Heat and Thermodynamics — Page 15 | JEE Physics

Q141

A heat engine has an efficiency of

{1 \over 6}

. When the temperature of sink is reduced by 62

^\circ

C, its efficiency get doubled. The temperature of the source is :

A 124

^\circ

C

B 37

^\circ

C

C 62

^\circ

C

D 99

^\circ

C

Correct Answer

Option D

Solution

\eta = 1 - {{{T_L}} \over {{T_H}}}

..... (i)

2\eta = 1 - {{({T_L} - 62)} \over {{T_H}}} = 1 - {{{T_L}} \over {{T_H}}} + {{62} \over {{T_H}}}

\Rightarrow \eta = {{62} \over {{T_H}}} \Rightarrow {1 \over 6} = {{62} \over {{T_H}}} \Rightarrow {T_H} = 6 \times 62 = 372

K In

^\circ

C $\Rightarrow$ 372 $-$ 273 = 99

^\circ

C

Q142

The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2

\times

10

-

9 J per molecules is :

A 0.75

\times

1011

B 3

\times

1011

C 1.5

\times

1011

D 6

\times

1011

Correct Answer

Option C

Solution

KE =

{3 \over 2}kT

PV =

{N \over {{N_A}}}RT

N =

{{PV} \over {kT}}

= N = 1.5 $\times$ 1011

Q143

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27

^\circ

C to 37

^\circ

C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol

-

1 k

-

1]

A work done by the gas is close to 332 J

B work done on the gas is close to 582 J

C work done by the gas is close to 582 J

D work done on the gas is close to 332 J

Correct Answer

Option B

Solution

Since, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3 (rot.) + 4 $\times$ 2 (vib.) = 14 &

\gamma = 1 + {2 \over f}

\gamma = 1 + {2 \over {14}} = {8 \over 7}

W = {{nR\Delta T} \over {\gamma - 1}} = - 582

As W < 0. work is done on the gas.

Q144

Two Carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T3. When workdone in both the cases is equal, to value of T is :

A

{2 \over 3}{T_1} + {3 \over 2}{T_3}

B

{1 \over 3}{T_1} + {2 \over 3}{T_3}

C

{3 \over 2}{T_1} + {1 \over 3}{T_3}

D

{2 \over 3}{T_1} + {1 \over 3}{T_3}

Correct Answer

Option D

Solution

{W_A} = 1 - {{{Q_2}} \over {{Q_1}}} = 1 - {T \over {{T_1}}} \Rightarrow {{{Q_2}} \over {{Q_1}}} = {T \over {{T_1}}}

{W_B} = 1 - {{{Q_3}} \over {({Q_2}/2)}} = 1 - {{{T_3}} \over T} \Rightarrow {{2{Q_3}} \over {{Q_2}}} = {{{T_3}} \over T}

Now, WA = WB

{Q_1} - {Q_2} = {{{Q_2}} \over 2} - {Q_3}

\Rightarrow {{2{Q_1}} \over {{Q_2}}} + {{2{Q_3}} \over {{Q_2}}} = 3

\Rightarrow {{2{T_1}} \over T} + {{{T_3}} \over T} = 3

$\Rightarrow$

{{2{T_1}} \over 3} + {{{T_3}} \over 3} = T

Q145

The temperature of equal masses of three different liquids x, y and z are 10

^\circ

C, 20

^\circ

C and 30

^\circ

C respectively. The temperature of mixture when x is mixed with y is 16

^\circ

C and that when y is mixed with z is 26

^\circ

C. The temperature of mixture when x and z are mixed will be :

A 28.32

^\circ

C

B 25.62

^\circ

C

C 23.84

^\circ

C

D 20.28

^\circ

C

Correct Answer

Option C

Solution

when x and y are mixed, Tf1 = 16

^\circ

C m1s1T + m2s2T2 = (m1s1 + m2s2)Tf1 s1 $\times$ 10 + s2 $\times$ 20 = (s1 + s2) $\times$ 16 s1 =

{2 \over 3}

s2 .... (i) when y and z are mixed, Tf2 = 26

^\circ

C m2s2T + m3s3T3 = (m3s3 + m3s3)Tf2 s2 $\times$ 20 + s3 $\times$ 30 = (s2 + s3) $\times$ 26 s3 =

