Class 11th

12.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ?

(Take 1 cal = 4.19 J)

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Contributor-Level 10

The work done, W = 22.3 J

Being an adiabatic process, $?$ Q = 0

$?$ W = -22.3 J : since the work is done on the system

From the 1st law of thermodynamics, we know $?$ Q = $? U +$ $?$ W, where $? U$ is the change of internal energy of the gas

$?$ U = 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

$?$ Q = 9.35 cal = 9.35 $* 4.19$ J = 39.1765 J

Heat absorbed $?$ Q = $? U +$ $?$ W

$?$ W = $?$ Q - $? U$ = 39.1765 – 22.3 = 16.8765 J

Therefore, work done by the system is 16.8765 J

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6 months ago

12.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?

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The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus the process is 'Adiabatic'.

Let the initial and final pressure inside the cylinder be $P_{1}$ & $P_{2}$ and volume be $V_{1}$ & $V_{2}$ .

Ratio of specific heat, $γ$ = 1.4

For an adiabatic process, we know $P_{1} V_{1}^{γ}$ = $P_{2} V_{2}^{γ}$

It is given $V_{2} =$ $\frac{V_{1}}{2}$

Hence $P_{1} V_{1}^{γ}$ = $P_{2} {\frac{V_{1}}{2}}^{γ}$ or $\frac{P_{2}}{P_{1}}$ = ( $V_{1}^{γ}) /$ ( ${\frac{V_{1}}{2}}^{γ})$ = $2^{γ}$ = $2^{1.4}$ = 2.639

Hence the pressure increases by a factor of 2.639

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6 months ago

12.3 Explain why

(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2.

(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

(c) Air pressure in a car tyre increases during driving.

(d) The climate of a harbor town is more temperate than that of a town in a desert at the same latitude.

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(a) When the two bodies at different temperatures brought in contact, heat flows from the body of higher temperature to the body with lower temperature till the thermal equilibrium is achieved and both the body attains the temperature of (T1 + T2 )/2. But only when the thermal capacities of both the bodies are equal.

(b) The coolant used in Chemical or in Nuclear plant should have high specific heat. Higher specific heat allows coolant to absorb more heat.

(c) In motion, the air temperature inside the tyre increases due to the motion of the air molecules. According to Charles's law, temperature is directly proportional to

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12.2 What amount of heat must be supplied to 2.0 * $10^{- 2}$ kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)

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Mass of Nitrogen, m = 2.0 * $10^{- 2}$ kg = 20 g

Rise in temperature, $? T$ = 45 °C

Molecular mass of $N_{2}$ . M = 28

Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles, n = $\frac{m}{M}$ = $\frac{20}{28}$ = 0.714

Molar specific heat at constant pressure for nitrogen, $C_{p}$ = $\frac{7}{2}$ R = 29.05 J/mol/K

The total amount of heat to be supplied is given by the relation

$?$ Q = n $C_{p} ? T$ = 0.714 $* 29.05 * 45$ = 933.38 J

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6 months ago

12.1 A geyser heats water flowing at the rate of 3.0 liters per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 * $10^{4}$ J/g?

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Initial temperature, $T_{1}$ = 27 °C

Final temperature, $T_{2}$ = 77 °C

Rise in temperature, $? T = T_{2} - T_{1} = 50 ?$

Heat of combustion = 4.0 * $10^{4}$ J/g

Specific heat of water, c = 4.21 J/g/ $?$

Mass of flowing water, m = 3 lit/min = 3000 g/min

Total heat used, $? Q = m c ? T$ = 3000 $* 4.21 * 50$ = 6.315 $* 10^{5}$ J/min

Rate of consumption = $\frac{6.315 * 10^{5}}{4.0 * 10^{4}}$ g/min = 15.79 g/min

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6 months ago

14.25 A mass attached to a spring is free to oscillate, with angular velocity $ω$ , in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters $ω, x_{0}$ and $v_{0}$ .

[Hint : Start with the equation x = a cos ( $ω$ t + $θ$ ) and note that the initial velocity is negative.]

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The displacement equation for an oscillating mass is given by : x = Acos ( $ω t + θ)$ , where

A = the amplitude

x = the displacement

$θ = p h a s e c o n s t a n t$

Velocity, V = $\frac{d x}{d t}$ = -A $ω s i n ? (ω$ t + $θ)$

At t = 0, x = $x_{0}$ , $x_{0} = A \cos ? θ$ = $x_{0}$ ….(i)

And $\frac{d x}{d t}$ = $- v_{0}$ = A $ω \sin ? θ$ …….(ii)

Squaring and adding, we get

$A^{2} ({c o s}^{2} θ + {s i n}^{2} θ) = x_{0}^{2} + (\frac{v_{0}^{2}}{ω^{2}})$ , A = $\sqrt{x_{0}^{2} + ({\frac{v_{0}}{ω})}^{2}}$

