Q: The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4. It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine

aligned & =0.4 \\ & ( COP )_ HP =10 \\ & ~W = Q _1=0.4 150=60 ~J \\ & COP = Dissered Effect Work input \\ & 10= Q _3 ~W \\ & Q _3=600 ~J \\ & Also \\ & =0.4=1- T _2 ~T _1 \\ & ~T _2=600 ~K \\ & COP = T _3 ~T _3- T _4 =10 aligned aligned & = 300 300- T _4 =10 \\ & ~T _4=270 ~K aligned

Q: An ideal monatomic gas of n moles is taken through a cycle W X Y Z W consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V-T diagram. The volume of the g

aligned & nRT _ W = P _ W ~V _ W =1 ~J \\ & P _ W = 1 64 10^6 ~Pa aligned For WX process aligned & P_X V_X^Y=P_W V_W^y \\ & P_X=P_W ( V_W V_X )^y aligned amount of heat absorbed in XY process aligned Q & = nCP T = n 5 2 R [ T _ Y - T _ X ] [ For monoatomic gas C _ P = 5 R 2 ] \\ Q & = 5 2 [ nRT _ Y - nRT T _ X ] \\ & = 5 2 [ P _ Y V _ Y - P _ X V _ X ] \\ & = 5 2 P _ X [ V _ Y - V _ X ] [ P _ X = P _ Y ; Isobaric process ] \\ & = 5 2 P _ W [ V _ W V _ X ]^ y [ V _ Y - V _ X ] aligned Putting val

Q: The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting, movable piston of area A. The left and right compartments are filled with 3 2 and

Extension in spring aligned & x=0.5 ~L -0.4 ~L \\ & =0.1 ~L aligned FBD of piston aligned & kx + P _2 ~A = P _1 ~A \\ & P _2 ~A = P _1 ~A - kx aligned P_2=P_1- k L A(10) ....(i) aligned & P _1 ~V = n _1 RT \\ & P _2 ~V = n _2 RT \\ & P _1 P _2 = n _1 n _2 = 3 2 aligned P _1= 3 2 P _2 ....(ii) aligned & P_2= 3 2 P_2- k L 10 A \\ & P_2 2 = k L 10 A \\ & P_2= k L 5 A = k L A \\ & = 1 5 =0.2 aligned

Q: One mole of a monatomic ideal gas undergoes the cyclic process J K L M J , as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct opt

aligned & J ( P _0, ~V _0, ~T _0 ) \\ & K ( P _0, 3 ~V _0, 3 ~T _0 ) \\ & M (2 P _0, V _0 2 , ~T _0 ) \\ & L (2 P _0, 3 ~V _0 2 , 3 ~T _0 ) \\ & P _0 ~V _0= nRT _0 \\ & JK isobaric W = P _0 (2 ~V _0 )=2 nRT _0 \\ & U = 3 2 nR (2 ~T _0 )=3 nRT _0 \\ & KL isothermal W = nR (3 ~T ) ( 1 2 )=-3 nRT _0 n 2 \\ & U =0 Q =-3 nRT _0 2 \\ & LM isobaric =2 P _0 (- V _0 )=-2 nRT e _0 \\ & MJ isothermal nRT n _0 2 ; U =0 \\ & WD net =-2 nRT e _0 2 \\ & P 4, Q 3, R 5, ~S 2 aligned

Q: One mole of an ideal gas expands adiabatically from an initial state ( T A , V 0 ) to final state ( T f , 5 V 0 ) . Another mole of the same gas expands isothermally from a different ini

As we know for an Adiabatic process T V γ - 1 = Constant . ⇒ T A V 0 γ - 1 = T f 5 V 0 γ - 1 . . . i For Isothermal process T B = T f . . . ii Dividing equation i by equation ii , we get ⇒ T A T B = 5 γ - 1

Question 1

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process ( ab ), followed by an isochoric process ( bc ) and an adiabatic process ( ca ) as shown in the figure. The volumes of the

