Heat and Thermodynamics — Page 31 | JEE Physics

Q301

Match with : .tg .tg

List - I		List - II
(A)	3 Translational degrees of freedom	(I)	Monoatomic gases
(B)	3 Translational, 2 rotational degrees of freedoms	(II)	Polyatomic gases
(C)	3 Translational, 2 rotational and 1 vibrational degrees of freedom	(III)	Rigid diatomic gases
(D)	3 Translational, 3 rotational and more than one vibrational degrees of freedom	(IV)	Nonrigid diatomic gases

A (A)-(I), (B)-(III), (C)-(IV), (D)-(II)

B (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

C (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

D (A)-(I), (B)-(IV), (C)-(III), (D)-(II)

Correct Answer

Option A

Solution

Monoatomic gases possess only translational degrees of freedom.

So, option (I) matches with (A).

Polyatomic gases have translational and rotational degrees of freedom.

So, option (II) matches with (D).

Rigid diatomic gases have translational, rotational and vibrational degrees of freedom.

So, option (III) matches with (B).

Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom.

So, option (IV) matches with (C).

Q302

Match the with .tg .tg

List - I		List - II
(A)	Pressure varies inversely with volume of an ideal gas.	(I)	Adiabatic process
(B)	Heat absorbed goes partly to increase internal energy and partly to do work.	(II)	Isochoric process
(C)	Heat is neither absorbed nor released by a system.	(III)	Isothermal process
(D)	No work is done on or by a gas.	(IV)	Isobaric process

A A-I, B-III, C-II, D-IV

B A-I, B-IV, C-II, D-III

C A-III, B-I, C-IV, D-II

D A-III, B-IV, C-I, D-II

Correct Answer

Option D

Solution

Let's analyze each statement step by step: For (A): “Pressure varies inversely with volume of an ideal gas.”

At constant temperature (isothermal process), Boyle's law applies:

PV = \text{constant}.

Thus, pressure and volume have an inverse relationship.

Hence, (A) corresponds to the Isothermal process, which is (III).

For (B): “Heat absorbed goes partly to increase internal energy and partly to do work.”

In a constant pressure (isobaric) process, when heat is supplied, part of it is used to do work (expansion, for example) and the rest increases the internal energy.

Therefore, (B) matches with the Isobaric process, which is (IV).

For (C): “Heat is neither absorbed nor released by a system.”

This implies that the process is adiabatic (no heat exchange).

Thus, (C) matches with the Adiabatic process, which is (I).

For (D): “No work is done on or by a gas.”

Work done by a gas is given by

W = P\Delta V.

If the volume does not change (constant volume), then no work is done.

So, (D) identifies with the Isochoric process, which is (II).

Putting it all together: (A) → (III) Isothermal process (B) → (IV) Isobaric process (C) → (I) Adiabatic process (D) → (II) Isochoric process Thus, the correct answer is: Option D A-III, B-IV, C-I, D-II.

Q303

A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion

\alpha

/oC. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by

\Delta

TK. Young's modulus, Y, for this metal is :

A

{F \over {A\alpha \Delta T}}

B

{F \over {A\alpha (\Delta T - 273)}}

C

{F \over {2A\alpha \Delta T}}

D

{{2F} \over {A\alpha \Delta T}}

Correct Answer

Option A

Solution

We know, Young's Modulus, Y =

{{Stress} \over {Strain}}

Stress =

{F \over A}

Strain =

{{\Delta l} \over l}

$\therefore$ Y =

{{{F \over A}} \over {{{\Delta l} \over l}}}

We also know,

l

f =

l

i (1 + $\alpha$

\Delta

T) $\Rightarrow$

l

f =

l

i +

l

i $\alpha$

\Delta

T $\Rightarrow$

{{{l_f} - {l_i}} \over {{l_i}}}

= $\alpha$

\Delta T

$\Rightarrow$

{{\Delta l} \over {{l_i}}}

= $\alpha$

\Delta

T $\therefore$ y =

{{{F \over A}} \over {\alpha \Delta T}}

=

{F \over {A\alpha \Delta T}}

Q304

Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1 (= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in tum, rejects to a heat reservoir at T3 (=400 K). Calculate the temperature T2 if the work outputs of the two engines are equal :

A 600 K

B 400 K

C 300 K

D 500 K

Correct Answer

Option D

Solution

Here, Q1 = W1 + Q2 and Q2 = W2 + Q3 Given that, W1 = W2 $\therefore$ Q1 $-$ Q2 = Q2 $-$ Q3 $\Rightarrow$ nCv(T1 $-$ T2) = nCv(T2 $-$ T3) $\Rightarrow$ T1 $-$ T2 = T2 $-$ T3 $\Rightarrow$ T2 =

