Heat and Thermodynamics — Page 8 | JEE Physics

Q71

CP and Cv are specific heats at constant pressure and constant volume respectively. It is observed that CP – Cv = a for hydrogen gas CP – Cv = b for nitrogen gas The correct relation between a and b is

A a = 28 b

B a = 1/14 b

C a = b

D a = 14 b

Correct Answer

Option D

Solution

As we know, for 1 g mole of a gas, Cp – Cv = R where Cp and Cv are molar specific heat capacities.

So, when n gram moles are given, Cp – Cv =

{R \over n}

For hydrogen (n = 2), Cp – Cv =

{R \over 2}

=

a

For nitrogen (n = 28), Cp – Cv =

{R \over 28}

=

b

$\therefore$

{a \over b} = 14

$\Rightarrow$

a

= 14

b

Q72

An external pressure P is applied on a cube at 0oC so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and

\alpha

is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

A

{P \over {3\alpha K}}

B

{P \over {\alpha K}}

C

{3 \alpha \over {P K}}

D 3PK

\alpha

Correct Answer

Option A

Solution

As we know, Bulk modulus K =

{{\Delta P} \over {\left( {{{ - \Delta V} \over V}} \right)}}

$\Rightarrow$

{{{\Delta V} \over V} = {P \over K}}

V = V0(1 + $\gamma$

\Delta

t)

{{{\Delta V} \over {{V_0}}} = \gamma \Delta t}

$\therefore$

{{P \over K} = \gamma \Delta t}

$\Rightarrow$

{\Delta t = {P \over {\gamma K}}}

=

{{P \over {3\alpha K}}}

Q73

The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1

\times

105 Pa. If Ni and Nf are the number of molecules in the room before and after heating, then Nf – Ni will be :

A - 1.61

\times

1023

B 1.38

\times

1023

C 2.5

\times

1025

D - 2.5

\times

1025

Correct Answer

Option D

Solution

Given: Initial temperature Ti = 17 + 273 = 290 K Final temperature Tf = 27 + 273 = 300 K Atmospheric pressure, P0 = 1 × 105 Pa Volume of room, V0 = 30 m3 Difference in number of molecules, Nf – Ni = ?

We know PV = nRT =

{N \over {{N_A}}}

RT The number of molecules N =

{{PV{N_A}} \over {RT}}

$\therefore$ Nf – Ni =

{{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)

=

{{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)

= – 2.5 $\times$ 1025

Q74

In an experiment, a sphere of aluminium of mass 0.20 kg is heated upto 150oC. Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40oC. The specific heat of aluminium is : (take 4.2 Joule = 1 calorie)

A 378 J/kg

-

oC

B 315 J/kg

-

oC

C 476 J/kg

-

oC

D 434 J/kg

-

oC

Correct Answer

Option D

Solution

Let specific heat of aluminium = S, As we know from principle of calorimetry, Qgiven = Qused

\therefore\,\,\,

0.2 $\times$ S $\times$ (150 $-$ 40) = 150 $\times$ 1 $\times$ (40 $-$ 27) + 25 $\times$ (40 $-$ 27) $\Rightarrow$

\,\,\,

0.2 $\times$ S $\times$ 110 = 150 $\times$ 13 + 25 $\times$ 13 $\Rightarrow$

\,\,\,

S = 434 J/kg - oC

Q75

A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300K). The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm, and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule ?

A ~1013

B ~1012

C ~1011

D ~1010

Correct Answer

Option B

Solution

V = 25 × 10–3 m3, N = 1 mole of O2 T = 300 K Vrms = 200 m/s

\because \lambda = {1 \over {\sqrt 2 N\pi {r^2}}}

Average time

{1 \over \tau } = {{ < V > } \over \lambda } = 200\,.N\pi {r^2}.\sqrt 2

= {{\sqrt 2 \times 200 \times 6.023 \times {{10}^{23}}} \over {25 \times {{10}^{ - 3}}}}.\pi \times {10^{ - 18}} \times 0.09

Average no. of collision

\approx {10^{10}}

Q76

N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?

A

{1 \over 2}

nRT

B 0

C

{3 \over 2}

nRT

D

{5 \over 2}

nRT

Correct Answer

Option A

Solution

Initial kinetic energy of N mole of diatomic gas, Ki = N

{5 \over 2}

RT Kinetic energy of n mole of monoatomic gas = n

{3 \over 2}

RT When n mole of diatomic gas converted into monoatomic gas then remaining diatomic gas = (N $-$ n) Find kinetic energy, KF = (2m)

{3 \over 2}

RT + (N $-$ n)

{5 \over 2}

RT =

{1 \over 2}

nRT +

{5 \over 2}

NRT

\therefore\,\,\,

Change in kinetic energy,

\Delta

K = Kf $-$ Ki =

{1 \over 2}

nRT

Q77

A Carnot's engine works as a refrigerator between

250

K and

300

K. It receives

500

cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

A

420

J

B

772

J

C

2100

J

D

2520

J

Correct Answer

Option A

Solution

Given, Cold body temperature (T2) = 250 K Hot body temperature (T1) = 300 K Received heat (Q2) = 500 cal Let, required works done = $\omega$

