Heat and Thermodynamics — Page 18 | JEE Physics

Q171

A certain amount of gas of volume

\mathrm{V}

at

27^{\circ} \mathrm{C}

temperature and pressure

2 \times 10^{7} \mathrm{Nm}^{-2}

expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use

\gamma=1.5)

:

A

3.536 \times 10^{5} \mathrm{~Pa}

B

3.536 \times 10^{6} \mathrm{~Pa}

C

1.25 \times 10^{6} \mathrm{~Pa}

D

1.25 \times 10^{5} \mathrm{~Pa}

Correct Answer

Option B

Solution

Let AB is isothermal process and BC is adiabatic process then for AB process PAVA = PBVB $\Rightarrow$ PB = 107 Nm $-$ 2 For process BC PBV

_B^r

= PC V

_C^r

PC = 3.536 x $\times$ 106 Pa

Q172

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is

\sqrt2

times the speed of sound, then the value of n will be :

A 1

B 2

C 3

D 4

Correct Answer

Option B

Solution

Molar mass

M - {{2 \times 4 + n \times 1} \over {2 + n}}

..... (i) Also,

\gamma = {{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}} \over {{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}} = {{2 \times 5R + n \times 7R} \over {2 \times 3R + n \times 5R}}

\Rightarrow \gamma = {{10 + 7n} \over {6 + 5n}}

...... (ii) Given that

{V_{rms}} = \sqrt 2 \,{V_{sound}}

\Rightarrow \sqrt {{{3RT} \over M}} = \sqrt 2 \sqrt {{{\gamma RT} \over M}}

\Rightarrow \gamma = {3 \over 2}

\Rightarrow n = 2

Q173

Let

\eta_{1}

is the efficiency of an engine at

T_{1}=447^{\circ} \mathrm{C}

and

\mathrm{T}_{2}=147^{\circ} \mathrm{C}

while

\eta_{2}

is the efficiency at

\mathrm{T}_{1}=947^{\circ} \mathrm{C}

and

\mathrm{T}_{2}=47^{\circ} \mathrm{C}

The ratio

\dfrac{\eta_{1}}{\eta_{2}}

will be :

A 0.41

B 0.56

C 0.73

D 0.70

Correct Answer

Option B

Solution

{\eta _1} = 1 - {{420} \over {720}} = {{300} \over {720}}

And

{\eta _2} = 1 - {{320} \over {1220}} = {{900} \over {1220}}

\Rightarrow {{{\eta _1}} \over {{\eta _2}}} = {{300} \over {720}} \times {{1220} \over {900}}

\simeq 0.56

Q174

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of

40 \mathrm{~K}

at constant pressure. The increase in the internal energy of the gas in this process is : (Given

\mathrm{R}=8.3 \,\mathrm{JK}^{-1} \mathrm{~mol}^{-1}

)

A 5810 J

B 3486 J

C 11620 J

D 6972 J

Correct Answer

Option B

Solution

\Delta U = n{C_v}\Delta T

= 7 \times {{3R} \over 2} \times 40

= 3486\,J

Q175

A monoatomic gas at pressure

\mathrm{P}

and volume

\mathrm{V}

is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :

A P

B 8P

C 32P

D 64P

Correct Answer

Option C

Solution

P{V^\gamma }=

constant

\Rightarrow P{V^\gamma } = (P'){\left( {{v \over 8}} \right)^\gamma }

where

\gamma = 5/3

\Rightarrow P' = 32P

Q176

An ice cube of dimensions

60 \mathrm{~cm} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm}

is placed in an insulation box of wall thickness

1 \mathrm{~cm}

. The box keeping the ice cube at

0^{\circ} \mathrm{C}

of temperature is brought to a room of temperature

40^{\circ} \mathrm{C}

. The rate of melting of ice is approximately : (Latent heat of fusion of ice is

3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}

and thermal conducting of insulation wall is

0.05 \,\mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1}

)

A

61 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}

B

61 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}

C

208 \mathrm{~kg} \mathrm{~s}^{-1}

D

30 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}

Correct Answer

Option B

Solution

{{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}

\Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}

\Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}

\simeq 61.1 \times {10^{ - 5}}

kg/s

Q177

A gas has

n

degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :

A

\dfrac{n}{n+2}

B

\dfrac{n+2}{n}

C

\dfrac{n}{2n+2}

D

\dfrac{n}{n-2}

Correct Answer

Option A

Solution

{C_V} = {{nR} \over 2}

And

{C_P} = {{nR} \over 2} + R

\Rightarrow {{{C_V}} \over {{C_P}}} = {{{{nR} \over 2}} \over {{{nR} \over 2} + R}} = {n \over {n + 2}}

Q178

Read the following statements : A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice. B. Two bodies

P

and

Q

having equal surface areas are maintained at temperature

10^{\circ} \mathrm{C}

and

20^{\circ} \mathrm{C}

. The thermal radiation emitted in a given time by

\mathrm{P}

and

\mathrm{Q}

are in the ratio

1: 1.15

. C. A Carnot Engine working between

100 \mathrm{~K}

and

400 \mathrm{~K}

has an efficiency of

75 \%

. D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice. Choose the correct answer from the options given below :

A A, B, C only

B A, B only

C A, C only

D B, C, D only

Correct Answer

Option A

Solution

From Newton's cooling law

{{dQ} \over {dt}} = - k(T - {T_s})

the statement A is correct. For B

U = \sigma eA{T^4}

So,

{{{U_1}} \over {{U_2}}} = {\left( {{{283} \over {293}}} \right)^4} \simeq {1 \over {1.15}}

Statement B is correct For C

\eta = 1 - {{{T_1}} \over {{T_2}}} = 1 - {{100} \over {400}} = {3 \over 4}

So, efficiency is 75% C is correct For D From Newton's law of cooling

{{dQ} \over {dt}} = - k(T - {T_s})

The statement is wrong.

Q179

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is

1: 4

, then A. The r.m.s. velocity of gas molecules in two vessels will be the same. B. The ratio of pressure in these vessels will be

1: 4

. C. The ratio of pressure will be

1: 1

. D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of

1: 4

. Choose the correct answer from the options given below :

A A and C only

B B and D only

C A and B only

D C and D only

Correct Answer

Option C

Solution

{v_{rms}} = \sqrt {{{3RT} \over {{M_0}}}}

because T is same vrms will be same so, A is correct D is incorrect

{{{P_1}} \over {{P_2}}} = {{{n_1}R{T_1}/{V_1}} \over {{n_2}R{T_2}/{V_2}}} = {{{n_1}} \over {{n_2}}} = {1 \over 4}

B is correct C is incorrect

Q180

Which statements are correct about degrees of freedom ? (A) A molecule with n degrees of freedom has n

^{2}

different ways of storing energy. (B) Each degree of freedom is associated with

\dfrac{1}{2}

RT average energy per mole. (C) A monatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom. (D)

\mathrm{CH}_{4}

has a total of 6 degrees of freedom. Choose the correct answer from the options given below :

A (B) and (C) only

B (B) and (D) only

C (A) and (B) only

D (C) and (D) only

Correct Answer

Option B

Solution

Statement A is incorrect, statement B is correct by equipartition of energy.

Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH4 is a polyatomic gas so it has 6 degree of freedom.

So only B and D are correct.