Heat and Thermodynamics — Page 9 | JEE Physics

Q81

A thermally insulated vessel contains 150g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to : (Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 105 J kg–1)

A 35 g

B 130 g

C 20 g

D 150 g

Correct Answer

Option C

Solution

Let x grams of water is evaporated.

According to the principle of calorimetry, Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water Given, mass of water = 150 g

\Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \times 2.10 \times {10^6}

\Rightarrow (150 - x) \times 3.36 = 21x

\Rightarrow x = {{150} \over {7.25}} = 20.6

$\therefore$

x \approx 20g

Q82

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to : [Boltzmann Constant kB = 1.38 × 10–23 J/K Avogadro Number NA = 6.02 × 1026 /kg Radius of Earth : 6.4 × 106 m Gravitational acceleration on Earth = 10ms–2]

A 3 × 105 K

B 104 K

C 650 K

D 800 K

Correct Answer

Option B

Solution

{V_{rms}} = \sqrt {{{3RT} \over M}} = 11.2 \times {10^3}m/s

$\Rightarrow$

T = {M \over {3R}} \times {\left( {11.2 \times {{10}^3}} \right)^2}

=

{{2 \times {{10}^{ - 3}}} \over {3 \times 8.3}} \times 125.44 \times {10^6} = {10^4}K

Q83

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is

\overline v

, m is its mass and kB is Boltzmann constant, then its temperature will be :

A

{{m{{\overline v }^2}} \over {5{k_B}}}

B

{{m{{\overline v }^2}} \over {6{k_B}}}

C

{{m{{\overline v }^2}} \over {7{k_B}}}

D

{{m{{\overline v }^2}} \over {3{k_B}}}

Correct Answer

Option C

Solution

An HCl molecule, being diatomic, has: 3 translational degrees of freedom 2 rotational degrees of freedom 2 vibrational degrees of freedom The total number of degrees of freedom is $3 + 2 + 2 = 7$ .

According to the equipartition theorem, each degree of freedom contributes $\dfrac{1}{2} k_B T$ to the total energy.

So the total energy is given by: $\dfrac{7}{2} k_B T$ The translational kinetic energy is related to the root-mean-square (rms) speed $\overline{v}$ by: $\dfrac{1}{2} m \overline{v}^2 = \dfrac{7}{2} k_B T$ Rearranging to solve for the temperature, we find: $T = \dfrac{m \overline{v}^2}{7k_B}$

Q84

A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely ? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)

A 10–3 K

B 10–1 K

C 10–5K

D 10–4 K

Correct Answer

Option C

Solution

By law of conservation of energy

{1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T

\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}}

K

Q85

The specific heats, CP and CV of a gas of diatomic molecules, A, are given (in units of J mol–1 K–1) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then :-

A A is rigid but B has a vibrational mode

B A has a vibrational mode but B has none

C A has one vibrational mode and B has two

D Both A and B have a vibrational mode each

Correct Answer

Option B

Solution

For A:

{{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{29} \over {22}}

It gives f = 6.3 $\approx$ 6 (3 translational, 2 rotational and 1 vibrational) For B:

{{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{30} \over {21}}

$\therefore$ f = 4.67 $\approx$ 5 (3 translational, 2 rotational, no vibrational)

Q86

A vessel contains

14 \mathrm{~g}

of nitrogen gas at a temperature of

27^{\circ} \mathrm{C}

. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be : Take

\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1}

.

A 2229 J

B 5616 J

C 9360 J

D 13,104 J

Correct Answer

Option C

Solution

n = 0.5 T = 300 For vrms to be doubled T' = 4 $\times$ 300 = 1200 $\Rightarrow$ Heat transferred

= (0.5)\left( {{5 \over 2}} \right)(8.32)(900)

= 9360

J

Q87

One mole of ideal gas passes through a process where pressure and volume obey the relation

P = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

. Here P0 and V0 are constants. Calculate the change in the temperature of the gas if its volume changes form V0 to 2V0

A

{3 \over 4}{{{P_0}{V_0}} \over R}

B

{1 \over 2}{{{P_0}{V_0}} \over R}

C

{5 \over 4}{{{P_0}{V_0}} \over R}

D

{1 \over 4}{{{P_0}{V_0}} \over R}

Correct Answer

Option C

Solution

Given

P = {P_o}\left\{ {1 - {1 \over 2}{{\left( {{{{V_o}} \over V}} \right)}^2}} \right\};

...(i) As n = 1 mole $\therefore$ PV = nRT = RT $\Rightarrow$ P =

{{RT} \over V}

....(ii) From (i) and (ii), we get

{{RT} \over V} = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

$\Rightarrow$ T =

{V \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]

Case 1 : when V = V0 then Ti =

{{{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {{V_0}}}} \right)}^2}} \right]

=

{{{V_0}{P_0}} \over {2R}}

Case 2 : when V = 2V0 then Tf =

{{2{V_0}} \over R} \times {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over {2{V_0}}}} \right)}^2}} \right]

=

{7 \over 4}{{{P_0}{V_0}} \over R}

Then

\Delta

T = Tf - Ti =

{7 \over 4}{{{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over {2R}}

=

{{{P_0}{V_0}} \over R}\left( {{7 \over 4} - {1 \over 2}} \right)

=

{5 \over 4}{{{P_0}{V_0}} \over R}

Q88

When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by

\Delta

T. the heat required to produce the same change in temperature, at a constant pressure is :

A

{7 \over 5}Q

B

{3 \over 2}Q

C

{2 \over 3}Q

D

{5 \over 3}Q

Correct Answer

Option A

Solution

Heat supplied at constant volume Q = nCV

\Delta

T and heat supplied at constant pressure Q' = nCP

\Delta

T

Q' = {{{C_P}} \over {{C_V}}}Q = \left( {1 + {2 \over 5}} \right)Q = {7 \over 5}Q

Q89

At 40o C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40oC to 20oC it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young’s modulus of brass are 10–5 /oC and 1011 N/m 2 , respectively; g= 10 ms–2 )

A 1.5 kg

B 0.5 kg

C 9 kg

D 0.9 kg

Correct Answer

Option C

Solution

Mg = \left( {{{Ay} \over \ell }} \right)\Delta \ell

Mg = \left( {Ay} \right)\alpha \Delta T = 2\pi

It is closest to 9

Q90

Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume ? (R = 8.3 J/mol K)

A 21.6 J/mol K

B 17.4 J/mol K

C 15.7 J/mol K

D 19.7 J/mol K

Correct Answer

Option B

Solution

{f_{mix}} = {{{n_1}{f_1} + {n_2}{f_2}} \over {{n_1} + {n_2}}}

\Rightarrow {{2 \times 3 + 3 \times 5} \over 5} = {{21} \over 5}

{C_v} = {{fR} \over 5} = {{21} \over 5} \times {R \over 2} = 17.4

J/mol K