{{{T_1} + {T_3}} \over 2}

=

{{600 + 400} \over 2}

= 500 K

Q305

An unknown metal of mass 192 g heated to a temperature of 100oC was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. (Specific heat of brass is 394 J kg–1 K–1)

A 458 J kg–1 K–1

B 1232 J kg–1 K–1

C 654 J kg–1 K–1

D 916 J kg–1 K–1

Correct Answer

Option D

Solution

192 $\times$ S $\times$ (100 $-$ 21.5) = 128 $\times$ 394 $\times$ (21.5 $-$ 8.4) + 240 $\times$ 4200 $\times$ (21.5 $-$ 8.4) $\Rightarrow$ S = 916

Q306

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is :

A

{5 \over 3}

B

{2 \over 5}

C

{3 \over 5}

D

{2 \over 3}

Correct Answer

Option B

Solution

For adiabatic process : TV $\gamma$ $-$ 1 = constant For diatomic process : $\gamma$ $-$ 1 =

{7 \over 5} - 1

$\therefore$ x =

{2 \over 5}

Q307

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is

\alpha

then the value of S is :

A

\alpha

B

{\alpha \over 2}

C 2

\alpha

D

{1 \over \alpha }

Correct Answer

Option A

Solution

Change of length of wire with temperature,

\Delta

\ell

=

\alpha \ell \Delta \theta

Time period of pendulum at temperature $\theta$ , T $\theta$ = 2 $\pi$

\sqrt {{{\ell + \Delta \ell } \over g}}

= 2 $\pi$

\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}}

=

2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}

=

2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}

\simeq

T0

\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)

Here T0 = time period at temperature 0oC. $\therefore$ Change in time period,

\Delta

T = T $\theta$ $-$ T0 =

{{{T_0}\Delta \ell } \over {2\ell }}

=

{{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}

$\therefore$

{{\Delta T} \over {\Delta \theta }}

=

{{{T_0}\alpha } \over 2}

Given that T0 = 2, $\therefore$

{{\Delta T} \over {\Delta \theta }}

=

{{2\alpha } \over 2}

= $\alpha$

{{\Delta T} \over {\Delta \theta }}

is the shape of

\Delta

T and

{\Delta \theta }

curve = S (given) $\therefore$ S = $\alpha$

Q308

200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :

A 8.4 kJ

B 4.2 kJ

C 16.7 kJ

D 167.4 kJ

Correct Answer

Option C

Solution

According to the first law of thermodynamics, Q =

\Delta

u + w For isochoric process Q =

\Delta

U = ms

\Delta

t

\Delta

T = (233 $-$ 213) = 20 k $\therefore$

\Delta

u = 0.2 $\times$ 4184 $\times$ 20 = 16.7 kJ

Q309

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by

\Delta

T. The net change in its length is zero. Let

\ell

be the length of the rod, A its area of cross-section,Y its Young’s modulus, and

\alpha

its coefficient of linear expansion. Then, F is equal to :

A

\ell

2 Y

\alpha

\Delta

T

B

\ell

A Y

\alpha

\Delta

T

C A Y

\alpha

\Delta

T

D

{{AY} \over {\alpha \,\Delta T}}

Correct Answer

Option C

Solution

Because of thermal expansion, change in length (

\Delta

\ell

) =

\ell

$\alpha$

\Delta

T . . . . .(1) Because of compressive force, the compansion is

\Delta

\ell

' ,

\therefore\,\,\,

Young's Modulus (y) =

{{{F \over A}} \over {{{\Delta \ell} \over \ell }}}

$\Rightarrow$

\,\,\,

F = YA

{{\Delta \ell '} \over \ell }

As net change in length is 0 . So,

\Delta \ell '

=

\Delta \ell

=

\ell \alpha \,\Delta T

\therefore\,\,\,

F = YA $\times$

{{\ell \alpha \,\Delta T} \over \ell }

= AY $\alpha$

\Delta

T

Q310

A steel rail of length 5 m and area of cross section 40cm2 is prevented from expanding along its length while the temperature rises by 10oC. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately :

A 2

\times

107 N

B 1

\times

105 N

C 2

\times

109 N

D 3

\times

10

-

5 N

Correct Answer

Option B

Solution

Young's modulus (Y) =

{{{F \over A}} \over {{{\Delta L} \over L}}}

as

{{{\Delta L} \over L}}

= $\alpha$

\Delta

$\theta$

\therefore\,\,\,

Y =

{{F \over {A\alpha \Delta \theta }}}

$\Rightarrow$

\,\,\,

F = YA $\alpha$

\Delta

$\theta$ = 2 $\times$ 1011 $\times$ 40 $\times$ 10 $-$ 4 $\times$ 1.2 $\times$ 10 $-$ 5 $\times$ 10 = 9.6 $\times$ 104 N

\simeq

1 $\times$ 105 N