Heat and Thermodynamics — Page 20 | JEE Physics

Q191

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A: Efficiency of a reversible heat engine will be highest at

-273^{\circ} \mathrm{C}

temperature of cold reservoir. Reason R: The efficiency of Carnot's engine depends not only on the temperature of the cold reservoir but it depends on the temperature of the hot reservoir too and is given as

\eta=\left(1-\dfrac{T_{2}}{T_{1}}\right)

In the light of the above statements, choose the correct answer from the options given below

A Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

C A is false but

\mathbf{R}

is true

D A is true but

\mathbf{R}

is false

Correct Answer

Option A

Solution

\eta = 1 - {{{T_{\mathrm{cold}}}} \over {{T_{\mathrm{hot}}}}}

{T_{\mathrm{cold}}} = \mathrm{OK} \Rightarrow \eta = 1

(max) A is correct. R is also correct and explains A.

Q192

The pressure

(\mathrm{P})

and temperature (

\mathrm{T})

relationship of an ideal gas obeys the equation

\mathrm{PT}^{2}=

constant. The volume expansion coefficient of the gas will be :

A

3 T^{2}

B

\dfrac{3}{T^2}

C

\dfrac{3}{T^3}

D

\dfrac{3}{T}

Correct Answer

Option D

Solution

PT^2

= constant From

PV = nRT \Rightarrow {{{T^3}} \over V} =

constant

{T^3} \propto V

..... (1)

3{T^2}dT \propto dV

..... (2) From (1) and (2)

{{3dT} \over T} = {{dV} \over V}

$\therefore$

\gamma = {1 \over V}{{dV} \over {dT}} = {3 \over T}

Q193

Heat is given to an ideal gas in an isothermal process. A. Internal energy of the gas will decrease. B. Internal energy of the gas will increase. C. Internal energy of the gas will not change. D. The gas will do positive work. E. The gas will do negative work. Choose the correct answer from the options given below :

A B and D only

B C and E only

C A and E only

D C and D only

Correct Answer

Option D

Solution

Isothermal process

\Delta T=0

\Delta U=\frac{f}{2}nR\Delta T

\Delta U=0

No change in internal energy

\Delta Q=\Delta W

(1

^{st}

law)

\Delta Q=+\mathrm{ve}

\Delta W=+\mathrm{ve}

Q194

Heat energy of 184 kJ is given to ice of mass 600 g at

-12^\circ \mathrm{C}

. Specific heat of ice is

\mathrm{2222.3~J~kg^{-1^\circ}~C^{-1}}

and latent heat of ice in 336

\mathrm{kJ/kg^{-1}}

A. Final temperature of system will be 0

^\circ

C. B. Final temperature of the system will be greater than 0

^\circ

C. C. The final system will have a mixture of ice and water in the ratio of 5 : 1. D. The final system will have a mixture of ice and water in the ratio of 1 : 5. E. The final system will have water only. Choose the correct answer from the options given below :

A A and E only

B B and D only

C A and C only

D A and D only

Correct Answer

Option D

Solution

Heat required to raise the temperature of ice to 0

^\circ

C is

= {{60} \over {1000}}(2222.3)(12)

= 16000.5

J

\approx 16

kJ Heat required to melt ice completely

= \left( {{{600} \over {1000}}} \right)(336)

kJ

= 201.6

kJ Energy left

= (184 - 16) = 168

kJ $\therefore$ Partial ice will melt $\therefore$

168 = ({m_{ice\,melted}})336

0.5

kg

= ({m_{ice\,melted}})

$\therefore$

{m_{ice}}:{m_{water}} = 1:5

Q195

At 300 K, the rms speed of oxygen molecules is

\sqrt {{{\alpha + 5} \over \alpha }}

times to that of its average speed in the gas. Then, the value of

\alpha

will be (used

\pi = {{22} \over 7}

)

A 27

B 28

C 24

D 32

Correct Answer

Option B

Solution

{v_{rms}} = \sqrt {{{\alpha + 5} \over \alpha }} {v_{avg}}

\sqrt {{{3RT} \over m}} = \sqrt {{{\alpha + 5} \over 5}} \sqrt {{{8RT} \over {\pi m}}}

{{3 \times \pi } \over 8} = {{\alpha + 5} \over \alpha }

{{33} \over {28}} = {{\alpha + 5} \over \alpha }

\alpha = 28

Q196

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If

d Q

and

d W

represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics

d Q=d U-d W

. Reason R: First law of thermodynamics is based on law of conservation of energy. In the light of the above statements, choose the correct answer from the options given below:

A Both A and R are correct but R is not the correct explanation of A

B Both A and R are correct and R is the correct explanation of A

C A is correct but R is not correct

D A is not correct but R is correct

Correct Answer

Option B

Solution

Heat supplied to the system $=d Q$ Work done on the system $=-d W$ By the first law of thermodynamics, $d Q=d U+d W$ We done the work, therefore,

d Q=d U+(-d W) = d U-d W

So, both A and R are correct and R is the correct explanation of A.

Q197

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

A

\dfrac{9}{2}R

B

\dfrac{5}{2}R

C

\dfrac{3}{2}R

D

\dfrac{7}{2}R

Correct Answer

Option D

Solution

Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom \& it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$

\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}{2}=\frac{7 \mathrm{R}}{2}

Q198

The root mean square velocity of molecules of gas is

A Proportional to temperature (

T

)

B Inversely proportional to square root of temperature

\left( {\sqrt {{1 \over T}} } \right)

C Proportional to square of temperature (

T^2

)

D Proportional to square root of temperature (

\sqrt T

)

Correct Answer

Option D

Solution

The rms speed of a gas molecule is

\mathrm{V}_{\mathrm{RMS}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}

$\mathrm{V}_{\mathrm{RMS}} \propto \sqrt{\mathrm{T}}$

Q199

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :

A 300 K

B 1000 K

C 900 K

D 360 K

Correct Answer

Option B

Solution

$\eta=1-\dfrac{T_{\text{sink }}}{T_{\text{source }}}$ $50 \%$ efficiency $\Rightarrow \dfrac{1}{2}=1-\dfrac{T_{\text{sink }}}{T_{\text{source }}}$ $\dfrac{1}{2}=1-\dfrac{T_{\text{sink }}}{600} \Rightarrow T_{\text{sink }}=300$ Now, $70 \%$ efficiency $\Rightarrow \dfrac{7}{10}=1-\dfrac{T_{\text{sink }}}{T_{\text{source }}}$ $\dfrac{300}{T_{\text{source }}}=\dfrac{3}{10}$ $T_{\text{source }}=1000 \mathrm{~K}$

Q200

Let

\gamma_1

be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and

\gamma_2

be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio,

\dfrac{\gamma_1}{\gamma_2}

is :

A

\dfrac{35}{27}

B

\dfrac{25}{21}

C

\dfrac{21}{25}

D

\dfrac{27}{35}

Correct Answer

Option B

Solution

For monoatomic gas $\gamma_1=\dfrac{5}{3}$ For diatomic gas at low temperatures

\begin{aligned} & \gamma_2=\frac{7}{5} \\\\ & \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21} \end{aligned}