Heat and Thermodynamics — Page 12 | JEE Physics

Q111

Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CP/CV value will be :

A 23/15

B 67/45

C 40/27

D 19/13

Correct Answer

Option D

Solution

{{{C_P}} \over {{C_V}}} = {{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}} \over {{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}}

=

{{n \times {{5R} \over 2} + 2n \times {{7R} \over 2}} \over {n \times {{3R} \over 2} + 2n \times {{5R} \over 2}}}

=

{{5 + 14} \over {3 + 10}}

=

{{19} \over {13}}

Q112

Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibrational mode, and have a mass

{m \over 4}

. The ratio of the specific heats (

C_V^A

and

C_V^B

) of gas A and B, respectively is :

A 7 : 9

B 5 : 7

C 3 : 5

D 5 : 9

Correct Answer

Option B

Solution

Degree of freedom of a diatomic molecule if vibration is absent = 5 Degree of freedom of a diatomic molecule if vibration is present = 7 $\therefore$

C_V^A

=

{5 \over 2}R

and

C_V^B

=

{7 \over 2}R

$\therefore$

{{C_V^A} \over {C_V^B}} = {5 \over 7}

Q113

When the temperature of a metal wire is increased from 0oC to 10oC, its length increases by 0.02%. The percentage change in its mass density will be closest to :

A 0.008

B 0.06

C 0.8

D 2.3

Correct Answer

Option B

Solution

Given,

{{\Delta L} \over L}

= 0.02% We know,

\Delta

L = L $\alpha$

\Delta

T $\Rightarrow$

{{\Delta L} \over L} = \alpha \Delta T

= 0.02 Also, $\beta$ = 2 $\alpha$ $\Rightarrow$

\beta \Delta T = 2\alpha \Delta T

= 0.04 Density( $\rho$ ) =

{M \over {AL}}

$\Rightarrow$

{{\Delta \rho } \over \rho } = {{\Delta M} \over M} - {{\Delta A} \over A} - {{\Delta L} \over L}

(

{{\Delta M} \over M}

= 0 as M = constant) $\Rightarrow$

{{\Delta \rho } \over \rho } = {{\Delta A} \over A} + {{\Delta L} \over L}

=

\beta \Delta T + \alpha \Delta T

= 0.04 + 0.02 = 0.06 %

Q114

The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4

\times

105 J kg–1. 100 grams of ice at 0oC is placed in 200 g of water at 25oC. The amount of ice that will melt as the temperature of water reaches 0oC is close to (in grams) :

A 63.8

B 61.7

C 69.3

D 64.6

Correct Answer

Option B

Solution

Heat loss by water

Q = {m_w}s\Delta \theta

= \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J

This heat will absorbed by the ice and let mass

\Delta

mi got melted. So

\Delta {m_i}L = 21000

\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}} \times {10^3}\,gm = 61.7\,grams

Q115

Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The raito of their respective mean free times is closest to :

A 2.3

B 1.83

C 4.67

D 3.67

Correct Answer

Option B

Solution

\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}

Mean free time, t =

{\lambda \over v}

Also v $\propto$

\sqrt {{T \over M}}

$\therefore$ t $\propto$

{{\sqrt M } \over d}

{{{t_{Ar}}} \over {{t_{xe}}}}

=

{{d_{Xe}^2} \over {d_{Ar}^2}} \times \sqrt {{{{M_{Ar}}} \over {{M_{Xe}}}}}

=

{\left( {{{0.1} \over {0.07}}} \right)^2} \times \sqrt {{{40} \over {140}}}

= 1.09 $\therefore$ Nearest possible answer is 1.83.

Q116

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules :

A continuously lose their energy till it reaches wall.

B are attracted by the walls of container.

C suffer change in momentum when impinge on the walls of container.

D continuously stick to the walls of container.

Correct Answer

Option C

Solution

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules contain uniform speed, random motion and perform elastic collision with each other, as well as with the walls of container.

As a result of which gaseous molecules suffer change in momentum when impinge on the walls of container

Q117

The correct relation between the degrees of freedom f and the ratio of specific heat

\gamma

is :

A

f = {2 \over {\gamma - 1}}

B

f = {2 \over {\gamma + 1}}

C

f = {{\gamma + 1} \over 2}

D

f = {1 \over {\gamma + 1}}

Correct Answer

Option A

Solution

\gamma = 1 + {2 \over f}

$\Rightarrow$

f = {2 \over {\gamma - 1}}

Q118

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : When a rod lying freely is heated, no thermal stress is developed in it. Reason R : On heating, the length of the rod increases. In the light of the above statements, choose the correct answer from the options given below :

A A is true but R is false

B A is false but R is true

C Both A and B are true but R is NOT the correct explanation of A

D Both A and R are true and R is the correct explanation of A

Correct Answer

Option C

Solution

When a rod is free and it is heated then there is no thermal stress produced in it.

The rod will expand due to increase in temperature.

So both A & R are true.

Q119

A diatomic gas, having

{C_p} = {7 \over 2}R

and

{C_v} = {5 \over 2}R

, is heated at constant pressure. The ratio dU : dQ : dW :

A 5 : 7 : 3

B 3 : 7 : 2

C 5 : 7 : 2

D 3 : 5 : 2

Correct Answer

Option C

Solution

dV = n{5 \over 2}R\Delta T

dQ = n{7 \over 2}R\Delta T

dW = dQ - dV

= n{2 \over 2}R\Delta T

$\therefore$

dV:dQ:dW

= n{5 \over 2}R\Delta T:n{7 \over 2}R\Delta T:n{2 \over 2}R\Delta T

= 5:7:2

Q120

Given below are two statements : Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false.

D Statement I is false but Statement II is true.

Correct Answer

Option C

Solution

The translational kinetic energy & rotational kinetic energy both obey Maxwell's distribution independent of each other.

T.K.E. of diatomic molecules =

{3 \over 2}kT

R.K.E. of diatomic molecules =

{2 \over 2}kT

So statement I is true but statement II is false.