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4 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

= nC_vdT= nC_v(T_B-T_A)

= $\frac{3}{2} R (T_{B} - T_{A})$

= $\frac{3}{2} (R T_{B} - R T_{A}) = \frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

Heat exchanged = $\frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

(b) For process BC , p =constant

dQ= dU+dW = $\frac{3}{2} R (T_{C} - T_{B}) + P_{B} (V_{C} - V_{B})$

heat exchanged = $\frac{5}{2} P_{B} (V_{C} {- V}_{A})$

(c) For process CD , because CD is adiabatic , dQ= heat exchanged =0

(d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

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Thermodynamics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig .

A to B : volume constant B to C : adiabatic C to D : volume constant D to A : adiabatic V_c=V_D=2V_A=2V_B

(a) In which part of the cycle heat is supplied to the engine from outside?

(b) In which part of the cycle heat is being given to the surrounding by the engine?

(c) What is the work done by the engine in one cycle? Write your answer in term of P_A, P_B, V_A

(d) What is the efficiency of the engine? [γ =5/3 for the gas], (C_v= 3/2R for one mole)

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) For the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{n R}{V} T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to the system in process AB.

(b) For the process CD volume is constant but the pressure decreases, hence temperature also decreases . so heat is also given to the surroundings.

(c) W_AB= $\int_{A}^{B} p d V = 0$ , W_CD= $\int_{V c}^{V d} p d V = 0$

W_BC= $\int_{V B}^{V c} p d V = k \int_{V B}^{V c} \frac{d V}{V^{Y}} = \frac{k}{1 - Y} [V^{1 - Y}]$

= $\frac{1}{1 - Y}$ [pV]= $\frac{P_{c} V_{c} - P_{B} V_{B}}{1 - Y}$

W_DA= $\frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

B and C lies on adiabatic curve BC

P_BV_B^Y= P_CV_C^Y

P_C = P_B( $\frac{V_{B}}{V_{C}}$ )^Y = P_B( $\frac{1}{2}$ )^Y= 2^-YP_B

Total work done by the engine in one cycle ABCD

...more

This is a long answer type question as classified in NCERT Exemplar

(a) For the process AB

dV=0 and dW=0

dQ=dU+dW=dU

dQ=dU= change in internal energy , so heat utilised is equal to change in internal energy.

Since p= $\frac{n R}{V} T,$ in adiabatic temperature is directly proportional to pressure. So heat is supplied to the system in process AB.

(b) For the process CD volume is constant but the pressure decreases, hence temperature also decreases . so heat is also given to the surroundings.

(c) W_AB= $\int_{A}^{B} p d V = 0$ , W_CD= $\int_{V c}^{V d} p d V = 0$

W_BC= $\int_{V B}^{V c} p d V = k \int_{V B}^{V c} \frac{d V}{V^{Y}} = \frac{k}{1 - Y} [V^{1 - Y}]$

= $\frac{1}{1 - Y}$ [pV]= $\frac{P_{c} V_{c} - P_{B} V_{B}}{1 - Y}$

W_DA= $\frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

B and C lies on adiabatic curve BC

P_BV_B^Y= P_CV_C^Y

P_C = P_B( $\frac{V_{B}}{V_{C}}$ )^Y = P_B( $\frac{1}{2}$ )^Y= 2^-YP_B

Total work done by the engine in one cycle ABCDA

W= W_AB+W_BC+W_CD+W_DA= W_BC+W_DA

= $\frac{P_{C} V_{C} - P_{B} V_{B}}{1 - Y} + \frac{P_{A} V_{A} - P_{D} V_{D}}{1 - Y}$

W = $\frac{1}{1 - Y} [2^{- Y} P_{B} 2 V_{B} - P_{B} V_{B} + P_{A} P_{A} - 2^{- Y} P_{B} 2 V_{B}]$

W = $\frac{1}{1 - Y} (2^{1 - Y} - 1) (P_{B} - P_{A}) V_{A}$

W= $\frac{3}{2} [1 - (\frac{1}{2})$ ^2/3]( $P_{B} - P_{A}$ )V_A

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Thermodynamics Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig.

(a) Find the work done when the gas is taken from state 1 to state 2.

(b) What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?