Accepted Answer

For the monoatomic ideal gas, = 5 3 . Process ab is isothermal, so T_a = T_b and V_a = V_1, V_b = V_2. Process bc is isochoric, so V_b = V_c = V_2. Process ca is adiabatic, so T_c V_c^ -1 = T_a V_a^ -1 . T_c V_2^ 2/3 = T_a V_1^ 2/3 T_c = T_a ( V_1 V_2 )^ 2/3 . If V_2 V_1 = 8, T_c = T_a ( 1 8 )^ 2/3 = T_a 4 T_a = 4 T_c. Thus, statement (C) is correct. Heat absorbed in isothermal process ab is Q_ ab = nRT_a ( V_2 V_1 ). For V_2 V_1 = 8, Q_ ab = nRT_a (8) = 3nRT_a 2 3 0.7 nRT_a = 2.1 nRT_a. Heat re

Question 2

As shown in the figure, five Carnot engines, each with efficiency and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engin

Accepted Answer

Let Q_0 be the heat absorbed by the first engine and Q_1 be the heat rejected. The efficiency of the first engine is = 1 - Q_1 Q_0 Q_1 = Q_0(1 - ) Similarly, for the second engine, the heat rejected is Q_2 = Q_1(1 - ) = Q_0(1 - )^2 Following this pattern, for the 5^ th engine, the heat rejected is Q_5 = Q_0(1 - )^5 The total work done by all five engines is W = W_1 + W_2 + W_3 + W_4 + W_5 = (Q_0 - Q_1) + (Q_1 - Q_2) + + (Q_4 - Q_5) = Q_0 - Q_5 The net efficiency of the system is _ net = W Q_0 =

Question 3

Ten moles of an ideal monoatomic gas, initially in state a at atmospheric pressure and temperature T_a = 27^ , is enclosed in a metal cylinder of volume V_0 fitted with a frictionless piston. The gas

Accepted Answer

For state a: n = 10, T_a = 27^ C = 300 K , P_a = 1 atm , V_a = V_0. For a monoatomic gas, C_v = 3 2 R and = 5 3 . Process a b is a sudden compression, which is an adiabatic process. V_b = V_0 3 Using T_a V_a^ -1 = T_b V_b^ -1 : T_b = T_a ( V_a V_b )^ -1 = 300 (3)^ 5/3 - 1 = 300 3^ 2/3 = 300 (9)^ 1/3 Given 9^ 1/3 = 2.08, we get T_b = 300 2.08 = 624 K . Using P_a V_a^ = P_b V_b^ : P_b = P_a ( V_a V_b )^ = 1 (3)^ 5/3 = 3 3^ 2/3 = 3 2.08 = 6.24 atm . Thus, the pressure at state b is 6.24 times the a

Question 4

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4. It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine

Accepted Answer

aligned & =0.4 \ & ( COP )_ HP =10 \ & ~W = Q _1=0.4 150=60 ~J \ & COP = Dissered Effect Work input \ & 10= Q _3 ~W \ & Q _3=600 ~J \ & Also \ & =0.4=1- T _2 ~T _1 \ & ~T _2=600 ~K \ & COP = T _3 ~T _3- T _4 =10 aligned aligned & = 300 300- T _4 =10 \ & ~T _4=270 ~K aligned

Question 5

An ideal monatomic gas of n moles is taken through a cycle W X Y Z W consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V-T diagram. The volume of the g

Accepted Answer

aligned & nRT _ W = P _ W ~V _ W =1 ~J \ & P _ W = 1 64 10^6 ~Pa aligned For WX process aligned & P_X V_X^Y=P_W V_W^y \ & P_X=P_W ( V_W V_X )^y aligned amount of heat absorbed in XY process aligned Q & = nCP T = n 5 2 R [ T _ Y - T _ X ] [ For monoatomic gas C _ P = 5 R 2 ] \ Q & = 5 2 [ nRT _ Y - nRT T _ X ] \ & = 5 2 [ P _ Y V _ Y - P _ X V _ X ] \ & = 5 2 P _ X [ V _ Y - V _ X ] [ P _ X = P _ Y ; Isobaric process ] \ & = 5 2 P _ W [ V _ W V _ X ]^ y [ V _ Y - V _ X ] aligned Putting val

Question 6

The left and right compartments of a thermally isolated container of length L are separated by a thermally conducting, movable piston of area A. The left and right compartments are filled with 3 2 and

