Heat and Thermodynamics — Page 29 | JEE Physics

Q281

During the melting of a slab of ice at 273 K at atmospheric pressure :

A Internal energy of ice-water system remains unchanged.

B Positive work is done by the ice-water system on the atmosphere.

C Positive work is done on the ice-water system by the atmosphere.

D Internal energy of the ice-water system decreases.

Correct Answer

Option C

Solution

During the melting of a slab of ice at 273 K under atmospheric pressure, the following occurs: As the ice melts, its volume decreases.

Because of this volume reduction, the atmosphere exerts a positive work on the ice-water system.

When heat is absorbed by the ice-water system, we denote the absorbed heat as $\Delta \mathrm{Q}$ , which is positive.

The work done by the ice-water system is negative because it is being compressed by the external pressure.

According to the first law of thermodynamics: $\Delta \mathrm{U} = \Delta \mathrm{Q} + \Delta \mathrm{W}$ In this equation, $\Delta \mathrm{U}$ represents the change in internal energy, $\Delta \mathrm{Q}$ is the heat absorbed by the system, and $\Delta \mathrm{W}$ is the work done by the system.

Since the work done is negative and the absorbed heat is positive, the overall internal energy ( $\Delta \mathrm{U}$ ) increases.

Q282

Match with .

List - I		List - II
(A)	Isothermal	(I)	ΔW (work done) = 0
(B)	Adiabatic	(II)	ΔQ (supplied heat) = 0
(C)	Isobaric	(III)	ΔU (change in internal energy) ≠ 0
(D)	Isochoric	(IV)	ΔU = 0

A (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

B (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

C (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

D (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

Correct Answer

Option C

Solution

To accurately match the terms from List-I with the corresponding statements in List-II, consider the properties of each thermodynamic process: Isothermal Process (A): During an isothermal process, the temperature remains constant ( $\Delta T = 0$ ), which means that the change in internal energy ( $\Delta U$ ) is zero.

Therefore, it corresponds to (IV).

Adiabatic Process (B): In an adiabatic process, no heat is exchanged with the surroundings ( $\Delta Q = 0$ ).

This matches with (II).

Isobaric Process (C): An isobaric process involves constant pressure conditions ( $\Delta P = 0$ ), but internal energy ( $\Delta U$ ) can change.

Thus, it's associated with (III).

Isochoric Process (D): For an isochoric process, volume remains constant ( $\Delta V = 0$ ), which means that no work is done ( $\Delta W = 0$ ).

This aligns with (I).

Based on these descriptions, the correct matching is: (A) Isothermal → (IV) (B) Adiabatic → (II) (C) Isobaric → (III) (D) Isochoric → (I)

Q283

Consider a rectangular sheet of solid material of length

l=9 \mathrm{~cm}

and width

\mathrm{d}=4 \mathrm{~cm}

. The coefficient of linear expansion is

\alpha=3.1 \times 10^{-5} \mathrm{~K}^{-1}

at room temperature and one atmospheric pressure. The mass of sheet

m=0.1 \mathrm{~kg}

and the specific heat capacity

C_{\mathrm{v}}=900 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}

. If the amount of heat supplied to the material is

8.1 \times 10^2 \mathrm{~J}

then change in area of the rectangular sheet is :

A

2.0 \times 10^{-6} \mathrm{~m}^2

B

6.0 \times 10^{-7} \mathrm{~m}^2

C

3.0 \times 10^{-7} \mathrm{~m}^2

D

4.0 \times 10^{-7} \mathrm{~m}^2

Correct Answer

Option A

Solution

Calculate the Change in Temperature: Use the formula: $\Delta Q = mC_v \Delta T$ Given: $\Delta Q = 8.1 \times 10^2 \, \text{J}$ $m = 0.1 \, \text{kg}$ $C_v = 900 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$ Substitute the values to find $\Delta T$ : $8.1 \times 10^2 = 0.1 \times 900 \times \Delta T$ Solving for $\Delta T$ : $\Delta T = \dfrac{8.1 \times 10^2}{0.1 \times 900} = 9 \, \text{K}$ Calculate the Change in Area: The change in area $\Delta A$ is calculated using: $\Delta A = A_0 \times 2 \alpha \Delta T$ Where: $A_0$ is the original area of the sheet.

$\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}$ .

$\Delta T = 9 \, \text{K}$ .

