Heat and Thermodynamics — Page 27 | JEE Physics

Q261

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.Reason (R) : In isothermal process, PV = constant, while in adiabatic process

PV^{\gamma}

= constant. Here

\gamma

is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.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 Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

C (A) is false but (R) is true

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

Correct Answer

Option D

Solution

We know, ideal gas equation, $\mathrm{Pv = nRT}$ For isothermal process, T = constant $\mathrm{\Rightarrow PV = const.}$ $\mathrm{\Rightarrow P \propto \dfrac{1}{V}}$ For an adiabatic process, $\mathrm{PV^r=constant}$ We can see, for the same pressure incresement,

{P_2} - {P_1}

,

\left( {{v_4} - {v_3}} \right) > \left( {{v_2} - {v_1}} \right)

i.e. the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Hence, option 4 is corret.

Q262

A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

A

\dfrac{13}{5}t

B

\dfrac{10}{13}t

C

\dfrac{5}{13}t

D

\dfrac{13}{10}t

Correct Answer

Option A

Solution

Here, we will use the Newton's law of cooling. We know,

{{{T_1} - {T_2}} \over {\Delta t}} = k\left( {{{{T_1} + {T_2}} \over 2} - {T_s}} \right)

.... (1) where, T $_1$ and T $_2$ are initial and final temperatures respectively. $\mathrm{T_s}$ = surrounding temperature $\Delta t$ = time taken $k$ = positive constant Given, Ist case :

{T_1} = 90^\circ C

{T_2} = 80^\circ C

\Delta t = t

,

{T_s} = 20^\circ C

IInd case :

{T_1} = 80^\circ C

{T_2} = 60^\circ C

{T_s} = 20^\circ C

,

\Delta t = ?

Now, for Ist case,

{{90 - 80} \over t} = k\left( {{{90 + 80} \over 2} - 20} \right)

...... (from (1))

\Rightarrow {{10} \over t} = k\left( {85 - 20} \right)

\Rightarrow {{10} \over t} = 65k

.... (2) For IInd case,

{{80 - 60} \over {\Delta t}} = k\left( {{{80 + 60} \over 2} - 20} \right)

.... (From (1))

\Rightarrow {{20} \over {\Delta t}} = k\left( {70 - 20} \right)

\Rightarrow {{20} \over {\Delta t}} = 50k

.... (3) Now, by (2) $\div$ (3),

{{{{10} \over t}} \over {{{20} \over {\Delta t}}}} = {{13} \over {10}}

\Rightarrow {{\Delta t} \over t}\left( {{1 \over 2}} \right) = {{13} \over 5}

\Rightarrow \Delta t = {{13} \over 5}t

Q263

The work done in an adiabatic change in an ideal gas depends upon only :

A change in its pressure

B change in its temperature

C change in its specific heat

D change in its volume

Correct Answer

Option B

Solution

$\Delta \mathrm{W}=-\Delta \mathrm{U}=-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$

Q264

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),

A 64 E

B 16 E

C E

D 256 E

Correct Answer

Option C

Solution

To determine the energy radiated by two spherical bodies, both made of the same material but with different radii and temperatures, we can use the Stefan-Boltzmann law.

This law states that the energy radiated by a black body per unit time is proportional to the product of the surface area $A$ and the fourth power of its temperature $T$ .

The formula for power $P$ is given by: $P = \sigma e A T^4$ Where: $\sigma$ is the Stefan-Boltzmann constant $e$ is the emissivity of the material $A$ is the surface area of the sphere $T$ is the temperature in Kelvin Expressing this relationship as a proportionality: $P \propto A T^4$ For a sphere, the surface area $A$ is: $A = 4\pi R^2$ Hence, the power radiated is proportional to: $P \propto R^2 T^4$ For two bodies with radii $R_1$ and $R_2$ , and temperatures $T_1$ and $T_2$ , the ratio of their powers is: $\dfrac{P_1}{P_2} = \left( \dfrac{R_1}{R_2} \right)^2 \left( \dfrac{T_1}{T_2} \right)^4$ Given: $R_1 = 0.2 \, \text{m}$ , $T_1 = 800 \, \text{K}$ $R_2 = 0.8 \, \text{m}$ , $T_2 = 400 \, \text{K}$ Substituting into the formula: $\dfrac{E}{E'} = \left( \dfrac{0.2}{0.8} \right)^2 \left( \dfrac{800}{400} \right)^4$ Calculating each part: $\left( \dfrac{0.2}{0.8} \right)^2 = \left( \dfrac{1}{4} \right)^2 = \dfrac{1}{16}$ $\left( \dfrac{800}{400} \right)^4 = 2^4 = 16$ Combining these calculations: $\dfrac{E}{E'} = \dfrac{1}{16} \times 16 = 1$ This implies that the energy radiated by the bigger body $E'$ equals the energy radiated by the smaller body $E$ .