Accepted Answer

Extension in spring aligned & x=0.5 ~L -0.4 ~L \ & =0.1 ~L aligned FBD of piston aligned & kx + P _2 ~A = P _1 ~A \ & P _2 ~A = P _1 ~A - kx aligned P_2=P_1- k L A(10) ....(i) aligned & P _1 ~V = n _1 RT \ & P _2 ~V = n _2 RT \ & P _1 P _2 = n _1 n _2 = 3 2 aligned P _1= 3 2 P _2 ....(ii) aligned & P_2= 3 2 P_2- k L 10 A \ & P_2 2 = k L 10 A \ & P_2= k L 5 A = k L A \ & = 1 5 =0.2 aligned

Question 7

One mole of a monatomic ideal gas undergoes the cyclic process J K L M J , as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct opt

Accepted Answer

aligned & J ( P _0, ~V _0, ~T _0 ) \ & K ( P _0, 3 ~V _0, 3 ~T _0 ) \ & M (2 P _0, V _0 2 , ~T _0 ) \ & L (2 P _0, 3 ~V _0 2 , 3 ~T _0 ) \ & P _0 ~V _0= nRT _0 \ & JK isobaric W = P _0 (2 ~V _0 )=2 nRT _0 \ & U = 3 2 nR (2 ~T _0 )=3 nRT _0 \ & KL isothermal W = nR (3 ~T ) ( 1 2 )=-3 nRT _0 n 2 \ & U =0 Q =-3 nRT _0 2 \ & LM isobaric =2 P _0 (- V _0 )=-2 nRT e _0 \ & MJ isothermal nRT n _0 2 ; U =0 \ & WD net =-2 nRT e _0 2 \ & P 4, Q 3, R 5, ~S 2 aligned

Question 8

One mole of an ideal gas undergoes two different cyclic processes I  and II , as shown in the P - V diagrams below. In cycle I , processes a , &#160; b , &#160; c and d are isobaric, isothermal,

Accepted Answer

For first case: W I = W a + W b + W c + W d = 4 P 0 2 V 0 - V 0 + n R T &#160; ln 4 V 0 2 V 0 + 2 P 0 V 0 - 4 V 0 + 0 = 4 P 0 V 0 + n R 8 P 0 V 0 n R ln 2 - 6 P 0 V 0 = 8 P 0 V 0 &#160; ln 2 - 2 P 0 V 0 For second case:  W II = W ' a + W ' b + W ' c + W ' d = n R T &#160; ln 2 V 0 V 0 + 0 + P 0 V 0 - 2 V 0 + 0 = n R 4 P 0 V 0 n R ln 2 - P 0 V 0 = 4 P 0 V 0 &#160; ln 2 - P 0 V 0 Therefore, required ratio is  W I W II = 8 P 0 V 0

Question 9

One mole of an ideal gas expands adiabatically from an initial state ( T A , &#160; V 0 ) to final state ( T f , &#160; 5 V 0 ) . Another mole of the same gas expands isothermally from a different ini

Accepted Answer

As we know for an Adiabatic process T V &#947; - 1 = Constant . &#8658; T A V 0 &#947; - 1 = T f 5 V 0 &#947; - 1 &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . i For Isothermal process T B = T f &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . ii Dividing equation i  by equation ii , we get &#8658; T A T B = 5 &#947; - 1

Question 10

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas &#947; = 5 3 and one mole of an ideal diatomic gas &#947; = 7 5 . Here, &#947;  is the ratio of the specif

Accepted Answer

Change in internal energy for the mixture can be written as, &#8710; U = n 1 C v 1 &#8710; T + n 2 C v 2 &#8710; T = n 1 C v 1 + n 2 C v 2 &#8710; T &#160; &#160; &#160; &#160; &#160; . . . i For isobaric process, work done  = P &#8710; V = n 1 + n 2 R &#8710; T &#160; &#160; &#160; &#160; &#160; &#160; . . . ii Divide i by ii , we get &#8710; U W = n 1 C v 1 + n 2 C v 2 &#8710; T n 1 + n 2 R &#8710; T &#8658; &#8710; U = W R n 1 C v 1 + n 2 C v 2 n 1 + n 2 = 66 R 2 &#215; 3 R 2 + 1 &#215;

