Heat and Thermodynamics — Page 23 | JEE Physics

Q221

Work done by a Carnot engine operating between temperatures

127^{\circ} \mathrm{C}

and

27^{\circ} \mathrm{C}

is

2 \mathrm{~kJ}

. The amount of heat transferred to the engine by the reservoir is :

A 8 kJ

B 2 kJ

C 4 kJ

D 2.67 kJ

Correct Answer

Option A

Solution

The efficiency of a Carnot engine is given by:

\eta = 1 - \frac{T_{c}}{T_{h}}

where: $\eta$ is the efficiency of the Carnot engine,

T_{c}

is the temperature of the cold reservoir,

T_{h}

is the temperature of the hot reservoir.

Note that the temperatures must be in Kelvin for this formula.

Here, the given temperatures are

127^{\circ}C

and

27^{\circ}C

. Converting these to Kelvin gives:

T_{h} = 127^{\circ}C + 273.15 = 400.15 K

T_{c} = 27^{\circ}C + 273.15 = 300.15 K

So the efficiency of the Carnot engine is:

\eta = 1 - \frac{300.15}{400.15} = 0.25

The efficiency can also be defined as the work done divided by the heat supplied:

\eta = \frac{W}{Q_{h}}

where:

W

is the work done by the engine,

Q_{h}

is the heat transferred to the engine by the hot reservoir. We can rearrange this equation to solve for

Q_{h}

:

Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ

Q222

Given below are two statements: Statement I: If heat is added to a system, its temperature must increase. Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. In the light of the above statements, choose the correct answer from the options given below

A Both Statement I and Statement II are true

B Statement I is false but Statement II is true

C Statement I is true but Statement II is false

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

Statement I: If heat is added to a system, its temperature must increase.

This statement is not necessarily true.

For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature.

The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).

Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase.

This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.

Q223

The number of air molecules per cm

^3

increased from

3\times10^{19}

to

12\times10^{19}

. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

A 1.25

B 0.25

C 0.50

D 0.75

Correct Answer

Option B

Solution

1. The collision frequency (f) is given by the formula :

f = \sqrt{2} \pi d^2 v n_v

Where: - $d$ is the diameter of the molecule, - $v$ is the average velocity of the molecules, and - $n_v$ is the number density (number of molecules per unit volume).

2.

From this equation, we can see that the collision frequency (f) is directly proportional to the number density ( $n_v$ ), because all the other variables ( $d$ and $v$ ) are constant :

f \propto n_v

3.

Therefore, the ratio of two different collision frequencies ( $f_1$ and $f_2$ ) is equal to the ratio of their corresponding number densities ( $n_{v1}$ and $n_{v2}$ ):

\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}}

4.

Given that the number density increased from $3 \times 10^{19}$ to $12 \times 10^{19}$ , we can substitute these values into the equation to find the ratio of the collision frequencies:

\frac{f_1}{f_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}}

5. Simplifying this equation gives:

\frac{f_1}{f_2} = 0.25

So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.

Q224

A body cools in 7 minutes from

60^{\circ} \mathrm{C}

to

40^{\circ} \mathrm{C}

. The temperature of the surrounding is

10^{\circ} \mathrm{C}

. The temperature of the body after the next 7 minutes will be:

A

34^{\circ} \mathrm{C}

B

28^{\circ} \mathrm{C}

C

32^{\circ} \mathrm{C}

D

30^{\circ} \mathrm{C}

Correct Answer

Option B

Solution

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings.

The average rate of cooling can be represented as:

\frac{T_1-T_2}{t} = K \left(\frac{T_1+T_2}{2} - T_s\right)

where: $T_1$ and $T_2$ are the initial and final temperatures of the body, $t$ is the time it takes for the body to cool from $T_1$ to $T_2$ , $T_s$ is the temperature of the surroundings, and $K$ is a constant of proportionality.

In the first 7 minutes, the body cools from $60^\circ C$ to $40^\circ C$ , and the surrounding temperature is $10^\circ C$ .

So, the first equation is:

\frac{60-40}{7} = K \left(\frac{60+40}{2} - 10\right) \tag{1}

In the next 7 minutes, the body cools from $40^\circ C$ to $T^\circ C$ , with the same surrounding temperature.

So, the second equation is:

\frac{40-T}{7} = K \left(\frac{40+T}{2} - 10\right) \tag{2}

Solving equation (1) for $K$ and substituting into equation (2) will give the temperature $T$ after the next 7 minutes:

T = 28^\circ C

Q225

The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:

A

1: 1

B

1: 2

C

1: 4

D

4: 1

Correct Answer

Option D

Solution

The speed of sound in a gas is given by the formula:

v = \sqrt{\frac{\gamma RT}{M}}

where

v

is the speed of sound, $\gamma$ is the adiabatic index,

R

is the universal gas constant,

T

is the temperature, and

M

is the molar mass of the gas.