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6 months ago

14.24 A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.

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Amplitude = 5 cm = 0.05 m

Time period, T = 0.2 s

(a) For displacement x = 5 cm = 0.05m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} * 0.05$ = $-$ 5 $π^{2}$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} - {0.05}^{2}}$ = 0

(b) For displacement x = 3 cm = 0.03m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} * 0.03$ = $-$ 3 $π^{2}$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} - {0.03}^{2}}$ = 0.4 $π m / s$

(c) For displacement x = 0 cm = 0 m

Acceleration is given by a = ${- ω}^{2} x$ = $- ({\frac{2 π}{T})}^{2} x$ = $- ({\frac{2 π}{0.2})}^{2} * 0$ = $0$ m/s

Velocity is given by V = $ω \sqrt{A^{2} - x^{2}}$ = $\frac{2 π}{T} \sqrt{{0.05}^{2} 0}$ = 0.5 $π m / s$

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14.23 A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant $α$ is defined by the relation J = – $α θ$ , where J is the restoring couple and $θ$ the angle of twist).

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Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillation of the disc have a time period, T = 1.5 s

The moment of inertia of the disc I = $\frac{1}{2}$ m $r^{2}$ = $\frac{1}{2} * 10 * {0.15}^{2} \frac{k g}{m^{2}}$ = 0.1125 $\frac{k g}{m^{2}}$

Time period, T = 2 $? \sqrt{\frac{I}{?}}$ ,

Torsional spring constant, $? = \frac{4 ?^{2} I}{T^{2}} = \frac{4 {* ?}^{2} * 0.1125}{{1.5}^{2}}$ = 1.974 Nm/rad

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14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

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The equation of displacement of a particle executing SHM at an instant t is given bas:

x = Asin $ω t$ , where A = Amplitude of oscillation and $ω$ = angular frequency = $\sqrt{\frac{k}{M}}$

The velocity of the particle, v = $\frac{d x}{d t}$ = A $ω \cos ? ω t$

The kinetic energy of the particle $E_{k}$ = $\frac{1}{2}$ M $v^{2}$ = $\frac{1}{2}$ M ${(A ω \cos ? ω t)}^{2}$ = $\frac{1}{2}$ M $A^{2} ω^{2} {c o s}^{2} ω t$

The potential energy of the particle $E_{p}$ = $\frac{1}{2}$ k $x^{2}$ = $\frac{1}{2}$ k $A^{2} {s i n}^{2} ω t$

For time period T, the average kinetic energy over a single cycle is given as :

${E_{k}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{k} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} {c o s}^{2} ω t d t$ = $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} {c o s}^{2} ω t d t$

= $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} \frac{(1 - c o s 2 ω t)}{2} d t$ = $\frac{M A^{2} ω^{2}}{2 T} ({t + \frac{s i n 2 ω t}{2 ω})}_{0}^{T}$ = $\frac{M A^{2} ω^{2}}{4 T} (T)$ = $\frac{1}{4} M A^{2} ω^{2}$ …….(i)

Average potential energy

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The equation of displacement of a particle executing SHM at an instant t is given bas:

x = Asin $ω t$ , where A = Amplitude of oscillation and $ω$ = angular frequency = $\sqrt{\frac{k}{M}}$

The velocity of the particle, v = $\frac{d x}{d t}$ = A $ω \cos ? ω t$

The kinetic energy of the particle $E_{k}$ = $\frac{1}{2}$ M $v^{2}$ = $\frac{1}{2}$ M ${(A ω \cos ? ω t)}^{2}$ = $\frac{1}{2}$ M $A^{2} ω^{2} {c o s}^{2} ω t$

The potential energy of the particle $E_{p}$ = $\frac{1}{2}$ k $x^{2}$ = $\frac{1}{2}$ k $A^{2} {s i n}^{2} ω t$

For time period T, the average kinetic energy over a single cycle is given as :

${E_{k}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{k} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} {c o s}^{2} ω t d t$ = $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} {c o s}^{2} ω t d t$

Average potential energy over one cycle is given as :

${E_{p}}_{a v g} = \frac{1}{T} \int_{0}^{T} E_{p} d t$ = $\frac{1}{T} \int_{0}^{T} \frac{1}{2} M A^{2} ω^{2} \sin ? ω t d t$

= $\frac{M A^{2} ω^{2}}{2 T} \int_{0}^{T} \frac{(1 - c o s 2 ω t)}{2} d t$ = $\frac{M A^{2} ω^{2}}{2 T} ({t - \frac{s i n 2 ω t}{2 ω})}_{0}^{T}$ = $\frac{M A^{2} ω^{2}}{4 T} (T)$ = $\frac{1}{4} M A^{2} ω^{2}$ …….(ii)

From equations (i) and (ii), it is proved that average kinetic energy for a given period of time is equal to the average potential energy for the same period of time.

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14.21 You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation.

Estimate the values of

(a) The spring constant k and

(b) The damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

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