Question 11

A bubble has surface tension S . The ideal gas inside the bubble has ratio of specific heats &#947; = 5 3 . The bubble is exposed to the atmosphere and it always retains its spherical shape. When the

Accepted Answer

When,  Pressure Radius Temperature P a 1 &#8594; r 1 &#8594; T 1 P a 2 &#8594; r 2 &#8594; T 2 For adiabatic process P 1 V 1 &#947; = P 2 V 2 &#947; P a 1 + 4 T r 1 4 3 &#960; r 1 3 5 3 = P a 2 + 4 T r 2 4 3 &#960; r 2 3 5 3 r 1 r 2 5 = P a 2 + 4 T r 2 P a 1 + 4 T r 1 Therefore, option A  is incorrect. Now, T 1 V 1 &#947; - 1 = T 2 V 2 &#947; - 1 T 2 T 1 = V 1 V 2 &#947; - 1 = r 1 r 2 3 2 3 T 2 T 1 = P a 2 + 4 T r 2 P a 1 + 4 T r 1 2 5 Therefore, option  D  is correct. For op

Question 12

In the given  P - V diagram, a monoatomic gas &#947; = 5 3  is first compressed adiabatically from state A state B . Then it expands isothermally from state B to state C . [Given : 1 3 0 . 6

Accepted Answer

For adiabatic process  P V &#947; = C &#8658; 100 0 . 8 5 3 = 300 V B 5 3 &#8658; V B = 0 . 8 3 3 / 5 = 0 . 4 &#160; m 3 Work done in an adiabatic process is given by, W = P 1 V 1 - P 2 V 2 &#947; - 1 &#8658; W A B = P A V A - P B V B 5 3 - 1 = 80 - 300 &#215; 0 . 4 2 / 3 &#160; kJ &#8658; W A B = - 60 kJ Therefore, option  C  is correct. C &#8594; A  is isochoric  &#8658; D  is correct. For process  B C W B C = n R T ln V 2 V 1 = P V ln V 2 V 1 &#8658; W B C =

Question 13

An ideal gas of density &#961; = 0 . 2 &#160; kg &#160; m - 3 enters a chimney of height h at the rate of &#945; = 0 . 8 &#160; kg &#160; s - 1 from its lower end, and escapes through the upper end as

Accepted Answer

Here the gas undergoes adiabatic expansion,  P 1 &#8722; &#947; T &#947; = constant &#8658; P 2 P 1 = T 1 T 2 &#947; 1 &#8722; &#947; &#8658; P 2 = 300 150 2 1 &#8722; 2 &#215; 600 &#8658; P 2 = 600 4 = 150 &#160; Pa Thus, pressure of the gas at the upper end of the chimney is  P 2 = 150 &#160; Pa From ideal gas equation,  &#961; = PM RT &#8658; &#961; &#8733; P T Then,  &#961; 1 &#961; 2 = P 1 P 2 T 1 T 2 = 150 600 300 150 = 1 2 &#8658; &#961; 2 = 0 . 2 1 2 = 0 . 1 &#16

Question 14

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system d

Accepted Answer

(I) According to the first law of thermodynamics, &#8710; Q = &#8710; U + W &#8658; &#8710; U = &#8710; Q - W &#8658; U = M L - P &#916; V = 10 - 3 &#215; 2250 - 10 2 kP &#215; 10 - 3 - 10 - 6 m 3 = 2 . 25 &#160; kJ - 0 . 1 &#160; kJ = 2 . 15 &#160; kJ Therefore, I-P (II) For isobaric process, V 1 V 2 = T 1 T 2 &#8658; T 2 = 3 &#215; 500 = 1500 &#160; K Now the change in the internal energy will be, &#8710; U = n C V &#8710; T = 0 . 2 &#215; 5 2 &#215; 8 &#215; 1000 = 4 &#160; kJ Therefore,&nbsp

