Heat and Thermodynamics Questions — JEE Physics

Q1

Identify the characteristics of an adiabatic process in a monoatomic gas. (A) Internal energy is constant. (B) Work done in the process is equal to the change in internal energy. (C) The product of temperature and volume is a constant. (D) The product of pressure and volume is a constant. (E) The work done to change the temperature from

\mathrm{T}_1

to

\mathrm{T}_2

is proportional to

\left(\mathrm{T}_2-\mathrm{T}_1\right)

. Choose the correct answer from the options given below :

A (B), (D) only

B (B), (E) only

C (A), (C), (E) only

D (A), (C), (D) only

Correct Answer

Option B

Solution

In an adiabatic process, the key characteristics for a monoatomic gas are: No Heat Exchange: The adiabatic process occurs without heat transfer, meaning $Q = 0$ .

Work Done and Internal Energy Change: The work done on or by the system is equal to the change in internal energy.

Mathematically, this is expressed as: $\Delta U = -W$ where $\Delta U$ is the change in internal energy and $W$ is the work done.

Work and Temperature Relationship: The work done when changing the temperature from $T_1$ to $T_2$ is proportional to the difference between these temperatures.

This relationship can be expressed as: $|\text{Work Done}| = nC_v \Delta T \propto (T_2 - T_1)$ Here, $n$ is the number of moles, $C_v$ is the molar heat capacity at constant volume, and $\Delta T$ is the change in temperature.

From these characteristics, the correct features of an adiabatic process in a monoatomic gas relate to the relationships involving work done and temperature change, specifically options (B) and (E).

Q2

Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '

\alpha

'. The metal sheet is heated uniformly, by a small temperature

\Delta

T, so that its new temperature is T +

\Delta

T. Calculate the increase in the volume of the metal box.

A 3a3

\alpha

\Delta

T

B 4

\pi

a3

\alpha

\Delta

T

C

{{4 \over 3}}

\pi

a3

\alpha

\Delta

T

D 4a3

\alpha

\Delta

T

Correct Answer

Option A

Solution

We know that,

\gamma = 3\alpha

.... (i) where, $\alpha$ is the coefficient of linear expansion and $\gamma$ is the coefficient of volume expansion.

We know that,

{{\Delta V} \over V} = \gamma \Delta T

\Rightarrow {{\Delta V} \over V} = 3\alpha \Delta T

[from Eq. (i)]

\Delta V = 3{a^3}\alpha \Delta T

[ $\because$ volume of cube = a3 ]

Q3

If

{C_p}

and

{C_v}

denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

A

{C_p} - {C_v} = 28R

B

{C_p} - {C_v} = R/28

C

{C_p} - {C_v} = R/14

D

{C_p} - {C_v} = R

Correct Answer

Option B

Solution

According to Mayer's relationship

{C_p} - {C_v} = R

$\therefore$

{{{C_p}} \over M} - {{{C_v}} \over M} = {R \over M}

\,\,\,\,\,\,

Here

M=28.

Q4

Two different wires having lengths L1 and L2, and respective temperature coefficient of linear expansion

\alpha

1 and

\alpha

2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

A

2\sqrt {{\alpha _1}{\alpha _2}}

B

4{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}{{{L_2}{L_1}} \over {{{\left( {{L_2} + {L_1}} \right)}^2}}}

C

{{{\alpha _1} + {\alpha _2}} \over 2}

D

{{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}

Correct Answer

Option D

Solution

L'1 = L1(1 + $\alpha$ 1

\Delta

T) L'2 = L2(1 + $\alpha$ 2

\Delta

T) L'eq = (L1 + L2) (1 + $\alpha$ avg

\Delta

T) $\therefore$ (L1 + L2) (1 + $\alpha$ avg

\Delta

T) = L1(1 + $\alpha$ 1

\Delta

T) + L2(1 + $\alpha$ 2

\Delta

T) $\Rightarrow$ (L1 + L2) $\alpha$ avg = L1 $\alpha$ 1 + L2 $\alpha$ 2 $\Rightarrow$ $\alpha$ avg =

{{{\alpha _1}{L_1} + {\alpha _2}{L_2}} \over {{L_1} + {L_2}}}

Q5

Assuming the Sun to be a spherical body of radius

R

at a temperature of

TK

, evaluate the total radiant powered incident of Earth at a distance

r

from the Sun Where r0 is the radius of the Earth and

\sigma

is Stefan's constant.