(c) Given the internal energy for one mole of gas at temperature T is (3/2) RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V₂ = 2V₁.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= $\int_{V 1}^{V 2} p d V = K \int_{V 1}^{V 2} \frac{d V}{\sqrt[]{V}} = k {[\frac{\sqrt[]{V}}{\frac{1}{2}}]}_{V 1}^{V 2} = 2 K (\sqrt[]{V_{2}} - \sqrt[]{V_{1}})$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p \sqrt[]{V} \sqrt[]{V}}{n R}$

T= $\frac{K \sqrt[]{V}}{n R}$

T₁= $\frac{K \sqrt[]{V 1}}{n R}$ , T₁= $\frac{K \sqrt[]{V 2}}{n R}$

$\frac{T_{1}}{T_{2}} = \frac{\frac{k \sqrt[]{V_{1}}}{n R}}{\frac{k \sqrt[]{V_{2}}}{n R}} = {\sqrt[]{\frac{V_{1}}{V_{2}}}}^{}$ = $\sqrt[]{\frac{V_{1}}{2 V_{1}}} = \frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2} R T$

$? U = U_{2} - U_{1} = \frac{3}{2} R (T_{1} - T_{2})$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V} - 1$ )

$? W$ =2p₁V₁^1/2( $\sqrt[]{V_{2}} - \sqrt[]{V_{2}}$ )

= 2p₁V₁^1/2(2 $\sqrt[]{V_{1}} - \sqrt[]{V_{1}}$ )

= 2p₁V₁( $\sqrt[]{2} - 1$ )= 2RT₁( $\sqrt[]{2} - 1$ )

$? Q$ = $? U + ? W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2} - 1$ )+ 2RT₁( $\sqrt[]{2} - 1$ )

= $\frac{7}{2} R T_{1} (\sqrt[]{2} - 1)$

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4 months ago

Thermodynamics Class 11th

12.10 A refrigerator is to maintain eatables kept inside at 9 $?$ . If room temperature is 36 $?$ , calculate the coefficient of performance.

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Vishal Baghel

Contributor-Level 10

Temperature inside the refrigerator, $T_{1}$ = 9 $?$ = 9 + 273 K = 282 K

Room temperature, $T_{2}$ = 36 $?$ = 36 + 273 = 309 K

Coefficient of performance = $\frac{T_{1}}{T_{2} - T_{1}}$ = $\frac{282}{309 - 282}$ = 10.44

Therefore, the coefficient of performance is 10.44

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Thermodynamics Class 11th

12.9 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F

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Vishal Baghel

Contributor-Level 10

Total work done by the gas from D to E to F = Area of $? D E F$ = $\frac{1}{2}$ EF $* D F$

Where DF = Change in pressure = 600 – 300 = 300 N/ $m^{2}$

FE = change in volume = 5-2 = 3 $m^{3}$

Area of $? D E F =$ $\frac{1}{2} *$ 3 $* 300$ = 450 J

Therefore work done by the gas from D to E to F is 450 J.

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Thermodynamics Class 11th

12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

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Vishal Baghel

Contributor-Level 10

Heat is supplied to the system at a rate of 100W

Hence, heat supplied, Q = 100 J/s

The system performs at the rate of 75 J/s

Hence, work done, W = 75 J/s

From the 1st law of Thermodynamics, we have Q = U + W, where U is the internal energy

U = Q – W = 100 – 75 = 25 J/s = 25 W

Therefore the internal energy of the given electric heater increases at a rate of 25 W.

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Thermodynamics Class 11th

12.7 A steam engine delivers 5.4*108J of work per minute and services 3.6 * 109J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

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Vishal Baghel

Contributor-Level 10

Work done by the steam engine per minute, W = 5.4 $* 10^{8}$ J

Heat supplied by the boiler, H = 3.6 $* 10^{9}$ J

Efficiency of the engine, $η$ = $\frac{O u t p u t e n e r g y}{I n p u t e n e r g y}$ = $\frac{5.4 * 10^{8}}{3.6 * 10^{9}}$ = 0.15

Amount of heat wasted = Input energy – Output energy

= 3.6 $* 10^{9} -$ 5.4 $* 10^{8} = 3.06 * 10^{9}$ J

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Thermodynamics Class 11th

12.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :

(a) What is the final pressure of the gas in A and B ?

(b) What is the change in internal energy of the gas ?

(c) What is the change in the temperature of the gas ?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?

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Vishal Baghel

Contributor-Level 10

(a) When the stopcock is opened, the volume became double between cylinders A and B. Since volume is inversely proportional to pressure, the pressure will become half. So the initial pressure of 1 atm in cylinder A will become ½ atm in cylinder A and B.

(b) The internal energy will change when there is work done by the gas. In absence of any work done, there will be no change in internal energy.

(c) In absence of any work done, there will be no change in the temperature.

(d) The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation an

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Thermodynamics 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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Vishal Baghel

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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Thermodynamics Class 11th

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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Vishal Baghel

Contributor-Level 10

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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