Heat and Thermodynamics — Page 22 | JEE Physics

Q211

If the r. m.s speed of chlorine molecule is

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

at

27^{\circ} \mathrm{C}

, the r. m. s speed of argon molecules at the same temperature will be (Atomic mass of argon

=39.9 \mathrm{u}

, molecular mass of chlorine

=70.9 \mathrm{u}

)

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

The correct relationship between the rms speeds of the two gases is:

\frac{v_{\mathrm{Ar}}}{v_{\mathrm{Cl}}} = \sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}

Given the molar masses for argon and chlorine:

M_{\mathrm{Ar}} = 39.9 \mathrm{u}

M_{\mathrm{Cl}_2} = 70.9 \mathrm{u}

And the rms speed of chlorine molecules:

v_{\mathrm{Cl}} = 490 \mathrm{~m} / \mathrm{s}

We can now solve for the rms speed of argon molecules:

v_{\mathrm{Ar}} = \sqrt{\frac{70.9}{39.9}} \times 490

v_{\mathrm{Ar}} \approx 651.7 \mathrm{~m} / \mathrm{s}

The rms speed of argon molecules at the same temperature as the chlorine molecules is approximately

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

.

Q212

The Thermodynamic process, in which internal energy of the system remains constant is

A Isobaric

B Isochoric

C Adiabatic

D Isothermal

Correct Answer

Option D

Solution

If the temperature (T) remains constant, the internal energy (U) also remains constant, since the internal energy of an ideal gas depends only on its temperature.

In this case, the thermodynamic process in which the internal energy of the system remains constant is an isothermal process.

Isothermal processes occur at constant temperature, and for an ideal gas, this means that the internal energy remains constant as well.

Q213

The root mean square speed of molecules of nitrogen gas at

27^{\circ} \mathrm{C}

is approximately : (Given mass of a nitrogen molecule

=4.6 \times 10^{-26} \mathrm{~kg}

and take Boltzmann constant

\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}

)

A 91 m/s

B 1260 m/s

C 27.4 m/s

D 523 m/s

Correct Answer

Option D

Solution

To find the root mean square speed of molecules of nitrogen gas, we can use the formula:

v_{rms} = \sqrt{\frac{3k_BT}{m}}

where

v_{rms}

is the root mean square speed,

k_B

is the Boltzmann constant,

T

is the temperature in Kelvin, and

m

is the mass of a nitrogen molecule. First, we need to convert the temperature from Celsius to Kelvin:

T = 27^{\circ}\mathrm{C} + 273 = 300\,\mathrm{K}

Now, substitute the given values of

k_B

,

T

, and

m

into the formula:

v_{rms} = \sqrt{\frac{3(1.4 \times 10^{-23}\, \mathrm{JK}^{-1})(300\,\mathrm{K})}{4.6 \times 10^{-26}\,\mathrm{kg}}}

Simplify and calculate the root mean square speed:

v_{rms} = 523\,\mathrm{m/s}

The root mean square speed of molecules of nitrogen gas at

27^{\circ}\mathrm{C}

is 523 m/s.

Q214

The temperature of an ideal gas is increased from

200 \mathrm{~K}

to

800 \mathrm{~K}

. If r.m.s. speed of gas at

200 \mathrm{~K}

is

v_{0}

. Then, r.m.s. speed of the gas at

800 \mathrm{~K}

will be:

A

v_{0}

B

2 v_{0}

C

4 v_{0}

D

\dfrac{v_{0}}{4}

Correct Answer

Option B

Solution

The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:

v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.

In this case, we are given that the initial temperature is

200 \, K

and the final temperature is

800 \, K

. Let the r.m.s speed at

200 \, K

be

v_0

, then:

v_0 = \sqrt{\frac{3R \cdot 200}{M}}

Now, we want to find the r.m.s speed at

800 \, K

, let's call this

v_1

:

v_1 = \sqrt{\frac{3R \cdot 800}{M}}

Now, divide

v_1

by

v_0

:

\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}

Simplify the expression:

\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2

So,

v_1 = 2v_0

. Hence, the r.m.s speed of the gas at

800 \, K

will be 2 times the r.m.s speed of the gas at

200 \, K

.

