Heat and Thermodynamics — Page 14 | JEE Physics

Q131

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (kB is Boltzmann constant)

A

{1 \over 2}{k_B}T

B

{2 \over 3}{k_B}T

C

{3 \over 2}{k_B}T

D

{k_B}T

Correct Answer

Option A

Solution

Energy associated with each digress of freedom is

{1 \over 2}{k_B}T

.

Q132

Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be : (Molecular weight of oxygen is 32g/mol; R = 8.3 J K

-

1 mol

-

1)

A

\sqrt {{{3\pi } \over 8}}

B

\sqrt {{3 \over 3}}

C

\sqrt {{8 \over 3}}

D

\sqrt {{{8\pi } \over 3}}

Correct Answer

Option A

Solution

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

{V_{avg}} = \sqrt {{8 \over \pi }{{RT} \over M}}

{{{V_{rms}}} \over {{V_{avg}}}} = \sqrt {{{3\pi } \over 8}}

Q133

For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where

\gamma

is the ratio of specific heats) :

A

- {1 \over \gamma }{{dV} \over V}

B

- \gamma {V \over {dV}}

C

- \gamma {{dV} \over V}

D

{{dV} \over V}

Correct Answer

Option C

Solution

for adiabatic expansion : PV $\gamma$ = const. $\Rightarrow$ ln P + $\gamma$ ln v = const. $\Rightarrow$ differentiating both sides;

{{dp} \over p} + \gamma {{dv} \over v} = 0

\Rightarrow {{dp} \over p} = - \gamma {{dv} \over V}

Q134

An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S1 and that of the other part is S2. Given that S1 > S2. If the piston is removed then the total entropy of the system will be :

A S1

-

S2

B

{{{S_1}} \over {{S_2}}}

C S1

\times

S2

D S1 + S2

Correct Answer

Option D

Solution

for gas 1, S1 =

{f \over 2}{n_1}R

for gas 2, S2 =

{f \over 2}{n_2}R

after removal of piston, S =

{f \over 2}({n_1} + {n_2})R = {S_1} + {S_2}

Q135

Consider a mixture of gas molecule of types A, B and C having masses mA < mB < mC. The ratio of their root mean square speeds at normal temperature and pressure is :

A

{v_A} = {v_B} \ne {v_C}

B

{1 \over {{v_A}}} > {1 \over {{v_B}}} > {1 \over {{v_C}}}

C

{1 \over {{v_A}}} < {1 \over {{v_B}}} < {1 \over {{v_C}}}

D

{v_A} = {v_B} = {v_C} = 0

Correct Answer

Option C

Solution

rms velocity of gas molecules is given as

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

..... (i) where, m = molar mass of the gas in kilograms per mole, R = molar gas constant, and T = temperature in kelvin.

According to question, mA B C From Eq. (i),

{v_{rms}} \propto {1 \over {\sqrt m }}

$\therefore$ We can write, vA > vB > vC or

{1 \over {{v_A}}} < {1 \over {{v_B}}} < {1 \over {{v_C}}}

Q136

The amount of heat needed to raise the temperature of 4 moles of rigid diatomic gas from 0

^\circ

C to 50

^\circ

C when no work is done is ___________. (R is the universal gas constant).

A 500 R

B 250 R

C 750 R

D 175 R

Correct Answer

Option A

Solution

According to first law of thermodynamics,

\Delta

Q =

\Delta

U +

\Delta

W ..... (i) where,

\Delta

Q = quantity of heat energy supplied to the system,

\Delta

U = change in the internal energy of a closed system and

\Delta

W = work done by the system on its surroundings. As per question, no work is done $\therefore$

\Delta

W = 0 ..... (iii) From Eqs. (i) and (ii), we get

\Delta

Q = 0 +

\Delta

U $\Rightarrow$

\Delta

Q =

\Delta

U or

\Delta

Q =

\Delta

U = nCV

\Delta

T where, CV = specific heat capacity at constant volume for diatomic gas =

{{5R} \over 2}

\Delta

T = change in temperature = (50 $-$ 0) = 50

^\circ

C n = number of moles = 4 $\Rightarrow$

\Delta

Q = nCV

\Delta

T =

4 \times {{5R} \over 2} \times (50)

