Heat and Thermodynamics — Page 21 | JEE Physics

Q201

1 g of a liquid is converted to vapour at 3

\times

10

^5

Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm

^3

during this phase change, then the increase in internal energy in the process will be :

A 4800 J

B 4320 J

C 432000 J

D 4.32

\times

10

^8

J

Correct Answer

Option B

Solution

Work done = P $\Delta$ V = 3 × 105 × 1600 × 10–6 = 480 J Only 10% of heat is used in work done.

Hence $\Delta$ Q = 4800 J The rest goes in internal energy, which is 90% of heat.

Change in internal energy = 0.9 × 4800 = 4320 J

Q202

Given below are two statements : Statement I : The temperature of a gas is

-73^\circ

C. When the gas is heated to

527^\circ

C, the root mean square speed of the molecules is doubled. Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. In the light of the above statements, choose the correct answer from the option given below :

A Statement I is true but Statement II is false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Both Statement I and Statement II are false

Correct Answer

Option A

Solution

$T_{i}=200 \mathrm{~K} \quad\quad v_{\mathrm{rms}} \propto \sqrt{T}$ $T_{f}=800 \mathrm{~K}$ $\dfrac{V_{i}}{V_{f}}=\sqrt{\dfrac{T_{i}}{T_{f}}}=\sqrt{\dfrac{200}{800}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$ $V_{f}=2 V_{i}$ Translational K.E.

$=\left(\dfrac{3}{2} P V\right)$

Q203

A flask contains Hydrogen and Argon in the ratio

2: 1

by mass. The temperature of the mixture is

30^{\circ} \mathrm{C}

. The ratio of average kinetic energy per molecule of the two gases (

\mathrm{K}

argon/K hydrogen) is : (Given: Atomic Weight of

\mathrm{Ar}=39.9

)

A

\dfrac{39.9}{2}

B 2

C 39.9

D 1

Correct Answer

Option D

Solution

The average kinetic energy per molecule of a gas is given by the expression $\dfrac{3}{2}kT$ , where $k$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvins.

Since the temperature of the mixture is given in Celsius, we need to convert it to kelvins by adding 273.15 to get $303.15$ K.

Let us assume that the total mass of the mixture is $3x$ (where $x$ is a constant), then the mass of hydrogen and argon in the mixture will be $2x$ and $x$ respectively, according to the given ratio.

The number of moles of hydrogen and argon can be calculated using their respective masses and molar masses, which are 1 g/mol and 39.9 g/mol respectively.

Therefore: Number of moles of hydrogen = $\dfrac{2x}{1~\mathrm{g/mol}} = 2x$ mol Number of moles of argon = $\dfrac{x}{39.9~\mathrm{g/mol}} = \dfrac{x}{39.9}$ mol The total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon: Total number of moles of gas = $2x + \dfrac{x}{39.9} = \dfrac{79.9x}{39.9}$ mol The average kinetic energy per molecule of hydrogen is: $\dfrac{3}{2}kT_{\mathrm{H_2}} = \dfrac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$ The average kinetic energy per molecule of argon is: $\dfrac{3}{2}kT_{\mathrm{Ar}} = \dfrac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$ Therefore, the ratio of the average kinetic energy per molecule of argon to hydrogen is: $\dfrac{K_{\mathrm{Ar}}}{K_{\mathrm{H_2}}} = \dfrac{\dfrac{3}{2}kT_{\mathrm{Ar}}}{\dfrac{3}{2}kT_{\mathrm{H_2}}} = \dfrac{6.12 \times 10^{-21}~\mathrm{J}}{6.12 \times 10^{-21}~\mathrm{J}} = 1$ Hence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$ .

Q204

The initial pressure and volume of an ideal gas are P

_0

and V

_0

. The final pressure of the gas when the gas is suddenly compressed to volume

\dfrac{V_0}{4}

will be : (Given

\gamma

= ratio of specific heats at constant pressure and at constant volume)

A P

_0

(4)

^{\dfrac{1}{\gamma}}

B P

_0

C 4P

_0

D P

_0

(4)

^{\gamma}

Correct Answer

Option D

Solution

When the gas is compressed suddenly, it undergoes an adiabatic process where no heat is exchanged with the surroundings.