Question 15

Paragraph A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of

Accepted Answer

Writing  C V  in terms of degree of freedom of the gas f , we get, C V = f 2 R = 2 R &#8658; f = 4 , Therefore,​ adiabatic index of gas will be given by,​​​​​​ &#947; = 1 + 2 f = 1 + 2 4 = 3 2 The gas of the right portion has undergone through an adiabatic process so, T i V i &#947; - 1 = T f V f &#947; - 1 &#8658; T 0 &#160; V 0 3 2 - 1 = T R V 0 2 3 2 - 1 &#8658; T R = 2 T 0 &#8658; T R T 0 = 2

Question 16

Paragraph A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of

Accepted Answer

Writing  C V  in terms of degree of freedom of the gas f , we get, C V = f 2 R = 2 R &#8658; f = 4 , Therefore,​ adiabatic index of gas will be given by,​​​​​​ &#947; = 1 + 2 f = 1 + 2 4 = 3 2 The gas of the right portion has undergone through an adiabatic process so, T i V i &#947; - 1 = T f V f &#947; - 1 &#8658; T 0 &#160; V 0 3 2 - 1 = T R V 0 2 3 2 - 1 &#8658; T R = 2 T 0 &#8658; T R T 0 = 2 Work done by the gas  2 ,  = - &#8710; U = - n C V T f - T i = - 1 2 R

Question 17

An ideal gas undergoes a four step cycle as shown in the  P - V  diagram below. During this cycle, heat is absorbed by the gas in:

Accepted Answer

&#8757; &#160; &#160; P V = n R T &#160; &#160; &#160; . . . 1 In Step 1 : P = constant V =  increasing So, T  also will be increase (From equation 1 ) Now,  &#8757; &#160; &#160; Q = n C p &#8710; T &#8757; T &#160; &#8593; &#8756; &#160; &#160; &#8710; T =  Positive So,   Q = + v e = absorbed Now In step 4 , P = &#8593; V =  constant So from equation 1 , T  will also be increasing. Now,  &#8757; &#160; &#160; Q = n C v &#8710; T &#8757; &#160; T &#8593;

Question 18

A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8 . 3 kg

Accepted Answer

Initially PA ( 4 ) = 0 . 1 &#215; R &#215; 300 &#160; &#160; &#160; &#160; &#160; &#160; &#160; . . . ( i ) for both part Let final distance of piston from top is ' y ' 0 . 1 &#215; R &#215; 300 = P 1 &#215; A &#183; y = P 2 A ( 8 - y ) &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#8230; ii 30 R y + 8 . 3 g = 30 R 8 - y 30 y + 10 = 30 8 - y 3 + y y = 3 8 - y 24 + 8 y - 3 y - y 2 = 3 y y 2 - 2 y - 24 = 0 y = 2 &#177; 4 + 96 2 &#215; 1 = 2 &#177; 10 2 = 6 &

Question 19

A spherical bubble inside water has radius  R . Take the pressure inside the bubble and the water pressure to be  p 0 . The bubble now gets compressed radially in an adiabatic manner so that

Accepted Answer

$ aligned & W =( P )_ avg 4 R ^2 a \ &= | dP 2 4 R ^2 a | aligned $ (for small change $( P )_ avg $ arithmetic mean) $= PV gamma = c dP =- P V dV =- P _0 ~V 4 R ^2 a $ $= P _0 2 ~V 4 R ^2 a 4 R ^2 a $ $= P _0 2 4 R ^3 4 R ^2 a 4 R ^2 a $ $= (4 pRP a ^2 ) 3 2 $ $ x =2.05$

Question 20

Consider one mole of helium gas enclosed in a container at initial pressure P 1 and volume V 1 . It expands isothermally to volume 4 V 1 . After this, the gas expands adiabatically and its volume beco

Accepted Answer

∴ γ = 5 3 In isothermal process, W 1 = n R T ln 4 V V = 2 n R T ln 2 In adiabatic process, T ( 4 V ) γ - 1 = T ' ( 32 V ) γ - 1 ⇒ T ' = T ( 8 ) 1 - γ = T × 8 - 2 3 = T 4 W 2 = - n 3 2 R T ' - T = + 3 2 nR 3 T 4 = 9 8 n R T W 1 W 2 = 2 n R T ln 2 9 8 n R T = 16 9 ln 2 = f ln 2 f = 16 9 = 1 . 78

Thermodynamics — Previous Year Questions & Solutions