{3 \over 2}

s2 ..... (ii) when x and z are mixed m1s1T1 + m3s3T3 = (m1s1 + m3s3)Tf

{2 \over 3}

s2 $\times$ 10 +

{2 \over 3}

s2 $\times$ 20 =

\left( {{2 \over 3}{s_2} + {3 \over 2}{s_2}} \right){T_f}

Tf = 23.84

^\circ

C

Q146

A cylindrical container of volume 4.0

\times

10

-

3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is : [Take gas constant as 8.3 J mol

-

1 K

-

1]

A 249

\times

101 Pa

B 24.9

\times

103 Pa

C 24.9

\times

105 Pa

D 24.9 Pa

Correct Answer

Option C

Solution

V = 4 $\times$ 10 $-$ 3 m3 n = 3 moles T = 400 K PV = nRT $\Rightarrow$ P =

{{nRT} \over V}

P =

{{3 \times 8.3 \times 400} \over {4 \times {{10}^{ - 3}}}}

= 24.9 $\times$ 105 Pa

Q147

A refrigerator consumes an average 35W power to operate between temperature

-

10

^\circ

C to 25

^\circ

C. If there is no loss of energy then how much average heat per second does it transfer?

A 263 J/s

B 298 J/s

C 350 J/s

D 35 J/s

Correct Answer

Option A

Solution

{{{T_L}} \over {{T_H} - {T_L}}} = C.O.P. = {{{{dH} \over {dt}}} \over {{{dW} \over {dt}}}}

{{263} \over {35}} \times 35 = {{dH} \over {dt}}

{{dH} \over {dt}} = 263

watts

Q148

A balloon carries a total load of 185 kg at normal pressure and temperature of 27

^\circ

C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is

-

7

^\circ

C. Assuming the volume constant?

A 181.46 kg

B 214.15 kg

C 219.07 kg

D 123.54 kg

Correct Answer

Option D

Solution

Pm = $\rho$ RT $\therefore$

{{{P_1}} \over {{P_2}}} = {{{\rho _1}{T_1}} \over {{\rho _1}{T_2}}}

{{{\rho _1}} \over {{\rho _2}}} \Rightarrow {{{P_1}{T_2}} \over {{P_2}{T_1}}} = \left( {{{76} \over {45}}} \right) \times {{266} \over {300}}

{{{\rho _1}} \over {{\rho _2}}} \Rightarrow {{{M_1}} \over {{M_2}}} = {{76 \times 266} \over {45 \times 300}}

$\therefore$

{M_2} \Rightarrow {{45 \times 300 \times 185} \over {76 \times 266}} = 123.54

kg

Q149

An ideal gas is expanding such that PT3 = constant. The coefficient of volume expansion of the gas is :

A

{1 \over T}

B

{2 \over T}

C

{4 \over T}

D

{3 \over T}

Correct Answer

Option C

Solution

PT3 = constant

\left( {{{nRT} \over v}} \right)

T3 = constant T4 V $-$ 1 = constant T4 = kV

\Rightarrow 4{{\Delta T} \over T} = {{\Delta V} \over V}

....... (1)

\Delta

V = V $\gamma$

\Delta

T ........ (2) comparing (1) and (2), we get

\gamma = {4 \over T}

Q150

if the rms speed of oxygen molecules at 0

^\circ

C is 160 m/s, find the rms speed of hydrogen molecules at 0

^\circ

C.

A 640 m/s

B 40 m/s

C 80 m/s

D 332 m/s

Correct Answer

Option A

Solution

{V_{rms}} = \sqrt {{{3KT} \over M}}

{{{{({V_{rms}})}_{{O_2}}}} \over {{{({V_{rms}})}_{{H_2}}}}} = \sqrt {{{{M_{{H_2}}}} \over {{M_{{O_2}}}}}} = \sqrt {{2 \over {32}}}

{({V_{rms}})_{{H_2}}} = 4 \times {({V_{rms}})_{{O_2}}}

= 4 \times 160

= 640

m/s