= 500 R = 500 R

Q137

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5

\times

103 J ?

A 2.5

\times

102 s

B 4.1

\times

101 s

C 2.4

\times

103 s

D 2.5

\times

101 s

Correct Answer

Option A

Solution

\Delta

Q =

\Delta

U +

\Delta

W

{{\Delta Q} \over {\Delta t}} = {{\Delta U} \over {\Delta t}} + {{\Delta W} \over {\Delta t}}

{{6000} \over {60}}{J \over {\sec }} = {{2.5 \times {{10}^3}} \over {\Delta t}} + 90

\Delta

t = 250 sec

Q138

The entropy of any system is given by

S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]

where

\alpha

and

\beta

are the constants.

\mu

, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively. [Take

S = {{dQ} \over T}

] Choose the incorrect option from the following :

A

\alpha

and J have the same dimensions.

B S and

\alpha

have different dimensions

C S,

\beta

, k and

\mu

R have the same dimensions

D

\alpha

and k have the same dimensions

Correct Answer

Option D

Solution

Since, entropy of the system is given by

S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]

.... (i) As,

S = {Q \over {\Delta T}}

[given]

\Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}

.... (ii) $\because$ Dimensions of Q = [ML2T $-$ 2] Dimension of T = [K] Boltzmann constant,

k = {{energy} \over T}

[ $\because$ Dimensions of energy = [ML2T $-$ 2]]

\Rightarrow [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}

..... (iii) From Eqs. (ii) and (iii), we can write,

[S] = [k] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}

..... (iv) $\because$ Gas constant,

[R] = {{[Energy]} \over {[nT]}} = {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}}

.... (v) and mechanical equivalent of heat [J] = [M0L0T0] .... (vi) As, [ $\mu$ kR] = [J $\beta$ ]2 Using Eqs. (iii), (v) and (vi), we get

\Rightarrow [mol] \times {{[M{L^2}{T^{ - 2}}]} \over {[K]}} \times {{[M{L^2}{T^{ - 2}}]} \over {[mol\,K]}} = [{\beta ^2}]

\Rightarrow [\beta ] = [M{L^2}{T^{ - 2}}{K^{ - 1}}]

..... (vii) Using Eq. (i), we can write,

[{\alpha ^2}] = {{[S]} \over {[\beta ]}} = {{[M{L^2}{T^{ - 2}}{K^{ - 1}}]} \over {[M{L^2}{T^{ - 2}}{K^{ - 1}}]}} \Rightarrow \alpha = [{M^0}{L^0}{T^0}]

.... (viii) So, from Eqs. (iii) and (viii), we can say that $\alpha$ and k have different dimensions.

Q139

What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T?

A

{3 \over 2}{k_B}T

B

{k_B}T

C

{2 \over 3}{k_B}T

D

{1 \over 2}{k_B}T

Correct Answer

Option A

Solution

For a monoatomic ideal gas, the average kinetic energy per molecule is determined by the equipartition theorem.

This theorem states that the energy is equally distributed among all the available degrees of freedom.

A monoatomic gas has three translational degrees of freedom, corresponding to motion in the x, y, and z directions.

Each degree of freedom contributes an average energy of $\dfrac{1}{2} k_BT$ , where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.

So for a monoatomic gas with three translational degrees of freedom, the average energy per molecule is:

\frac{3}{2} k_BT.

Q140

A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If l1 and l2 are the lengths of the gas column, before and after the expansion respectively, then the value of

{{{T_1}} \over {{T_2}}}

will be :

A

{\left( {{{{l_1}} \over {{l_2}}}} \right)^{{2 \over 3}}}

B

{\left( {{{{l_2}} \over {{l_1}}}} \right)^{{2 \over 3}}}

C

{{{l_2}} \over {{l_1}}}

D

{{{l_1}} \over {{l_2}}}

Correct Answer

Option B

Solution

PVr = const. TVr $-$ 1 = const.

T{(l)^{{5 \over 3} - 1}}

= const.

{{{T_1}} \over {{T_2}}} = {\left( {{{{l_2}} \over {{l_1}}}} \right)^{{2 \over 3}}}