Heat and Thermodynamics — Page 28 | JEE Physics

Q271

An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below:

A A, C Only

B A, D, E Only

C E Only

D A, B, C, D Only

Correct Answer

Option B

Solution

For an ideal gas, the equation of state is given by

pV = nRT.

If during a process the pressure increases linearly with temperature (i.e.,

p \propto T

), then from the ideal gas law (with a fixed amount of gas,

n

, and with the gas constant,

R

), it follows that

\frac{p}{T} = \frac{nR}{V}.

Since

nR

is constant, a linear relationship between

p

and

T

implies that

\frac{p}{T}

remains constant throughout the process. This can only be true if the volume

V

is constant.

Hence, the process is isochoric.

For an isochoric (constant volume) process: The work done by the gas is given by

W = \int p\, dV,

and since

dV = 0

, we have

W = 0.

(Statement A is true.) The internal energy change for an ideal gas depends only on temperature:

\Delta U = nC_V\Delta T.

If the temperature increases, then

\Delta U > 0.

(Statement D is true.)

For an isochoric process, no expansion work is done, so any heat added goes entirely into increasing the internal energy.

This means the heat added is not different from the change in internal energy (in fact,

Q = \Delta U

).

(Statement B is false.)

Since the process is isochoric, the volume does not change (and certainly is not increased).

(Statement C is false, and Statement E stating it is isochoric is true.)

Thus, the correct true statements are A, D, and E.

Q272

The temperature of a body in air falls from

40^{\circ} \mathrm{C}

to

24^{\circ} \mathrm{C}

in 4 minutes. The temperature of the air is

16^{\circ} \mathrm{C}

. The temperature of the body in the next 4 minutes will be :

A

\dfrac{28}{3}{ }^{\circ} \mathrm{C}

B

\dfrac{56}{3}{ }^{\circ} \mathrm{C}

C

\dfrac{42}{3}{ }^{\circ} \mathrm{C}

D

\dfrac{14}{3}{ }^{\circ} \mathrm{C}

Correct Answer

Option B

Solution

\begin{aligned} & \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text{avg }}-\mathrm{T}_{\mathrm{s}}\right] \\ & \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\ & \frac{40-24}{4}=\mathrm{K}[32-16] \\ & \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\ & \text{ Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\ & 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\ & \frac{3 \mathrm{~T}}{2}=28 \\ & \mathrm{~T}=\frac{56}{3} \mathrm{C} \end{aligned}

Q273

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases. Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process. In the light of the above statements, choose the correct answer from the options given below :

A Both

(\mathbf{A})

and

(\mathbf{R})

are true and

(\mathbf{R})

is the correct explanation of

(\mathbf{A})

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 (A) is true but

(\mathbf{R})

is false

Correct Answer

Option C

Solution

Let's analyze both statements step by step.

Assertion (A): "In an insulated container, a gas is adiabatically shrunk to half of its initial volume.

The temperature of the gas decreases."

• In an adiabatic process, no heat is exchanged with the surroundings.

• For a reversible adiabatic (compression) process, the relationship between temperature and volume is given by

T V^{\gamma-1} = \text{constant},

where $\gamma$ is the heat capacity ratio (greater than 1). • If the volume is reduced from

V

to

V/2

, then

T_{\text{final}} = T_{\text{initial}} \left(\frac{V}{V/2}\right)^{\gamma-1} = T_{\text{initial}}\,(2)^{\gamma-1}.

• Since

(2)^{\gamma-1} > 1,

the final temperature is higher than the initial temperature.

• Therefore, the assertion that the temperature decreases is incorrect.

Reason (R): "Free expansion of an ideal gas is an irreversible and an adiabatic process."

• In a free expansion, the gas expands into a vacuum without doing any work on the surroundings.

• Since the container is insulated, no heat is exchanged, making it adiabatic.

• However, the process is inherently irreversible due to the spontaneous expansion and mixing of states.

• Thus, this statement is true.

Since Assertion (A) is false and Reason (R) is true (but the reason does not explain the false assertion about temperature change during compression), the correct answer is: Option C (A) is false but (R) is true.

