Heat and Thermodynamics — Page 3 | NEET Physics

Q21

The average thermal energy for a mono-atomic gas is : (k B is Boltzmann constant and T absolute temperature)

A

{3 \over 2}{k_B}T

B

{5 \over 2}{k_B}T

C

{7 \over 2}{k_B}T

D

{1 \over 2}{k_B}T

Correct Answer

Option A

Solution

The degree of freedom for monoatomic gas is 3. So, average thermal energy per molecule, K.E. avg =

{3 \over 2}{k_B}T

Q22

The mean free path for a gas, with molecular diameter d and number density n can be expressed as :

A

{1 \over {\sqrt 2 n\pi {d^2}}}

B

{1 \over {\sqrt 2 {n^2}\pi {d^2}}}

C

{1 \over {\sqrt 2 {n^2}{\pi ^2}{d^2}}}

D

{1 \over {\sqrt 2 n\pi d}}

Correct Answer

Option A

Solution

As per the formula,

\lambda = {1 \over {\sqrt 2 n\pi {d^2}}}

Q23

A cylinder contains hydrogen gas at pressure 249 kPa and temperature 27

^\circ

C Its density is : (R = 8.3 J mol -1 K -1 )

A 0.2 kg/m 3

B 0.1 kg/m 3

C 0.02 kg/m 3

D 0.5 kg/m 3

Correct Answer

Option A

Solution

From the ideal gas equation, PV = nRT also Volume (V) =

{{mass(M)} \over {density(\rho )}}

So, PM = $\rho$ RT P = 249 $\times$ 10 3 N/m 2 M = 2 $\times$ 10 -3 kg T = 333 K $\rho$ =

{{\left( {249 \times {{10}^3}} \right)\left( {2 \times {{10}^{ - 3}}} \right)} \over {8.3 \times 300}}

= 0.2 kg/m 3

Q24

The quantities of heat required to raise the temperature of two solid copper spheres of radii r 1 and r 2 (r 1 = 1.5r 2 ) through 1 K are in the ratio :

A

{9 \over 4}

B

{3 \over 2}

C

{5 \over 3}

D

{27 \over 8}

Correct Answer

Option D

Solution

\Delta Q = ms\Delta T

\Delta Q = {4 \over 3}\rho s\Delta T

{{\Delta {Q_1}} \over {\Delta {Q_2}}} = {{r_1^3} \over {r_2^3}} = {1.5^3} = {{27} \over 8}

Q25

Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire systems is thermally insulated. The stop cock is suddenly opened. The Process is :

A adiabatic

B isochoric

C isobaric

D isothermal

Correct Answer

Option A

Solution

The system is insulated thermally. So no heat exchange takes place. Therefore the process is adiabatic.

Q26

In which of the following processes, heat is neither absorbed nor released by a system?

A adiabatic

B isobaric

C isochoric

D isothermal

Correct Answer

Option A

Solution

From the characteristic property of adiabatic process, there is no exchange in heat.

Q27

Increase in tempertaure of a gas filled in a container would lead to :

A increase in its kinetic energy

B decrease in intermolecular distance

C decrease in its pressure

D increase in its mass

Correct Answer

Option A

Solution

Since, the increase of temperature will increase the kinetic energy for the given gas. For the given ideal gas, U =

{F \over 2}nRT

Q28

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

A

{2 \over 5}

B

{2 \over 3}

C

{1 \over 3}

D

{2 \over 7}

Correct Answer

Option A

Solution

Gas is monatomic, so C p =

{5 \over 2}R

Given process is isobaric. $\therefore$ dQ = nC p dT or dQ = n

\left( {{5 \over 2}R} \right)

dT Also, dW = PdV = nRdT ( $\because$ PV = nRT) $\therefore$ Required ratio =

{{dW} \over {dQ}}

=

{{nRdT} \over {n\left( {{5 \over 2}R} \right)dT}}

=

{2 \over 5}

Q29

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 10 5 N m –2 ) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

A 104.3 J

B 208.7 J

C 42.2 J

D 84.5 J

Correct Answer

Option B

Solution

Using first law of thermodynamics,

\Delta

Q =

\Delta

U +

\Delta

W $\Rightarrow$ 54 × 4.18 =

\Delta

U + 1.013 × 10 5 (167.1 × 10 –6 – 0) $\Rightarrow$

\Delta

U = 208.7 J

Q30

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Given : Mass of oxygen molecule (m) = 2.76 × 10 –26 kg, Boltzmann’s constant k B = 1.38 × 10 –23 J K –1 )

A 2.508 × 10 4 K

B 8.360 × 10 4 K

C 5.016 × 10 4 K

D 1.254 × 10 4 K

Correct Answer

Option B

Solution

Escape velocity from the Earth’s surface is vescape = 11200 m s –1 Let at temperature T, rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere.

Also, V rms = V escape =

\sqrt {{{3{k_B}T} \over {{m_{{O_2}}}}}}

$\Rightarrow$ 11200 =

\sqrt {{{3 \times 1.38 \times {{10}^{ - 23}} \times T} \over {2.76 \times {{10}^{ - 26}}}}}

$\Rightarrow$ T = 8.360 × 10 4 K