkT}}{\text{PV}}$ is the mass of a single molecule of the gas,

Molecular mass is a property of the gas.

3. 2.9 × 10^-7m

Explanation:

For air at STP, n = 2.7 × 10²⁵m^-3

d = 2 × 10^-10m

$\Rightarrow\text{l}=\frac{1}{\sqrt{2}\text{n}\pi\text{d}^2}$

On putting values, l = 2.9 × 10^-7m

4. $\sqrt{\frac{\gamma}{3}}$

1. $\text{P}\propto\text{p}$

2. 4:3:2.

Explanation:

Their average kinetic energies will be the same. Thus, $\frac{1}{2}$mv² will be the same.

$\text{v}_1^2:\text{v}_2^2:\text{v}_3^2$

 = m₃:m₂:m₁

= 40:30:20

= 4:3:2.

1. $\frac{16}{21}$

Explanation:

The ratio of moles is the same as the ratio of partial pressures.

The ratio of densities:

$\frac{\text{d}_1}{\text{d}_2}= \frac{\Big(\frac{\text{m}_1}{\text{V}}\Big)}{\Big(\frac{\text{m}_2}{\text{V}}\Big)}$

$=\frac{\text{m}_1}{\text{m}_2}= \frac{\Big(\frac{\text{n}_1}{\text{M}_1}\Big)}{\Big(\frac{\text{n}_1}{\text{M}_2}\Big)}$

where n is the number of moles and M is the molecular mass.

$\frac{\text{d}_1}{\text{d}_2}=\Big(\frac{2}{3}\Big)\times\Big(\frac{16}{14}\Big)$

= $\frac{16}{21}$

1. Monoatomic.

Explanation:

$\frac{\text{R}}{\text{C}_{\text{V}}}=0.67=\frac{2}{3}$

$\text{C}_{\text{V}}=\frac{3}{2}\text{R}$

4. $\sqrt{2}$ times that of helium.

3. $2\text{v}$

Explanation:

Given, $\text{v}=\sqrt{\frac{3\text{RT}}{32}}$

Let the new rms speed be v'.

Molecule dissociate, M = 16

$\text{v}'=\sqrt{\frac{3\text{R}(2\text{T})}{16}}$

$=\sqrt{\frac{3\text{R}(4\text{T})}{32}}$

$=2\sqrt{\frac{3\text{R}\text{T}}{32}}=2\text{v}$

2. 6.023 × 10²²

Explanation:

According to Avogadro’s law 22.4L of any gas at STP is 6.023 × 10²³.

So, in 2.24L there will be $\frac{6.023\times10^{23}}{10}$

= 6.023 × 10²².

1. 0.8×10^-20J

Explanation:

Internal energy is given by: E = $\Big(\frac{1}{2}\Big)$Nmv²,

where N is the total number of molecules = nV.

E = 0.5×20×0.02×10^-20×4

= 0.8×10^-20J.

3. Temperature.

Explanation:

The energy of a given sample of an ideal gas depends only on temperature as average energy/ molecule/ degree of freedom $=\frac{1}{2}\text{k}_{\text{B}}\text{T}$

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.

4. 327°C

Explanation:

According to Charle's law, when P is constant, $\text{T}\propto\text{V}$ As V is doubled, T becomes twice i.e.,

T = 2 × (17 + 273)K = 600K

= 600 - 273 = 327°C

3. T

Explanation:

Root mean squared velocity is given by,

$\text{v}_\text{rms}=\sqrt{\frac{3\text{RT}}{\text{M}}}$

$\Rightarrow(\text{v}_\text{rms})^2=\frac{3\text{RT}}{\text{M}}$

$\Rightarrow(\text{v}_\text{rms})^2\alpha\text{T}$

2. $\frac{1}{4}$

Explanation:

$\frac{\text{C}_{\text{oxy}}}{\text{C}_{\text{H}}}=\sqrt{\frac{\text{m}_{\text{H}}}{\text{m}_{\text{oxy}}}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

1. T₁ > T₂.

Explanation:

The straight line T₁ has greater slope than T_​2. This means $\frac{\text{P}}{\rho}$ ratio is greater for T₁ than T₂. Now, rms velocity of a gas is given by $\sqrt{\frac{3\text{P}}{\rho}}.$ This means rms velocity of gas with T₁ molecules is greater than T₂ molecules. Again, gas with higher temperature has higher rms velocity.

3. Isochoric.

Explanation:

According to the graph, P is directly proportional to T.

Applying the equation of state, we get,

PV = nRT

$=\frac{\text{nR}}{\text{V}}\text{T}$

Given: $\text{P}\propto\text{T}$

This means $\frac{\text{nR}}{\text{V}}$ is a constant. So, V is also a constant.

Constant V implies the process is isochoric.

2. $\frac{1}{2}\text{KBT}$

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)

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.

2. $\frac{7}{5}$

Explanation:

$\text{C}_{\text{P}}=\frac{7}{2}\text{R}$

$\text{C}_{\text{V}}=\text{C}_{\text{P}}-\text{R}=\frac{7}{2}\text{R}-\text{R}=\frac{5}{2}\text{R}$

$\gamma=\frac{\text{C}_{\text{P}}}{\text{C}_{\text{V}}}=\frac{\frac{7}{2}\text{P}}{\frac{5}{2}\text{R}}=\frac{7}{5}$​​​​​​​

4. $\frac{7}{5}$

1. $\frac{5}{3}$

Explanation:

The value of Cv for a monatomic gas is $\frac{3}{2}$ R and

C_P is $\frac{5}{2}$ R.

Thus the value of γ is:

$\frac{\text{C}_\text{p}}{\text{C}_\text{V}}=\frac{5}{3}$

2. $\frac{4}{3}$

1. 100kPa.

Explanation:

Let the number of moles in the gas be n.

Applying equation of state, we get,

$\text{PV}=\text{nRT}$

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

$\Rightarrow2\times10^5=\frac{\text{nRT}}{\text{V}}\ ...(1)$

When half of the gas is removed, number of moles left behind $=\frac{\text{n}}{2}$

Let the pressure be P'.

$\text{P}'=\frac{\text{n}}{2}\frac{\text{RT}}{\text{V}}$

Now,

$\text{P}'=\frac{1}{2}\times2\times10^5=10^5$ [From eq. (1)]

$=100\text{kPa}$