Q274

A monoatomic gas having

\gamma = \dfrac{5}{3}

is stored in a thermally insulated container and the gas is suddenly compressed to

\left( \dfrac{1}{8} \right)^{\text{th}}

of its initial volume. The ratio of final pressure and initial pressure is: (

\gamma

is the ratio of specific heats of the gas at constant pressure and at constant volume)

A 16

B 32

C 28

D 40

Correct Answer

Option B

Solution

To find the ratio of final pressure ( $P_f$ ) to initial pressure ( $P_i$ ), we use the adiabatic condition: $P_i V_i^\gamma = P_f V_f^\gamma$ We aim to express the pressure ratio: $\dfrac{P_f}{P_i} = \left(\dfrac{V_i}{V_f}\right)^\gamma$ Given that the final volume $V_f$ is $\dfrac{1}{8}$ of the initial volume $V_i$ , this simplifies to: $\left(\dfrac{V_i}{V_f}\right) = 8$ Thus, the pressure ratio becomes: $\dfrac{P_f}{P_i} = 8^\dfrac{5}{3}$ Calculating $8^\dfrac{5}{3}$ gives: $\dfrac{P_f}{P_i} = 32$

Q275

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.(Take g = 10 m/s2, specific heat of water = 4200 J/(kg K))

A 0.36 K

B 0.23 K

C 0.48 K

D 0.14 K

Correct Answer

Option C

Solution

To find the rise in temperature of the water as it falls from a height, we start by equating the potential energy loss to the heat gained, assuming no heat is lost from the water in the pool.

The potential energy lost by the water can be expressed as: $mgh$ where: $m$ is the mass of water, $g$ is the acceleration due to gravity (10 m/s²), $h$ is the height (200 m).

The heat gained by the water when temperature changes is given by: $ms \Delta T$ where: $s$ is the specific heat capacity of water (4200 J/(kg K)), $\Delta T$ is the change in temperature.

Setting the potential energy loss equal to the heat gained: $mgh = ms \Delta T$ We can simplify this equation by canceling out the mass $m$ : $gh = s \Delta T$ Now, solve for $\Delta T$ : $\Delta T = \dfrac{gh}{s}$ Substitute the known values: $\Delta T = \dfrac{10 \times 200}{4200}$ Calculate $\Delta T$ : $\Delta T = \dfrac{2000}{4200}$ Simplify the fraction: $\Delta T = \dfrac{10}{21}$ Therefore, the rise in temperature is $\Delta T \approx 0.48 \, \text{K}$ .

Q276

In an adiabatic process, which of the following statements is true?

A The internal energy of the gas decreases as the temperature increases

B The molar heat capacity is zero

C Work done by the gas equals the increase in internal energy

D The molar heat capacity is infinite

Correct Answer

Option B

Solution

Option B is correct. • In an adiabatic process no heat is exchanged, so

\delta Q = 0.

• The (molar) heat capacity for any process is defined by

C = \frac{\delta Q}{n\,dT}\,.

Since $\delta Q=0$ even when $dT\neq0$ , it follows that

C_{\rm adiabatic}=0\,.

All other statements are therefore false.

Q277

The equation for real gas is given by

\left(\mathrm{P}+\dfrac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V}-\mathrm{b})=\mathrm{RT}

, where

\mathrm{P}, \mathrm{V}, \mathrm{T}

and R are the pressure, volume, temperature and gas constant, respectively. The dimension of

\mathrm{ab}^{-2}

is equivalent to that of :

A Compressibility

B Planck's constant

C Energy density

D Strain

Correct Answer

Option C

Solution

The equation for a real gas is given by: $\left(\mathrm{P} + \dfrac{\mathrm{a}}{\mathrm{V}^2}\right)(\mathrm{V} - \mathrm{b}) = \mathrm{RT}$ Here, $\mathrm{P}$ , $\mathrm{V}$ , $\mathrm{T}$ , and $R$ are the pressure, volume, temperature, and gas constant, respectively.

To find the dimension of $ab^{-2}$ , we proceed with the following steps: Rearrange the equation for dimensions: $[\mathrm{a}] = [\mathrm{P}][\mathrm{V}^2]$ Since $[\mathrm{P}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ and $[\mathrm{V}] = \mathrm{L}^3$ , we have: $[\mathrm{a}] = \mathrm{ML}^{-1}\mathrm{T}^{-2}(\mathrm{L}^6) = \mathrm{ML}^5\mathrm{T}^{-2}$ As derived from the equation: $[\mathrm{b}] = [\mathrm{V}] = \mathrm{L}^3$ Now, to find $[\mathrm{ab}^{-2}]$ : $[\mathrm{ab}^{-2}] = \dfrac{[\mathrm{a}]}{[\mathrm{b}]^2} = \mathrm{ML}^5\mathrm{T}^{-2} \cdot \mathrm{L}^{-6} = \mathrm{ML}^{-1}\mathrm{T}^{-2}$ This dimension, $\mathrm{ML}^{-1}\mathrm{T}^{-2}$ , is equivalent to the dimension of energy density.