A

4\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}

B

\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}

C

r_0^2{R^2}\sigma {{{T^4}} \over {4\pi {r^2}}}

D

{R^2}\sigma {{{T^4}} \over {{r^2}}}

Correct Answer

Option B

Solution

Total power radiated by Sun

= \sigma {T^4} \times 4\pi {R^2}

The intensity of power at earth's surface

= {{\sigma {T^4} \times 4\pi {R^2}} \over {4\pi {r^2}}}

Total power received by Earth

= {{\sigma {T^4}{R^2}} \over {{r^2}}}\left( {\pi r_0^2} \right)

Q6

An ideal gas occupies a volume of 2m3 at a pressure of 3

\times

106 Pa. The energy of the gas is :

A 6

\times

104 J

B 9

\times

106 J

C 3

\times

102 J

D 108 J

Correct Answer

Option B

Solution

Energy =

{1 \over 2}

nRT =

{f \over 2}

PV =

{f \over 2}

(3 $\times$ 106) (2) = f $\times$ 3 $\times$ 106 Considering gas is monoatomic i.e. f = 3 E. = 9 $\times$ 106 J

Q7

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27

^\circ

C. The ratio of their average kinetic energy per molecule respectively will be :

A 3 : 2

B 9 : 4

C 2 : 3

D 1 : 1

Correct Answer

Option D

Solution

K{E_{avg}} = {3 \over 2}kT

(At lower temperature) As temperature is same for both the gases.

$\Rightarrow$ Both gases will have same average kinetic energy.

\Rightarrow {{{{(K{E_{avg}})}_{\arg on}}} \over {{{(K{E_{avg}})}_{oxygen}}}} = {1 \over 1}

Q8

An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (Cp ) and at constant volume (Cv) is :

A 6

B

{7 \over 2}

C

{5 \over 2}

D

{7 \over 5}

Correct Answer

Option D

Solution

For ideal gas molecule with 5 degree of freedom, Cv =

{5 \over 2}

R and Cp =

{7 \over 2}

R

\therefore\,\,\,

{{{C_p}} \over {{C_v}}}

=

{{{7 \over 2}R} \over {{5 \over 2}R}}

=

{7 \over 5}

Q9

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature,

{27^ \circ }C.

The work done on the gas will be :

A

300

R

B

300

R

ln

6

C

300

R

ln

2

D

300

R

ln

7

Correct Answer

Option C

Solution

We know, Work done on gas = nRT

\ell

n

\left( {{{{P_f}} \over {{P_i}}}} \right)

Given Pf = 2Pi T = 27 + 273 = 300 K and for monoatomic gas, n = 1.

\therefore\,\,\,\,

Work done = 1 $\times$ R $\times$ 300

\ell

n(2) = 300 R

\ell

n(2)

Q10

For a given gas at 1 atm pressure, rms speed of the molecule is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be :

A 100 m/s

B 100

\sqrt 5

m/s

C 80

\sqrt 5

m/s

D 80 m/s

Correct Answer

Option B

Solution

{V_{rms}} = \sqrt {{{3RT} \over {{M_w}}}}

\Rightarrow {V_{rms}} \propto \sqrt T

Now,

{v \over {200}} = \sqrt {{{500} \over {400}}}

\Rightarrow {v \over {200}} = {{\sqrt 5 } \over 2}

\Rightarrow v = 100\sqrt 5

m/s