Therefore, we can use the adiabatic equation of state to relate the initial and final pressure and volume of the gas:

P_0V_0^\gamma=P_fV_f^\gamma

where

P_f

and

V_f

are the final pressure and volume of the gas, respectively. Since the gas is compressed to

\frac{V_0}{4}

, we have:

V_f=\frac{V_0}{4}

Substituting this into the adiabatic equation of state, we get:

P_f=P_0\left(\frac{V_0}{V_f}\right)^\gamma=P_0\left(\frac{4}{1}\right)^\gamma=4^\gamma P_0

Therefore, the final pressure of the gas when it is suddenly compressed to

\frac{V_0}{4}

is

4^\gamma P_0

.

Q205

The mean free path of molecules of a certain gas at STP is

1500 \mathrm{~d}

, where

\mathrm{d}

is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at

373 \mathrm{~K}

is approximately:

A

750 \mathrm{~d}

B

1500 \mathrm{~d}

C

\mathrm{2049~ d}

D

1098 \mathrm{~d}

Correct Answer

Option C

Solution

The mean free path (λ) of molecules in a gas is given by the formula:

\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}

where k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure.

At STP (standard temperature and pressure), the temperature is

273\mathrm{~K}

and the pressure is

1\mathrm{~atm}

. We are given that the mean free path at STP is

1500d

. Let's denote the mean free path at

373\mathrm{~K}

as λ':

\lambda' = \frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}

To find the ratio of the mean free path at

373\mathrm{~K}

to that at STP, we can divide λ' by λ:

\frac{\lambda'}{\lambda} = \frac{\frac{k(373\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}{\frac{k(273\mathrm{~K})}{\sqrt{2}\pi d^2 (1\mathrm{~atm})}}

The Boltzmann constant, pressure, and molecular diameter cancel out:

\frac{\lambda'}{\lambda} = \frac{373\mathrm{~K}}{273\mathrm{~K}}

Now, we can solve for λ':

\lambda' = \lambda \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}

Substituting the given value of λ as

1500d

:

\lambda' = 1500d \cdot \frac{373\mathrm{~K}}{273\mathrm{~K}}

\lambda' \approx 2049d

Thus, the mean free path of the molecules at

373\mathrm{~K}

while maintaining the standard pressure is approximately

2049d

.

Q206

Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V) and Temperature (T). The gas in A is compressed isothermally to

\dfrac{1}{8}

of its original volume while the gas in B is compressed adiabatically to

\dfrac{1}{8}

of its original volume. The ratio of final pressure of gas in B to that of gas in A is

A

\dfrac{1}{8}

B 8

^\dfrac{3}{2}

C 4

D 8

Correct Answer

Option C

Solution

The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$ , where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$ .

The final volume is $\dfrac{1}{8}$ of the initial volume, so $P_2 = P_1 \times \dfrac{V_1}{V_2} = P_1 \times 8 = 8P_1$ .

The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\gamma = \text{constant}$ , where $\gamma$ is the ratio of the heat capacities, which is $\dfrac{5}{3}$ for a monoatomic gas.

Since $V_2 = \dfrac{V_1}{8}$ , we have $P_2 = P_1 \times \left(\dfrac{V_1}{V_2}\right) ^ \gamma = P_1 \times 8^\dfrac{5}{3} = P_1 \times 2^5 = 32P_1$ .

The ratio of the final pressure of gas in B to that of gas in A is therefore $\dfrac{32P_1}{8P_1} = 4$ .

Q207

A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms

-

1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron = 0.42 Jg

-

1

^\circ

C

-

1]

A 675

^\circ

C

B 1600

^\circ

C

C 16.07

^\circ

C

D 6.75

^\circ

C

Correct Answer

Option C

Solution

{1 \over 2} \times 1.5 \times {60^2} \times {1 \over 4} = 100 \times 0.42 \times \Delta T

\Delta T = {{1.5 \times {{60}^2}} \over {8 \times 100 \times 0.42}} = 16.07^\circ C

Q208

The rms speed of oxygen molecule in a vessel at particular temperature is

\left(1+\dfrac{5}{x}\right)^{\dfrac{1}{2}} v

, where

v

is the average speed of the molecule. The value of

x

will be:

\left(\right.

