Heat and Thermodynamics — Page 5 | JEE Physics

Q41

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will

A increase

B decrease

C remain same

D decrease for some, while increase for others

Correct Answer

Option C

Solution

Since pressure and volume are not changing, so temperature remains same.

Q42

Even Carnot engine cannot give

100\%

efficiency because we cannot

A prevent radiation

B find ideal sources

C reach absolute zero temperature

D eliminate friction.

Correct Answer

Option C

Solution

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

For

\eta = 1

or

100\%

,

{T_2} = 0K.

The temperature of

0

K

(absolute zero) can not be obtained.

Q43

At what temperature is the

r.m.s

velocity of a hydrogen molecule equal to that of an oxygen molecule at

{47^ \circ }C?

A

80K

B

-73

K

C

3

K

D

20

K

Correct Answer

Option D

Solution

{v_{rms}} =

\sqrt {{{RT} \over M}}

For

{v_{rms}}

to be equal

{{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}

Here

{M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;

{T_{{O_2}}} = 47 + 273 = 320K

$\therefore$

{{{T_{{H_2}}}} \over 2} = {{320} \over {32}}

\Rightarrow {T_{{H_2}}} = 20K

Q44

A mixture of hydrogen and oxygen has volume 500 cm3, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be :-

A 3 : 8

B 3 : 16

C 16 : 3

D 8 : 3

Correct Answer

Option C

Solution

PV = nRT 400 $\times$ 103 $\times$ 500 $\times$ 10 $-$ 6 = n

\left( {{{25} \over 3}} \right)

(300) n =

{{2 \over {25}}}

n = n1 + n2

{{2 \over {25}}}

=

{{{M_1}} \over 2} + {{{M_2}} \over {32}}

Also, M1 + M2 = 0.76 gm

{{{M_2}} \over {{M_1}}} = {{16} \over 3}

Q45

A Carnot engine, whose efficiency is

40\%

, takes in heat from a source maintained at a temperature of

500

K.

It is desired to have an engine of efficiency

60\% .

Then, the intake temperature for the same exhaust (sink) temperature must be :

A efficiency of Carnot engine cannot be made larger than

50\%

B

1200

K

C

750

K

D

600

K

Correct Answer

Option C

Solution

0.4 = 1 - {{{T_2}} \over {500}}\,\,\,\,

and

\,\,\,\,0.6 = 1 - {{{T_2}} \over {{T_1}}}

on solving we get

{T_2} = 750\,K

Q46

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0

^\circ

C will be : (Given gas constant R = 8.3 JK

-

1-mol

-

1)

A 249 J

B 415 J

C 498 J

D 830 J

Correct Answer

Option C

Solution

\Delta Q = n{C_v}\Delta T

(Isochoric process)

= 2 \times {{3R} \over 2} \times 20

= 498

J

Q47

In van der Waal equation

\left[ {P + {a \over {{V^2}}}} \right]

[V

-

b] = RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants

{a \over b}

is dimensionally equal to :

A

{P \over V}

B

{V \over P}

C PV

D PV3

Correct Answer

Option C

Solution

From the equation

[a] \equiv [P{V^2}]

[b] \equiv [V]

\Rightarrow \left[ {{a \over b}} \right] \equiv [PV]

Q48

Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180oC and B upto ToC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is

A 200oC

B 270oC

C 230oC

D 250oC

Correct Answer

Option C

Solution

\Delta {\ell _1} = \Delta {\ell _2}

\ell {\alpha _1}\Delta {T_1} = \ell {\alpha _2}\Delta {T_2}

{{{\alpha _1}} \over {{\alpha _2}}} = {{\Delta {T_1}} \over {\Delta {T_2}}}

{4 \over 3} = {{T - 30} \over {180 - 30}}

T = {230^o}C

Q49

Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surface tension is

T,

density of liquid is

\rho

and

L

is its latent heat of vaporization.

A

\rho L/T

B

\sqrt {T/\rho L}

C

T/\rho L

D

2T/\rho L

Correct Answer

Option D

Solution

When radius is decrease by

\Delta R,

4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]

\Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]

\Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,

[

\Delta R

is very small ]

\Rightarrow R = {{2T} \over {\rho L}}

Q50

A heat engine is involved with exchange of heat of 1915 J, – 40J, + 125 J and –Q J, during one cycle achieving an efficiency of 50.0%. The value of Q is

A 980 J

B 40 J

C 400 J

D 640 J

Correct Answer

Option A

Solution

$\eta$ = Work done Heat supplied $\Rightarrow$

\frac{1}{2}

=

\frac{1915-40+125-Q}{1915+125}

$\Rightarrow$

\frac{1}{2}

=

\frac{2000-Q}{2040}

$\Rightarrow$ 2040 = 4000 – 2Q $\Rightarrow$ 2Q = 1960 $\Rightarrow$ Q = 980 J