First, calculate the original area $A_0$ : $A_0 = l \times d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 0.0036 \, \text{m}^2$ Now, substitute into the formula for $\Delta A$ : $\Delta A = 0.0036 \times 2 \times 3.1 \times 10^{-5} \times 9$ Simplifying: $\Delta A = 2.0 \times 10^{-6} \, \text{m}^2$ Therefore, the change in area of the rectangular sheet is $\boxed{2.0 \times 10^{-6} \, \text{m}^2}$ .

Q284

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K . If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K , at steady state the pressure in the vessels will be (in kPa ).

A 24

B 4.4

C 18

D 6

Correct Answer

Option D

Solution

\begin{aligned} &\text{ Number of masses will remain constant }\\ &\begin{aligned} & \mathrm{n}_1+\mathrm{n}_2=\mathrm{n}_{\mathrm{f}} \\ & \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}+\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{~V}_{\mathrm{f}}}{\mathrm{RT}_{\mathrm{f}}} \\ & \frac{8 \times 2 \mathrm{~V}}{\mathrm{R} \times 1000}+\frac{7 \times \mathrm{V}}{\mathrm{R} \times 500}=\frac{\mathrm{P}_{\mathrm{f}}(3 \mathrm{~V})}{\mathrm{R} \times 600} \\ & \frac{16}{1000}+\frac{14}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{R} \times 600} \\ & \frac{30}{1000}=\frac{\mathrm{P}_{\mathrm{f}}}{200} \\ & \mathrm{P}_{\mathrm{f}}=6 \mathrm{kPa} \end{aligned} \end{aligned}

Q285

Consider the sound wave travelling in ideal gases of

\mathrm{He}, \mathrm{CH}_4

, and

\mathrm{CO}_2

. All the gases have the same ratio

\dfrac{P}{\rho}

, where

P

is the pressure and

\rho

is the density. The ratio of the speed of sound through the gases

\mathrm{V}_{\mathrm{He}}: \mathrm{V}_{\mathrm{CH}_4}: \mathrm{V}_{\mathrm{CO}_2}

is given by

A

\sqrt{\dfrac{7}{5}}: \sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}

B

\sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{4}{3}}

C

\sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{7}{5}}

D

\sqrt{\dfrac{4}{3}}: \sqrt{\dfrac{5}{3}}: \sqrt{\dfrac{7}{5}}

Correct Answer

Option C

Solution

The speed of sound in an ideal gas is given by

v = \sqrt{\gamma \frac{P}{\rho}},

where: $\gamma$ is the ratio of specific heats,

P

is the pressure, and $\rho$ is the density. Since the problem states that

\frac{P}{\rho}

is the same for all the gases, the speed of sound in each gas is determined solely by the factor

\sqrt{\gamma}

. Let's determine the appropriate $\gamma$ for each gas: Helium (He): Helium is a monatomic gas. For a monatomic gas,

\gamma = \frac{5}{3}

. Therefore,

v_{\mathrm{He}} \propto \sqrt{\frac{5}{3}}

.

Methane (CH $_4$ ): Methane is a polyatomic gas (a tetrahedral molecule with three rotational degrees of freedom).

For nonlinear polyatomic gases, $\gamma$ is typically taken as

\frac{4}{3}

. Thus,

v_{\mathrm{CH_4}} \propto \sqrt{\frac{4}{3}}

. Carbon Dioxide (CO $_2$ ): Carbon dioxide is a linear molecule. For linear molecules,

\gamma = \frac{7}{5}

. Hence,

v_{\mathrm{CO_2}} \propto \sqrt{\frac{7}{5}}

. Therefore, the ratio of the speeds of sound in the gases is:

v_{\mathrm{He}} : v_{\mathrm{CH_4}} : v_{\mathrm{CO_2}} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{7}{5}}.

Comparing this expression with the given options, we see that it matches Option C.

Q286

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are

3 \times 10^{-7} \mathrm{~m}

and

600 \mathrm{~m} / \mathrm{s}

, respectively. Find the frequency of its collisions.

A

5 \times 10^8 / \mathrm{s}

B

9 \times 10^5 / \mathrm{s}

C

2 \times 10^{10} / \mathrm{s}

D

2 \times 10^9 / \mathrm{s}

Correct Answer

Option D

Solution

To determine the frequency of collisions for oxygen molecules, we use the formula for frequency: $\text{Frequency} = \dfrac{1}{T} = \dfrac{V_{\text{avg}}}{\lambda}$ where $V_{\text{avg}}$ is the average speed of the molecules and $\lambda$ is the mean free path.