Thus, $E' = E$ .

Q265

An amount of ice of mass

10^{-3} \mathrm{~kg}

and temperature

-10^{\circ} \mathrm{C}

is transformed to vapour of temperature

110^{\circ} \mathrm{C}

by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice

=2100 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}

, specific heat of water

=4180 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}

, specific heat of steam

=1920 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}

, Latent heat of ice

=3.35 \times 10^5 \mathrm{Jkg}^{-1}

and Latent heat of steam

=2.25 \times 10^6

\mathrm{Jkg}^{-1}

)

A 3022 J

B 3043 J

C 3003 J

D 3024 J

Correct Answer

Option B

Solution

\Delta {Q_1} = m{s_i}\Delta {T_2},\,\Delta {Q_2} = m{L_i}

\Delta {Q_3} = m{S_w}\Delta {T_3},\,\Delta {Q_4} = m{L_s},\,\Delta {Q_5} = m{S_s}\Delta {T_5}

\Delta {Q_{net}} = m({S_i}\Delta {T_1} + {L_i} + {S_w}\Delta {T_3} + {L_s} + {S_s}\Delta {T_s})

= {10^{ - 3}}(2100 \times 10 + 335000 + 4180 \times 100 + 2250000 + 1920 \times 10)

= {10^{ - 3}}(3043200) = 3043.2\,J

So,

\Delta {Q_{net}} \approx 3043\,J

Q266

Given are statements for certain thermodynamic variables, (A) Internal energy, volume

(\mathrm{V})

and mass

(\mathrm{M})

are extensive variables. (B) Pressure (P), temperature ( T ) and density (

\rho

) are intensive variables. (C) Volume (V), temperature (T) and density (

\rho

) are intensive variables. (D) Mass (M), temperature (T) and internal energy are extensive variables. Choose the correct answer from the options given below :

A (C) and (D) Only

B (A) and (B) Only

C (D) and (A) Only

D (B) and (C) Only

Correct Answer

Option B

Solution

Check Statement (A): “Internal energy (U), volume (V), and mass (M) are extensive variables.”

Internal energy is extensive.

Volume is extensive.

Mass is extensive.

Hence, statement (A) is correct.

Check Statement (B): “Pressure (P), temperature (T), and density (ρ) are intensive variables.”

Pressure is intensive.

Temperature is intensive.

Density is intensive.

Hence, statement (B) is correct.

Check Statement (C): “Volume (V), temperature (T), and density (ρ) are intensive variables.”

Volume is not intensive; it is an extensive property.

(T) and (ρ) are indeed intensive.

Since volume is miscategorized here, (C) is incorrect.

Check Statement (D): “Mass (M), temperature (T), and internal energy are extensive variables.”

Mass is extensive.

Temperature is not extensive (it is intensive).

Internal energy is extensive.

Because temperature is miscategorized, (D) is incorrect.

Only (A) and (B) are correct statements.

Therefore, the correct choice is: Option B: (A) and (B) Only.

Q267

For a diatomic gas, if

\gamma_1=\left(\dfrac{C p}{C v}\right)

for rigid molecules and

\gamma_2=\left(\dfrac{C p}{C v}\right)

for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct? (Cp and Cv are specific heats of the gas at constant pressure and volume)

A

\gamma_2<\gamma_1

B

\gamma_2>\gamma_1

C

\gamma_2=\gamma_1

D

2 \gamma_2=\gamma_1

Correct Answer

Option A

Solution

Let's analyze the problem step by step: For a rigid diatomic molecule (without vibrational modes): Degrees of freedom: Translation: 3 Rotation: 2 Total: 5 The molar specific heat at constant volume is given by:

C_v = \frac{5}{2}R

At constant pressure, it is:

C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R

Therefore, the ratio is:

\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4

For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited): Additional vibrational mode contributions: Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).