\therefore\,\,\,\,

For the refrigerator, Efficiency =

1 - {{T2} \over {T1}} = {\omega \over {{Q_2} + \omega }}

$\Rightarrow$

\,\,\,\,

1 $-$

{{250} \over {300}} = {\omega \over {{Q_2} + \omega }}

$\Rightarrow$

\,\,\,\,

{{{Q_2} + \omega } \over \omega }

=

{{300} \over {50}}

= 6 $\Rightarrow$

\,\,\,\,

\omega = {{{Q_2}} \over 5}

$\Rightarrow$

\,\,\,\,

\omega = {{500 \times 4.2} \over 5}

$\Rightarrow$

\,\,\,\,

\omega = 420\,J

Q78

Two moles of an ideal monatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

A (a) 195 K (b) 2.7 kJ

B (a) 189 K (b) 2.7 kJ

C (a) 195 K (b) –2.7 kJ

D (a) 189 K (b) – 2.7 kJ

Correct Answer

Option D

Solution

For adiabatic process, pv $\gamma$ = constant. and we know, pv = nRT

\therefore\,\,\,

p =

{{nRT} \over v}

\therefore\,\,\,

{{nRT} \over v} \times {v^\gamma }

= constant $\Rightarrow$

\,\,\,

T v $\gamma$ $-$ 1 = constant.

\therefore\,\,\,

T1 v1 $\gamma$ $-$ 1 = T2 v2 $\gamma$ $-$ 1 Given that, This is a monoatomic gas. So, degree of frequency f = 3

\therefore\,\,\,

$\gamma$ =

{{{c_p}} \over {{c_v}}}

= 1 +

{2 \over f}

= 1 +

{2 \over 3}

=

{5 \over 3}

T1 = 27 + 273 = 300 K v1 = v and v2 = 2v

\therefore\,\,\,

T2 (2V)

^{{2 \over 3}}

= 300(v)

^{{2 \over 3}}

$\Rightarrow$

\,\,\,

T2 =

{{300} \over {{2^{{2 \over 3}}}}}

= 189 K Change in internal energy,

\Delta

U =

{1 \over 2}

nfR

\Delta

T =

{1 \over 2}

$\times$ 2 $\times$ 3 $\times$ 8.31 $\times$ (189 $-$ 300) = $-$ 2.7 KJ

Q79

The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in ms-1 is : [kB =1.4

\times

10-23 J/K; mHe = 7

\times

10 -27 kg ]

A 1.3

\times

104

B 1.3

\times

103

C 1.3

\times

105

D 1.3

\times

102

Correct Answer

Option B

Solution

We know that

{v_{rms}} = \sqrt {{{3{k_B}T} \over m}}

. Given,

{k_B} = 1.4 \times {10^{ - 23}}

J/K; T = 300 K; m = 7 $\times$ 10 $-$ 27 kg. Therefore,

{v_{rms}} = \sqrt {{{3 \times 1.4 \times {{10}^{ - 23}} \times 300} \over {7 \times {{10}^{ - 27}}}}}

= \sqrt {{{3 \times 300 \times 14 \times {{10}^{ - 23}}} \over {7 \times 10 \times {{10}^{ - 27}}}}}

= \sqrt {{{{3^2} \times {{10}^2} \times 2 \times {{10}^4}} \over {10}}} = 3 \times 10 \times {10^2} \times \sqrt {{2 \over {10}}}

= 3 \times \sqrt {10} \times {10^2} \times \sqrt 2 = 3 \times \sqrt 2 \times \sqrt 5 \times {10^2} \times \sqrt 2

= 3 \times 2 \times \sqrt 5 \times {10^2}

{v_{rms}} = 6 \times {10^2} \times \sqrt 5 = 13.41 \times {10^2}

or

{v_{rms}} = 1.34 \times {10^3}

Q80

Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two energies A and B are represented by

{\eta _A}

and

{\eta _B},

respectively, then what is the value of

{{{\eta _B}} \over {{\eta _A}}}

?

A

{{12} \over 7}

B

{{7} \over 12}

C

{{12} \over 5}

D

{{5} \over 12}

Correct Answer

Option A

Solution

Given that two Carnot engines A and B are operated in series. Efficiency of a Carnot engine

\eta = 1 - {{{T_2}} \over {{T_1}}}

; where T1 is temperature of source and T2 is temperature of sink. Therefore,

{\eta _A} = 1 - {{{T_{2A}}} \over {{T_{1A}}}} = 1 - {T \over {600}}

; T is heat rejected by A and

{\eta _B} = 1 - {{{T_{2B}}} \over {{T_{1B}}}} = 1 - {{100} \over T}

; T is heat rejected by A and received by B Now,

{{{\eta _B}} \over {{\eta _A}}}{{1 - {{100} \over T}} \over {1 - {T \over {600}}}} = \left( {{{T - 100} \over T}} \right) \times {{600} \over {600 - T}}

Let T = 350 K, therefore,

{{{\eta _B}} \over {{\eta _A}}} = {{350 - 100} \over {350}} \times {{600} \over {600 - 350}} = {{250} \over {350}} \times {{600} \over {25}} = {{600} \over {350}}

\Rightarrow {{{\eta _B}} \over {{\eta _A}}} = {{12} \over 7}