Given: Average speed ( $V_{\text{avg}}$ ) = 600 m/s Mean free path ( $\lambda$ ) = $3 \times 10^{-7}$ m Substitute these values into the formula: $\text{Frequency} = \dfrac{600}{3 \times 10^{-7}} = 2 \times 10^9 \, \text{s}^{-1}$ Therefore, the frequency of collisions is $2 \times 10^9 \, \text{collisions per second}$ .

Q287

An ideal gas exists in a state with pressure

P_0

, volume

V_0

. It is isothermally expanded to 4 times of its initial volume

\left(\mathrm{V}_0\right)

, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is

A

\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.75)

B

\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.75)

C

\mathrm{P}_0 \mathrm{~V}_0(2 \ln 2-0.25)

D

\mathrm{P}_0 \mathrm{~V}_0(\ln 2-0.25)

Correct Answer

Option B

Solution

\begin{aligned} & \omega_1=\mathrm{P}_0 \mathrm{v}_0 \ell \mathrm{n} 4 \\ & \omega_2=\frac{\mathrm{P}_0}{4}\left(-3 \mathrm{v}_0\right)=-\frac{3 \mathrm{P}_0 \mathrm{v}_0}{4} \\ & \omega_3=0 \\ & \mathrm{Q}_{\mathrm{T}}=\Delta \mathrm{U}_{\text{cyclic }}+\omega \\ & \mathrm{Q}_{\mathrm{T}}=\omega \quad\left(\Delta \mathrm{U}_{\text{cyclic }}=0\right) \\ & \mathrm{Q}_{\mathrm{T}}=\mathrm{P}_0 \mathrm{v}_0\left(\ln 4-\frac{3}{4}\right) \\ & =\mathrm{P}_0 \mathrm{v}_0(2 \ln 2-0.75) \end{aligned}

Q288

Pressure of an ideal gas, contained in a closed vessel, is increased by

0.4 \%

when heated by

1^{\circ} \mathrm{C}

. Its initial temperature must be:

A 2500 K

B

25^{\circ} \mathrm{C}

C

250^{\circ} \mathrm{C}

D 250 K

Correct Answer

Option D

Solution

In this scenario, we're dealing with an isochoric process, meaning the volume of the gas remains constant.

For an ideal gas in an isochoric process, the pressure ( $P$ ) is directly proportional to the temperature ( $T$ ) in Kelvin.

The relationship can be expressed as: $P \propto T$ Given the percentage increase in pressure is $0.4\%$ when the temperature is increased by $1^{\circ} \text{C}$ , we can use the relationship: $\dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$ Substituting the given values into the equation, we have: $\dfrac{0.4}{100} = \dfrac{1}{T}$ Solving for $T$ , we find: $T = 250 \, \text{K}$ Therefore, the initial temperature of the gas in the vessel must be $250 \, \text{K}$ .

Q289

''Heat cannot by itself flow from a body at lower temperature to a body at higher temperature'' is a statement or consequence of :

A second law of thermodynamics

B conservation of momentum

C conservation of mass

D first law of thermodynamics

Correct Answer

Option A

Solution

The statement "Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement or consequence of the second law of thermodynamics.

This law essentially states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible.

In simpler terms, it implies that heat energy cannot spontaneously transfer from a colder body to a hotter one.

Therefore, the correct option is : Option A : second law of thermodynamics.

Q290

Statement - 1: The temperature dependence of resistance is usually given as

R = {R_0}\left( {1 + \alpha \,\Delta t} \right).

The resistance of wire changes from

100\Omega

to

150\Omega

when its temperature is increased from

{27^ \circ }C

to

{227^ \circ }C

. This implies that

\alpha = 2.5 \times {10^{ - 3}}/C.

Statement - 2:

R = {R_0}\left( {1 + \alpha \,\Delta t} \right)

is valid only when the change in the temperature

\Delta T

is small and

\Delta T = \left( {R - {R_0}} \right) < < {R_0}.

A Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1

B Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1

C Statement - 1 is false, Statement - 2 is true

D Statement - 1 is true, Statement - 2 is false

Correct Answer

Option C

Solution

The relation

R = {R_0}\left( {1 + \alpha \,\Delta t} \right)

is valid for small values of

\Delta t

and

{R_0}

is resistance at

{0^ \circ }C

and also

\left( {R - {R_0}} \right)

should be much smaller than

{R_0}.

So, statement

(1)

is wrong but statement

(2)

is correct.