Take

\left.\pi=\dfrac{22}{7}\right)

A 4

B 8

C 28

D 27

Correct Answer

Option C

Solution

The relationship between the root-mean-square (rms) speed (

v_{rms}

) and the average speed (

v_{avg}

) of molecules in a gas can be found using the Maxwell-Boltzmann distribution.

The rms speed and average speed are related as follows:

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

v_{avg} = \sqrt{\frac{8RT}{\pi M}}

Where:

R

is the ideal gas constant

T

is the temperature in Kelvin

M

is the molar mass of the gas $\pi$ is the mathematical constant pi In this problem, the rms speed of the oxygen molecule is given by:

v_{rms} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}} v_{avg}

Now, let's divide the expression for

v_{rms}

by the expression for

v_{avg}

:

\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{\frac{3RT}{M}}}{\sqrt{\frac{8RT}{\pi M}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}

By simplifying the expression, we get:

\frac{v_{rms}}{v_{avg}} = \frac{\sqrt{3}}{\sqrt{\frac{8}{\pi}}} = \left(1+\frac{5}{x}\right)^{\frac{1}{2}}

Square both sides of the equation:

\frac{3}{\frac{8}{\pi}} = 1 + \frac{5}{x}

Now we will substitute the provided value of

\pi = \frac{22}{7}

:

\frac{3}{\frac{8}{\frac{22}{7}}} = 1 + \frac{5}{x}

By simplifying the expression, we get:

\frac{3 \cdot \frac{22}{7}}{8} = 1 + \frac{5}{x}

Now let's solve for

x

:

\frac{66}{56} - 1 = \frac{5}{x}

\frac{10}{56} = \frac{5}{x}

Multiplying both sides by

x

:

\frac{10}{56}x = 5

Finally, solving for

x

:

x = \frac{5 \cdot 56}{10} = 28

So, the value of

x

is

\boxed{28}

.

Q209

An engine operating between the boiling and freezing points of water will have A. efficiency more than 27%. B. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures. C. efficiency equal to

27 \%

D. efficiency less than

27 \%

Choose the correct answer from the options given below:

A B and C only

B B and D only

C B, C and D only

D A and B only

Correct Answer

Option B

Solution

To answer this question, we need to find the efficiency of a Carnot engine operating between the boiling and freezing points of water.

The boiling point of water is 100°C (373 K) and the freezing point is 0°C (273 K).

The efficiency of a Carnot engine is given by the formula: Efficiency =

1 - \frac{T_{cold}}{T_{hot}}

Plugging in the values for the boiling and freezing points of water: Efficiency =

1 - \frac{273}{373} = 1 - 0.732 = 0.268

So, the efficiency of a Carnot engine operating between these temperatures is approximately 26.8%.

Now, let's analyze the given options: A. efficiency more than 27% - This is incorrect, as the Carnot efficiency is 26.8% and no engine can be more efficient than a Carnot engine.

B. efficiency less than the efficiency of a Carnot engine operating between the same two temperatures - This is correct, as real engines are less efficient than Carnot engines operating between the same temperatures.

C. efficiency equal to

27 \%

- This is incorrect, as the Carnot efficiency is 26.8%, not 27%. D. efficiency less than

27 \%

- This is correct, as the efficiency is less than 27% (26.8%). Thus, the correct answer is "B and D only."

Q210

A source supplies heat to a system at the rate of

1000 \mathrm{~W}

. If the system performs work at a rate of

200 \mathrm{~W}

. The rate at which internal energy of the system increases is

A 600 W

B 1200 W

C 500 W

D 800 W

Correct Answer

Option D

Solution

The rate of increase of internal energy of a system can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

In this case, the heat being supplied to the system is 1000 W and the work being done by the system is 200 W.

Therefore, the rate at which the internal energy of the system increases is: 1000 W (heat supplied) - 200 W (work done) = 800 W So, the correct answer is 800 W.