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Page 1 JEE Solved Example on Gaseous State JEE Mains Q1: Gas ?? ( ?? ?????? ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction ?? ?? ( ?? )? ?? ?? ( ?? )+ ???? ( ?? ) If degree of dissociation of ?? is 0.4 and remains constant in entire range of temperature, then the correct ?? vs ?? graph is [Given ?? = ?? . ???? lit-atm mol /?? ] Ans: ?? = 0.16l AIM ? to find slope of P -T curve ???? = ?????? ?? ?? = ???? ?? 2Ag( g)? 3 B( g)+ 2C ( g) 1 - 0.4 3 2 ( 0.4) 2 2 × 0.4 0.6 0.6 0.4 Total moles ? 0.6 + 0.6 + 0.4 ? 1.6 Slope tan ?? = ?? ?? = ???? ?? = 1.6 × 0.08 0.16 tan ?? = 0.8 ?? = tan -1 0.8 Page 2 JEE Solved Example on Gaseous State JEE Mains Q1: Gas ?? ( ?? ?????? ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction ?? ?? ( ?? )? ?? ?? ( ?? )+ ???? ( ?? ) If degree of dissociation of ?? is 0.4 and remains constant in entire range of temperature, then the correct ?? vs ?? graph is [Given ?? = ?? . ???? lit-atm mol /?? ] Ans: ?? = 0.16l AIM ? to find slope of P -T curve ???? = ?????? ?? ?? = ???? ?? 2Ag( g)? 3 B( g)+ 2C ( g) 1 - 0.4 3 2 ( 0.4) 2 2 × 0.4 0.6 0.6 0.4 Total moles ? 0.6 + 0.6 + 0.4 ? 1.6 Slope tan ?? = ?? ?? = ???? ?? = 1.6 × 0.08 0.16 tan ?? = 0.8 ?? = tan -1 0.8 Q2: Calculate the ratio of rate of effusion of ?? ?? and ?? ?? from a container containing ???????? ?? ?? and ?????? ?? ?? : (A) ?? : ?? (B) ?? : ?? (C) ?? : ?? (D) 4: 1 Ans: (A) ?? O 2 ? ?? O 2 v M O 2 ? ?? O 2 v ?? O 2 ?? H 2 ? ?? H 2 v ?? H 2 ? ?? H 2 v ?? H 2 ?? O 2 ?? H 2 = v ?? H 2 ?? O 2 ?? O 2 ?? H 2 = v 2 32 ( 16 /32 2 /3 ) = 1 8 Q3: Kinetic energy of one mole of ???? ?? at ' ?? ' ?? is: (A) 3 RT (B) ?? /?? RT (C) ?? /?? RT (D) None of these Ans: (C) Kinetic energy of CO 2 at T K Total K.E. of energy ? 3 2 RT Q4: The mean kinetic energy of 1 mole of ???? ?? at ?????? ?? is ?? . The average kinetic energy at ???????? ?? will be: (A) ?? /?? (B) ???? (C) ?? /???? (D) ?? Ans: (B) K.E. = 3 2 ???? So at T = 300 K.E. = ?? = 3 2 ?? ( 300 ) At ?? = 1500 ?? . ?? . = 3 2 ?? ( 1500 )= ( 3 2 ?? ( 300 ) )( 5)= ?? 5 = 5?? Answer = B Page 3 JEE Solved Example on Gaseous State JEE Mains Q1: Gas ?? ( ?? ?????? ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction ?? ?? ( ?? )? ?? ?? ( ?? )+ ???? ( ?? ) If degree of dissociation of ?? is 0.4 and remains constant in entire range of temperature, then the correct ?? vs ?? graph is [Given ?? = ?? . ???? lit-atm mol /?? ] Ans: ?? = 0.16l AIM ? to find slope of P -T curve ???? = ?????? ?? ?? = ???? ?? 2Ag( g)? 3 B( g)+ 2C ( g) 1 - 0.4 3 2 ( 0.4) 2 2 × 0.4 0.6 0.6 0.4 Total moles ? 0.6 + 0.6 + 0.4 ? 1.6 Slope tan ?? = ?? ?? = ???? ?? = 1.6 × 0.08 0.16 tan ?? = 0.8 ?? = tan -1 0.8 Q2: Calculate the ratio of rate of effusion of ?? ?? and ?? ?? from a container containing ???????? ?? ?? and ?????? ?? ?? : (A) ?? : ?? (B) ?? : ?? (C) ?? : ?? (D) 4: 1 Ans: (A) ?? O 2 ? ?? O 2 v M O 2 ? ?? O 2 v ?? O 2 ?? H 2 ? ?? H 2 v ?? H 2 ? ?? H 2 v ?? H 2 ?? O 2 ?? H 2 = v ?? H 2 ?? O 2 ?? O 2 ?? H 2 = v 2 32 ( 16 /32 2 /3 ) = 1 8 Q3: Kinetic energy of one mole of ???? ?? at ' ?? ' ?? is: (A) 3 RT (B) ?? /?? RT (C) ?? /?? RT (D) None of these Ans: (C) Kinetic energy of CO 2 at T K Total K.E. of energy ? 