Heat and Thermodynamics — Page 25 | JEE Physics

Q241

The temperature of a gas having

2.0 \times 10^{25}

molecules per cubic meter at

1.38 \mathrm{~atm}

(Given,

\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}

) is :

A 500 K

B 300 K

C 200 K

D 100 K

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{PV}=\mathrm{nRT} \\ & \mathrm{PV}=\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{A}}} \mathrm{RT} \\ & \mathrm{N}=\text{ Total no. of molecules } \\ & \mathrm{P}=\frac{\mathrm{N}}{\mathrm{V}} \mathrm{kT} \\ & 1.38 \times 1.01 \times 10^5=2 \times 10^{25} \times 1.38 \times 10^{-23} \times \mathrm{T} \\ & 1.01 \times 10^5=2 \times 10^2 \times \mathrm{T} \\ & \mathrm{T}=\frac{1.01 \times 10^3}{2} \approx 500 \mathrm{~K} \end{aligned}

Q242

The temperature of a gas is

-78^{\circ} \mathrm{C}

and the average translational kinetic energy of its molecules is

\mathrm{K}

. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes

2 \mathrm{~K}

is :

A

-78^{\circ} \mathrm{C}

B

127^{\circ} \mathrm{C}

C

-39^{\circ} \mathrm{C}

D

117^{\circ} \mathrm{C}

Correct Answer

Option D

Solution

The average translational kinetic energy ( $K_{avg}$ ) of a molecule is directly proportional to the absolute temperature (T) of the gas, as described by the equation: $K_{avg} = \dfrac{3}{2}kT$ Where: $K_{avg}$ is the average kinetic energy, $k$ is the Boltzmann's constant, and $T$ is the temperature in Kelvin.

From the given problem, if the temperature of the gas is $-78^{\circ}C$ , which in Kelvin is $T_1 = -78 + 273 = 195K$ , and the average translational kinetic energy is $K$ , when the energy becomes $2K$ , we need to find the new temperature $(T_2)$ .

Using the direct proportionality relation, we can set up the following equation: $K \propto T$ $2K = K_2 = \dfrac{3}{2}kT_2$ Given that at $T_1$ , the kinetic energy is $K$ , and at $T_2$ , it's $2K$ , we can use the ratio as follows: $\dfrac{K_2}{K_1} = \dfrac{2K}{K} = 2 = \dfrac{T_2}{T_1}$ Thus, we have: $T_2 = 2T_1 = 2 \times 195K = 390K$ To find the temperature in Celsius, we convert $390K$ back to Celsius: $T_{2(Celsius)} = 390 - 273 = 117^{\circ}C$ Therefore, the temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is $117^{\circ}C$ .

The correct answer is: Option D

117^{\circ} \mathrm{C}

Q243

The volume of an ideal gas

(\gamma=1.5)

is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is :

A

\dfrac{4}{5}

B

\dfrac{8}{5 \sqrt{5}}

C

\dfrac{2}{\sqrt{5}}

D

\dfrac{16}{25}

Correct Answer

Option B

Solution

To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows: $P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$ Here, $P_{1}$ and $V_{1}$ are the initial pressure and volume, respectively, $P_{2}$ and $V_{2}$ are the final pressure and volume, respectively, and $\gamma$ is the heat capacity ratio of the gas.

Given in the question, $\gamma = 1.5$ , $V_{1} = 5$ litres and $V_{2} = 4$ litres.

We need to find the ratio $\dfrac{P_{1}}{P_{2}}$ .

Rearranging the adiabatic equation for the ratio $\dfrac{P_{1}}{P_{2}}$ , we get: $P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma} \Rightarrow \dfrac{P_{1}}{P_{2}} = \left(\dfrac{V_{2}}{V_{1}}\right)^{\gamma}$ Substituting the given values: $\dfrac{P_{1}}{P_{2}} = \left(\dfrac{V_{2}}{V_{1}}\right)^{\gamma} = \left(\dfrac{4}{5}\right)^{1.5}$ To calculate the value: $\dfrac{P_{1}}{P_{2}} = \left(\dfrac{4}{5}\right)^{1.5} = \left(\dfrac{4}{5}\right)^{\dfrac{3}{2}}$ Simplifying further: $\dfrac{P_{1}}{P_{2}} = \left(\dfrac{2^2}{5}\right)^{\dfrac{3}{2}} = \left(\dfrac{2^3}{5^{\dfrac{3}{2}}}\right) = \dfrac{8}{5\sqrt{5}}$ Therefore, the ratio of the initial pressure to the final pressure is $\dfrac{8}{5\sqrt{5}}$ , which corresponds to Option B.