Q215

1 \mathrm{~kg}

of water at

100^{\circ} \mathrm{C}

is converted into steam at

100^{\circ} \mathrm{C}

by boiling at atmospheric pressure. The volume of water changes from

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

as a liquid to

1.671 \mathrm{~m}^{3}

as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisaiton

=2257 \mathrm{~kJ} / \mathrm{kg}

, Atmospheric pressure =

\left.1 \times 10^{5} \mathrm{~Pa}\right)

A + 2090 kJ

B

-

2426 kJ

C + 2476 kJ

D

-

2090 kJ

Correct Answer

Option A

Solution

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ( $Q$ ) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

Q = m \times L

where

m = 1 \mathrm{~kg}

and

L = 2257 \mathrm{~kJ/kg}

:

Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}

Next, let's calculate the work done ( $W$ ) on the system during the process. The work done is given by:

W = -P \Delta V

where

P

is the atmospheric pressure and

\Delta V

is the change in volume. We are given that the atmospheric pressure is

P = 1 \times 10^5 \mathrm{~Pa}

, and the change in volume is

\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3

. Now, we can calculate the work done:

W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}

Finally, we can find the change in internal energy ( $\Delta U$ ) using the first law of thermodynamics:

\Delta U = Q + W

Substitute the values of

Q

and

W

:

\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}

The change in internal energy of the system during the process is +2090 kJ.

Q216

On a temperature scale '

\mathrm{X}

', the boiling point of water is

65^{\circ} \mathrm{X}

and the freezing point is

-15^{\circ} \mathrm{X}

. Assume that the

\mathrm{X}

scale is linear. The equivalent temperature corresponding to

-95^{\circ} \mathrm{X}

on the Farenheit scale would be:

A

-148^{\circ} \mathrm{F}

B

-48^{\circ} \mathrm{F}

C

-63^{\circ} \mathrm{F}

D

-112^{\circ} \mathrm{F}

Correct Answer

Option A

Solution

We are given two temperature scales: the X-scale and the Celsius scale.

The relationship between two linear temperature scales can be expressed as follows:

\frac{X - X_{\text{freeze}}}{X_{\text{boil}} - X_{\text{freeze}}} = \frac{C - C_{\text{freeze}}}{C_{\text{boil}} - C_{\text{freeze}}}

Here,

X_{\text{freeze}}

and

X_{\text{boil}}

are the freezing and boiling points of water on the X-scale, while

C_{\text{freeze}}

and

C_{\text{boil}}

are the freezing and boiling points of water on the Celsius scale. We are given the following values: X-scale:

X_{\text{boil}} = 65^{\circ} \mathrm{X}

X_{\text{freeze}} = -15^{\circ} \mathrm{X}

Celsius scale:

C_{\text{boil}} = 100^{\circ} \mathrm{C}

C_{\text{freeze}} = 0^{\circ} \mathrm{C}

Our goal is to find the equivalent temperature of

-95^{\circ} \mathrm{X}

on the Fahrenheit scale. Step 1: Convert

-95^{\circ} \mathrm{X}

to Celsius: Use the relationship between X-scale and Celsius scale:

\frac{-95 - (-15)}{65 - (-15)} = \frac{C - 0}{100 - 0}

Simplify and solve for C:

\frac{-80}{80} = \frac{C}{100}

C = -100^{\circ} \mathrm{C}

Step 2: Convert

-100^{\circ} \mathrm{C}

to Fahrenheit: Use the conversion formula between Celsius and Fahrenheit:

T_F = \frac{9}{5}T_C + 32

Substitute the Celsius temperature:

T_F = \frac{9}{5} \times (-100) + 32

T_F = -180 + 32

T_F = -148^{\circ} \mathrm{F}

So, the equivalent temperature corresponding to

-95^{\circ} \mathrm{X}

on the Fahrenheit scale is

-148^{\circ} \mathrm{F}

.