For diatomic gases, such as hydrogen (H₂) and oxygen (O₂), the adiabatic index $\gamma$ is the same, approximately equal to

\frac{7}{5}

, and the temperature is given as the same for both gases. Let's denote the speed of sound in hydrogen as

v_\text{H}

and in oxygen as

v_\text{O}

. The ratio of the speeds can be calculated as:

\frac{v_\text{H}}{v_\text{O}} = \sqrt{\frac{M_\text{O}}{M_\text{H}}}

The molar mass of hydrogen (H₂) is

2\, \text{g/mol}

, and the molar mass of oxygen (O₂) is

32\, \text{g/mol}

. Now, we can calculate the ratio of the speeds:

\frac{v_\text{H}}{v_\text{O}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

This means the ratio of the speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is

4:1

. Therefore, the correct answer is

4:1

.

Q226

A diatomic gas

(\gamma=1.4)

does

200 \mathrm{~J}

of work when it is expanded isobarically. The heat given to the gas in the process is :

A

800 \mathrm{~J}

B

600 \mathrm{~J}

C

700 \mathrm{~J}

D

850 \mathrm{~J}

Correct Answer

Option C

Solution

$\begin{aligned} & \gamma=1+\dfrac{2}{\mathrm{f}}=1.4 \Rightarrow \dfrac{2}{\mathrm{f}}=0.4 \\\\ & \Rightarrow \mathrm{f}=5 \\\\ & \mathrm{~W}=\mathrm{nR} \Delta \mathrm{T}=200 \mathrm{~J} \\\\ & \mathrm{Q}=\left(\dfrac{\mathrm{f}+2}{2}\right) \mathrm{nR} \Delta \mathrm{T} \\\\ & =\dfrac{7}{2} \times 200=700 \mathrm{~J}\end{aligned}$

Q227

If the root mean square velocity of hydrogen molecule at a given temperature and pressure is

2 \mathrm{~km} / \mathrm{s}

, the root mean square velocity of oxygen at the same condition in

\mathrm{km} / \mathrm{s}

is :

A 1.0

B 1.5

C 2.0

D 0.5

Correct Answer

Option D

Solution

Here is your text with LaTeX notation and paragraph tags converted as requested: To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:

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

where:

v_{\text{rms}}

is the root mean square velocity of the gas molecules,

k

is the Boltzmann constant,

T

is the absolute temperature in Kelvin, and

m

is the mass of one molecule of the gas.

Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.

Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen.

The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.

We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules.

Using the rms velocity formula and considering the ratio of masses:

\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{m_{\text{O₂}}}{m_{\text{H₂}}}}

We know the mass m is proportional to the molecular weight for each gas, so we can substitute:

\frac{v_{\text{rms, H₂}}}{v_{\text{rms, O₂}}} = \sqrt{\frac{M_{\text{O₂}}}{M_{\text{H₂}}}}

Now we plug in the values:

\frac{2 \text{ km/s}}{v_{\text{rms, O₂}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

Solving for

v_{\text{rms, O₂}}

:

v_{\text{rms, O₂}} = \frac{2 \text{ km/s}}{4} = 0.5 \text{ km/s}

Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.

The correct answer is Option D : 0.5.

Q228

Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

A

\dfrac{3}{2} \mathrm{R}

B

\dfrac{7}{4} \mathrm{R}

C

\dfrac{5}{2} \mathrm{R}

D

\dfrac{9}{4} \mathrm{R}

Correct Answer

Option D

Solution

To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).

Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.

For a monoatomic ideal gas, the molar specific heat at constant volume is:

C_{V,m1} = \frac{3}{2}R

For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:

C_{V,m2} = \frac{5}{2}R

The weighted average of the molar specific heat for the mixture is:

C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2}

Putting the values into the equation, we get:

C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6}

C_{V,m} = \frac{(3R + 15R)}{8}

C_{V,m} = \frac{18R}{8}

C_{V,m} = \frac{9}{4}R

Therefore, the molar specific heat of the mixture at constant volume is $\dfrac{9}{4}R$ . The correct answer is: Option D

\frac{9}{4}R

Q229

If three moles of monoatomic gas

\left(\gamma=\dfrac{5}{3}\right)

is mixed with two moles of a diatomic gas

\left(\gamma=\dfrac{7}{5}\right)

, the value of adiabatic exponent

\gamma

for the mixture is

A 1.35

B 1.52

C 1.40

D 1.75

Correct Answer

Option B

Solution

\begin{array}{ll} \mathrm{f}_1=3, & \mathrm{f}_2=5 \\ \mathrm{n}_1=3, & \mathrm{n}_2=2 \end{array}

\begin{aligned} & \mathrm{f}_{\text{mixture }}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{9+10}{\mathrm{f}}=\frac{19}{5} \\ & \gamma_{\text{mixture }}=1+\frac{2 \times 5}{19}=\frac{29}{19}=1.52 \end{aligned}

Q230

At which temperature the r.m.s. velocity of a hydrogen molecule equal to that of an oxygen molecule at

47^{\circ} \mathrm{C}

?

A 20 K

B 80 K

C 4 K

D

-73

K

Correct Answer

Option A

Solution

\begin{aligned} & \sqrt{\frac{3 R T}{2}}=\sqrt{\frac{3 R(320)}{32}} \\ & T=\frac{320}{16}=20 \mathrm{~K} \end{aligned}