Q278

$\text{ Match with . }$ $$ \begin{array}{lll} & \text { } & { }\\ \text { } \\ \text {

List - I		List - II
(A)	} & \text { Heat capacity of body } & \text {	(I)	} \mathrm{J} \mathrm{~kg}^{-1} \\ \text {
(B)	} & \text { Specific heat capacity of body } & \text {	(II)	} \mathrm{J} \mathrm{~K}^{-1} \\ \text {
(C)	} & \text { Latent heat } & \text {	(III)	} \mathrm{J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \\ \text {
(D)	} & \text { Thermal conductivity } & \text {	(IV)	} \mathrm{J} \mathrm{~m}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1} \end{array} \text {

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

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

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

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

Correct Answer

Option D

Solution

To match the entries in List-I with their corresponding units in List-II, we can use the definitions and formulas related to each concept in thermal physics: Heat Capacity of a Body (C') is defined as: $C' = \dfrac{\Delta Q}{\Delta T}$ The unit for heat capacity (C') is Joules per Kelvin (J K $^{-1}$ ).

Specific Heat Capacity of a Body (S) indicates how much heat energy is required to change the temperature of a unit mass of a substance by one degree Kelvin: $S = \dfrac{\Delta Q}{m \Delta T}$ The unit for specific heat capacity (S) is Joules per kilogram per Kelvin (J kg $^{-1}$ K $^{-1}$ ).

Latent Heat (L) is the heat energy absorbed or released during a phase change of a substance at a constant temperature: $L = \dfrac{\Delta Q}{m}$ The unit for latent heat (L) is Joules per kilogram (J kg $^{-1}$ ).

Thermal Conductivity (K) measures the ability of a material to conduct heat: $\Delta Q = \dfrac{KA \Delta T}{L} \quad \Rightarrow \quad K = \dfrac{\Delta Q \cdot L}{A \Delta T}$ The unit for thermal conductivity (K) is Joules per meter per Kelvin per second (J m $^{-1}$ K $^{-1}$ s $^{-1}$ ).

Using these formulas and their corresponding units, the correct matches are: (A) Heat capacity of body $\rightarrow$ (II) J K $^{-1}$ (B) Specific heat capacity of body $\rightarrow$ (III) J kg $^{-1}$ K $^{-1}$ (C) Latent heat $\rightarrow$ (I) J kg $^{-1}$ (D) Thermal conductivity $\rightarrow$ (IV) J m $^{-1}$ K $^{-1}$ s $^{-1}$

Q279

The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is: (Give: Molar mass of helium = 4 g/mol, Molar mass of argon = 40 g/mol)

A 1 :

\sqrt{10}

B 10 : 1

C 1 : 10

D 1 : 1

Correct Answer

Option D

Solution

Given the same room temperature of 300 K for helium and argon, we are tasked with finding the ratio of their average kinetic energies per molecule.

Formula for Kinetic Energy per Molecule: $\mathrm{K.E.} = \dfrac{\mathrm{f}}{2} \mathrm{kT}$ In this equation: $\mathrm{f}$ is the degrees of freedom, which is 3 for both helium (He) and argon (Ar) since they are monatomic gases. $k$ is the Boltzmann constant. $T$ is the temperature in Kelvin.

Kinetic Energy Comparison: For both helium and argon, since $\mathrm{f} = 3$ , the expression simplifies: $\dfrac{\mathrm{K} \cdot \mathrm{E}_{\mathrm{He}}}{\mathrm{K} \cdot \mathrm{E}_{\mathrm{Ar}}} = \dfrac{1}{1}$ Thus, the average kinetic energy per molecule for both gases is the same at the same temperature, leading to a ratio of: $1:1$ This result implies that despite their differences in molar masses (helium being 4 g/mol and argon 40 g/mol), the average kinetic energy per molecule remains equal at the same temperature.

Q280

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of

800 \mathrm{~cm}^3

and temperature

27^{\circ} \mathrm{C}

. The change in temperature when the gas is adiabatically compressed to

200 \mathrm{~cm}^3

is: (Take

\gamma=1.5 ; \gamma

is the ratio of specific heats at constant pressure and at constant volume)

A 300 K

B 600 K

C 327 K

D 522 K

Correct Answer

Option A

Solution

To determine the change in temperature when a gas is adiabatically compressed, we are given the following initial conditions: Initial volume, $V_1 = 800 \, \text{cm}^3$ Final volume, $V_2 = 200 \, \text{cm}^3$ Initial temperature, $T_1 = 27^\circ \text{C} = 300 \, \text{K}$ Adiabatic index (ratio of specific heats) $\gamma = 1.5$ In an adiabatic process, the relationship between temperature and volume is given by the equation: $TV^{\gamma-1} = \text{constant}$ Applying this to the initial and final states: $300 \times (800)^{0.5} = T_2 \times (200)^{0.5}$ Solving for the final temperature $T_2$ : $T_2 = 300 \times \left( \dfrac{800}{200} \right)^{0.5} = 300 \times (4)^{0.5}$ $T_2 = 300 \times 2 = 600 \, \text{K}$ Thus, the change in temperature is: $\Delta T = T_2 - T_1 = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K}$ Therefore, when the gas is adiabatically compressed, the temperature increases by 300 K.