Q217

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed

\left(v_{\mathrm{rms}}\right)

and choose the correct answer from the options given below:

A

\mathrm{v}_{\mathrm{rms}}(

mono

)=\mathrm{v}_{\mathrm{rms}}(

dia

)=\mathrm{v}_{\mathrm{rms}}(

poly

)

B

\mathrm{v}_{\mathrm{rms}}

(mono)

> \mathrm{v}_{\mathrm{rms}}(

dia

) > \mathrm{v}_{\mathrm{rms}}

(poly)

C

\mathrm{v}_{\mathrm{rms}}

(dia)

< \mathrm{v}_{\mathrm{rms}}

(poly)

< \mathrm{v}_{\text{rms }}

(mono)

D

\mathrm{v}_{\mathrm{rms}}

(mono)

< \mathrm{v}_{\mathrm{rms}}

(dia)

< \mathrm{v}_{\mathrm{rms}}

(poly)

Correct Answer

Option B

Solution

The root mean square speed (

v_{rms}

) of a gas is given by the formula:

v_{rms} = \sqrt{\frac{3kT}{m}}

where

k

is the Boltzmann constant,

T

is the temperature, and

m

is the molar mass of the gas molecules.

The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases.

Their molar masses are: Neon:

20.18\,\text{g/mol}

(monoatomic) Chlorine:

2 \times 35.45 = 70.90\,\text{g/mol}

(diatomic) Uranium hexafluoride:

238.03 + 6 \times 18.998 = 352.03\,\text{g/mol}

(polyatomic) Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:

v_{rms} \propto \frac{1}{\sqrt{m}}

The lighter the gas, the higher its root mean square speed.

Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest.

Therefore, the root mean square speeds will be:

v{rms}(\text{mono}) > v{rms}(\text{dia}) > v_{rms}(\text{poly})

Q218

A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be,

A 4RT

B 16RT

C 8RT

D 11RT

Correct Answer

Option D

Solution

The internal energy (U) of a gas depends on its degrees of freedom (f).

For a monatomic gas like neon, the degrees of freedom are f = 3 (translational).

For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).

The internal energy for each component of the gas mixture can be calculated using the formula:

U = \frac{f}{2}nRT

where n is the number of moles, R is the universal gas constant, and T is the temperature.

For the 4 moles of neon (monatomic):

U_{Ne} = \frac{3}{2} \cdot 4RT = 6RT

For the 2 moles of oxygen (diatomic):

U_{O_2} = \frac{5}{2} \cdot 2RT = 5RT

Now, to find the total internal energy, we sum the internal energies of the individual components:

U_{total} = U_{Ne} + U_{O_2} = 6RT + 5RT = 11RT

Thus, the total internal energy of the system is 11RT.

Q219

A gas is compressed adiabatically, which one of the following statement is NOT true.

A There is no heat supplied to the system

B The temperature of the gas increases.

C There is no change in the internal energy

D The change in the internal energy is equal to the work done on the gas.

Correct Answer

Option C

Solution

An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings.

This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.

Given the options: Option A: There is no heat supplied to the system.

This statement is TRUE.

In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.

Option B: The temperature of the gas increases.

This statement is TRUE.

When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.

Option C: There is no change in the internal energy.

This statement is NOT TRUE.

In an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

When the gas is compressed, work is done on the gas, which increases its internal energy.

Option D: The change in the internal energy is equal to the work done on the gas.

This statement is TRUE.

As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.

So, Option C is the statement that is not true for an adiabatic process.

Q220

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at

27^{\circ} \mathrm{C}

is

A

627^{\circ} \mathrm{C}

B

927^{\circ} \mathrm{C}

C

327^{\circ} \mathrm{C}

D

1227^{\circ} \mathrm{C}

Correct Answer

Option C

Solution

The kinetic energy of an ideal gas is given by the equation: $KE = \dfrac{3}{2} kT$ where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins.

Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double.

The original temperature is given as ( $27^\circ C$ ), which is equal to (300 K) in absolute terms.

Therefore, the final temperature ( $T_f$ ) in kelvins is: $T_f = 2 \cdot 300 K = 600 K$ Converting this back to degrees Celsius gives: $T_f = 600K - 273 = 327 ^\circ C$