Therefore, the degrees of freedom become:

3\ (\text{translation}) + 2\ (\text{rotation}) + 2\ (\text{vibration}) = 7

The molar specific heat at constant volume now is:

C_v = \frac{7}{2}R

And at constant pressure:

C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R

Thus, the ratio is:

\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857

Comparison: For the rigid molecule,

\gamma_1 \approx 1.4

. For the molecule with vibrational modes,

\gamma_2 \approx 1.2857

. Clearly,

\gamma_2 Therefore, the correct option is: Option A:

\gamma_2

Q268

The ratio of vapour densities of two gases at the same temperature is

\dfrac{4}{25}

, then the ratio of r.m.s. velocities will be :

A

\dfrac{5}{2}

B

\dfrac{25}{4}

C

\dfrac{4}{25}

D

\dfrac{2}{5}

Correct Answer

Option A

Solution

We are given that the ratio of the vapour densities of the two gases is

\frac{4}{25}.

Since vapour density is proportional to the molecular mass, we can write

\frac{M_1}{M_2} = \frac{4}{25},

where $M_1$ and $M_2$ are the molecular masses of the gases. The root mean square (r.m.s.) velocity of a gas is given by

v_{\text{rms}} = \sqrt{\frac{3RT}{M}},

where: $R$ is the universal gas constant, $T$ is the temperature, $M$ is the molecular mass of the gas.

Since both gases are at the same temperature, the ratio of their r.m.s. velocities is

\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{M_2}{M_1}}.

Substitute the ratio of molecular masses:

\frac{v_{\text{rms,1}}}{v_{\text{rms,2}}} = \sqrt{\frac{25}{4}} = \frac{5}{2}.

Thus, the ratio of their r.m.s. velocities is

\frac{5}{2}.

The correct answer is Option A:

\frac{5}{2}.

Q269

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J , then the mass of the bullet is _______ grams. (Latent heat of fusion of lead

=2.5 \times 10^4 \mathrm{JKg}^{-1}

and specific heat capacity of lead

=125 \mathrm{JKg}^{-1}

\left.\mathrm{K}^{-1}\right)

A 20

B 15

C 5

D 10

Correct Answer

Option D

Solution

Let's determine the mass step by step: The bullet is heated from 300 K to 600 K, so the increase in temperature is:

\Delta T = 600\, \text{K} - 300\, \text{K} = 300\, \text{K}.

The heat required to raise the temperature of the bullet (using the specific heat capacity of lead) is:

Q_1 = m \times c \times \Delta T = m \times 125\, \text{J/kg K} \times 300\, \text{K} = 37500\, m \, \text{J},

where $m$ is the mass in kilograms.

After reaching 600 K, the bullet melts.

The heat required for the phase change (melting) is given by:

Q_2 = m \times L = m \times 2.5 \times 10^4\, \text{J/kg} = 25000\, m \, \text{J}.

The total heat required for the process is:

Q = Q_1 + Q_2 = 37500\, m + 25000\, m = 62500\, m \, \text{J}.

We are given that the total heat is 625 J, so:

62500\, m = 625.

Solving for $m$ :

m = \frac{625}{62500} = 0.01\, \text{kg}.

Finally, converting the mass from kilograms to grams:

0.01\, \text{kg} = 0.01 \times 1000\, \text{g} = 10\, \text{g}.

Thus, the mass of the bullet is 10 grams. This corresponds to Option D.

Q270

Water of mass

m

gram is slowly heated to increase the temperature from

T_1

to

T_\gamma

. The change in entropy of the water, given specific heat of water is

1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}

, is :

A

\mathrm{m}\left(\mathrm{T}_2-\mathrm{T}_1\right)

B zero

C

\mathrm{m} \ln \left(\dfrac{\mathrm{T}_1}{\mathrm{~T}_2}\right)

D

\mathrm{m} \ln \left(\dfrac{\mathrm{T}_2}{\mathrm{~T}_1}\right)

Correct Answer

Option D

Solution

To determine the change in entropy when water is heated from temperature

T_1

to

T_2

(here,

T_\gamma

is equivalent to

T_2

), we use the definition of entropy change for a reversible process:

\Delta S = \int_{T_1}^{T_2} \frac{dQ}{T}

Since the water is being heated slowly, the process is reversible, and the heat added is given by:

dQ = mc\,dT

where: •

m

is the mass of the water, •

c

is the specific heat capacity (given as

1\,\mathrm{J\,kg^{-1}\,K^{-1}}

). Substitute

dQ

into the integral:

\Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}.

Since

c = 1

, the equation simplifies to:

\Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}.

Evaluating the integral, we have:

\int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right).

Thus, the change in entropy is:

\Delta S = m \ln\left(\frac{T_2}{T_1}\right).

Comparing with the options provided, the correct answer is: Option D:

m \ln\left(\frac{T_2}{T_1}\right).