Q244

A sample of 1 mole gas at temperature

T

is adiabatically expanded to double its volume. If adiab constant for the gas is

\gamma=\dfrac{3}{2}

, then the work done by the gas in the process is :

A

\mathrm{R} \mathrm{T}[2+\sqrt{2}]

B

\mathrm{RT}[2-\sqrt{2}]

C

\dfrac{\mathrm{R}}{\mathrm{T}}[2-\sqrt{2}]

D

\dfrac{T}{R}[2+\sqrt{2}]

Correct Answer

Option B

Solution

For an adiabatic process, the work done by the gas can be found using the formula:

W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}

Given that the volume is doubled (

V_2 = 2V_1

) and the adiabatic constant

\gamma = \frac{3}{2}

, we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $\gamma$ .

Recall, for an adiabatic process,

PV^{\gamma} = \text{constant}

, so we can write:

P_1V_1^{\gamma} = P_2V_2^{\gamma}

Using the fact that

V_2 = 2V_1

, we can express

P_2

in terms of

P_1

and

V_1

as follows:

P_1V_1^{\gamma} = P_2(2V_1)^{\gamma}

P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma}

P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma}

The work done then becomes:

W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1}

W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1}

Plugging in

\gamma = \frac{3}{2}

, we get:

W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1}

W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}}

W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}})

Since

P_1V_1 = nRT

for 1 mole (

n = 1

) of gas at temperature

T

, we can further simplify:

W = 2RT(1 - \sqrt{\frac{1}{2}})

W = 2RT(1 - \frac{1}{\sqrt{2}})

W = RT(2 - \sqrt{2})

Therefore, the correct option is: Option B

\mathrm{RT}[2-\sqrt{2}]

.

Q245

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are

8 \Omega

and

10 \Omega

respectively. After inserting in a hot bath of temperature

400^{\circ} \mathrm{C}

, the resistance of platinum wire is :

A 10

\Omega

B 16

\Omega

C 8

\Omega

D 2

\Omega

Correct Answer

Option B

Solution

The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

R_t = R_0(1 + \alpha t)

where:

R_t

is the resistance at temperature

t

,

R_0

is the resistance at 0°C (ice point), $\alpha$ is the temperature coefficient of resistance, and

t

is the temperature in degrees Celsius. In this question, we are given: Resistance at the ice point (

R_{0} = 8 \Omega

), Resistance at the steam point (

R_{100} = 10 \Omega

).

To find the temperature coefficient of resistance ( $\alpha$ ), we use the resistance values at the ice and steam points:

\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C

Now, to find the resistance

R_t

at

400 ^{\circ}C

, we substitute $\alpha$ ,

R_0

, and

t = 400

into the formula:

R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega

Hence, the resistance of the platinum wire at

400^{\circ}C

is

16 \Omega

. The correct option is: Option B: 16

\Omega

Q246

A Carnot engine

(\mathrm{E})

is working between two temperatures 473 K and 273 K . In a new system two engines - engine

E_1

works between 473 K to 373 K and engine

E_2

works between 373 K to 273 K . If

\eta_{12}, \eta_1

and

\eta_2

are the efficiencies of the engines

E, E_1

and

E_2

, respectively, then

A

\eta_{12}=\eta_1 \eta_2

B

\eta_{12}=\eta_1+\eta_2

C

\eta_{12} \geq \eta_1+\eta_2

D

\eta_{12}<\eta_1+\eta_2

Correct Answer

Option D

Solution

We know, efficiency of a carnot engine, $\eta$ or

e = 1 - {{{T_C}} \over {{T_H}}}

where, T $_C$ = temperature of cold sink T $_H$ = temperature of hot source So,

{\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423

{\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211

{\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268

Here,

{\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423

So, $${\eta _{12}}

Q247

On celcius scale the temperature of body increases by

40^{\circ} \mathrm{C}

. The increase in temperature on Fahrenheit scale is :

A

75^{\circ} \mathrm{F}

B

70^{\circ} \mathrm{F}

C

72^{\circ} \mathrm{F}

D

68^{\circ} \mathrm{F}

Correct Answer

Option C

Solution

To find the increase in temperature on the Fahrenheit scale, we use the relationship between the Celsius and Fahrenheit temperature scales.

The formula to convert Celsius to Fahrenheit is:

F = \frac{9}{5}C + 32

However, since we are interested in the increase in temperature, we can ignore the "+ 32" part of the formula, because this constant does not affect the change in temperature, only the absolute temperatures.

Thus, to find the increase in temperature on the Fahrenheit scale, we can use:

\Delta F = \frac{9}{5}\Delta C

Given that the increase in temperature is

40^{\circ}C

, we can substitute this value into the equation:

\Delta F = \frac{9}{5} \times 40^{\circ}C

\Delta F = 9 \times 8

\Delta F = 72^{\circ}F

Therefore, the increase in temperature on the Fahrenheit scale is

72^{\circ}F

. So, the correct answer is: Option C

72^{\circ} \mathrm{F}

Q248

A sample of gas at temperature

T

is adiabatically expanded to double its volume. Adiabatic constant for the gas is

\gamma=3 / 2

. The work done by the gas in the process is:

(\mu=1 \text{ mole })

A

R T[2 \sqrt{2}-1]

B

R T[2-\sqrt{2}]

C

R T[1-2 \sqrt{2}]

D

R T[\sqrt{2}-2]

Correct Answer

Option B

Solution

\begin{aligned} & w=\frac{-n R}{\gamma-1}(\Delta T) \\ &=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\ &=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\ &=R T(2-\sqrt{2}) \\ & \therefore \quad T V_\gamma^{-1}=\text{ cons. } \\ & T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\ & T_f=\frac{T}{\sqrt{2}} \end{aligned}

Q249

The translational degrees of freedom

\left(f_t\right)

and rotational degrees of freedom

\left(f_r\right)

of

\mathrm{CH}_4

molecule are:

A

f_t=2

and

f_r=2

B

f_t=3

and

f_r=3

C

f_t=3

and 4

f_r=2

$

D

f_t=2

and

f_r=3

Correct Answer

Option B

Solution

For non-linear polyatomic molecules, both translational and rotational degree of freedom have same value and is equal to 3.

Q250

A diatomic gas

(\gamma=1.4)

does

100 \mathrm{~J}

of work in an isobaric expansion. The heat given to the gas is :

A 150 J

B 490 J

C 350 J

D 250 J

Correct Answer

Option C

Solution

To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system.

The first law of thermodynamics states that:

\Delta Q = \Delta U + W

where: $\Delta Q$ is the heat added to the system, $\Delta U$ is the change in internal energy of the system, and $W$ is the work done by the gas.

For an isobaric process, the work done $W$ is given by:

W = P \Delta V

We're given that $W = 100 \ \mathrm{J}$ for this process, so:

W = 100 \ \mathrm{J}

The change in internal energy $\Delta U$ for an ideal gas can also be related to the temperature change and the specific heat capacity at constant volume $C_v$ .

Using the equation:

\Delta U = nC_v\Delta T

However, without direct values for $n$ , $\Delta T$ , or $C_v$ , we need to rely on the relation between the provided work and the heat capacity ratio $\gamma$ to find the heat added.

For a diatomic gas, $\gamma = C_p/C_v$ .

The heat added at constant pressure can also be described as:

\Delta Q = nC_p\Delta T

Since we know that for an ideal gas, the work done on the gas during an isobaric process is related to the heat added by the ratio of the specific heats ( $\gamma$ ), we can use the fact that $C_p = \gamma C_v$ , and relating that to the work done, we get:

\Delta Q = \frac{\gamma}{\gamma - 1} W

Substituting the known values, with $\gamma = 1.4$ , and $W = 100 \ \mathrm{J}$ , we find:

\Delta Q = \frac{1.4}{1.4 - 1} \times 100 \ \mathrm{J} = \frac{1.4}{0.4} \times 100 \ \mathrm{J} = 3.5 \times 100 \ \mathrm{J} = 350 \ \mathrm{J}

Therefore, the heat given to the gas is $\Delta Q = 350 \ \mathrm{J}$ , so the correct option is: Option C 350 J