3 2 RT Q4: The mean kinetic energy of 1 mole of ???? ?? at ?????? ?? is ?? . The average kinetic energy at ???????? ?? will be: (A) ?? /?? (B) ???? (C) ?? /???? (D) ?? Ans: (B) K.E. = 3 2 ???? So at T = 300 K.E. = ?? = 3 2 ?? ( 300 ) At ?? = 1500 ?? . ?? . = 3 2 ?? ( 1500 )= ( 3 2 ?? ( 300 ) )( 5)= ?? 5 = 5?? Answer = B Q5: At what temperature the RMS velocity of oxygen molecules is equal to that of ???? ?? at ?????? ?? ? (A) +?????? ° ?? (B) -?????? ° ?? (C) ???? ° ?? (D) ?????? ?? Ans: (B) RMS velocity = v 3???? ?? Let temperature be to RMS O 2 = RMS SO 2 ? v 3RT MO 2 = v 3RT( 300 ) M SO 2 ? T 0 = 300× 32 64 = 150 K = -123 ° C Q6: Temperature at which most probable speed of ?? ?? becomes equal to root mean square speed of ?? ?? is [Given: ?? ?? at ?????? ° ?? ]: (A) ?????? ?? (B) ???????? ?? (C) ?????? ?? (D) ?????? ?? Ans: (B) v 2?? ?? 0 ?? O 2 = v 3???? ?? N 2 = 2×?? 0 32 = 3×700 K 28 ? T 0 = 1200 K Q7: A rigid container containing ???????? gas at some pressure and temperature. The gas has been allowed to escape (do not consider any effusion or diffusion) from the container due to which pressure of the gas becomes half of its initial pressure and temperature become ( ?? /?? rd of its initial. The mass of gas (in gms) escaped is: (A) 7.5 (B) 1.5 (C) 2.5 (D) 3.5 Ans: (C) Final pressure, temperature = ?? ?? , ?? ?? Initial pressure, temperature = P 0 'T i Page 4 JEE Solved Example on Gaseous State JEE Mains Q1: Gas ?? ( ?? ?????? ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction ?? ?? ( ?? )? ?? ?? ( ?? )+ ???? ( ?? ) If degree of dissociation of ?? is 0.4 and remains constant in entire range of temperature, then the correct ?? vs ?? graph is [Given ?? = ?? . ???? lit-atm mol /?? ] Ans: ?? = 0.16l AIM ? to find slope of P -T curve ???? = ?????? ?? ?? = ???? ?? 2Ag( g)? 3 B( g)+ 2C ( g) 1 - 0.4 3 2 ( 0.4) 2 2 × 0.4 0.6 0.6 0.4 Total moles ? 0.6 + 0.6 + 0.4 ? 1.6 Slope tan ?? = ?? ?? = ???? ?? = 1.6 × 0.08 0.16 tan ?? = 0.8 ?? = tan -1 0.8 Q2: Calculate the ratio of rate of effusion of ?? ?? and ?? ?? from a container containing ???????? ?? ?? and ?????? ?? ?? : (A) ?? : ?? (B) ?? : ?? (C) ?? : ?? (D) 4: 1 Ans: (A) ?? O 2 ? ?? O 2 v M O 2 ? ?? O 2 v ?? O 2 ?? H 2 ? ?? H 2 v ?? H 2 ? ?? H 2 v ?? H 2 ?? O 2 ?? H 2 = v ?? H 2 ?? O 2 ?? O 2 ?? H 2 = v 2 32 ( 16 /32 2 /3 ) = 1 8 Q3: Kinetic energy of one mole of ???? ?? at ' ?? ' ?? is: (A) 3 RT (B) ?? /?? RT (C) ?? /?? RT (D) None of these Ans: (C) Kinetic energy of CO 2 at T K Total K.E. of energy ? 3 2 RT Q4: The mean kinetic energy of 1 mole of ???? ?? at ?????? ?? is ?? . The average kinetic energy at ???????? ?? will be: (A) ?? /?? (B) ???? (C) ?? /???? (D) ?? Ans: (B) K.E. = 3 2 ???? So at T = 300 K.E. = ?? = 3 2 ?? ( 300 ) At ?? = 1500 ?? . ?? . = 3 2 ?? ( 1500 )= ( 3 2 ?? ( 300 ) )( 5)= ?? 5 = 5?? Answer = B Q5: At what temperature the RMS velocity of oxygen molecules is equal to that of ???? ?? at ?????? ?? ? (A) +?????? ° ?? (B) -?????? ° ?? (C) ???? ° ?? (D) ?????? ?? Ans: (B) RMS velocity = v 3???? ?? Let temperature be to RMS O 2 = RMS SO 2 ? v 3RT MO 2 = v 3RT( 300 ) M SO 2 ? T 0 = 300× 32 64 = 150 K = -123 ° C Q6: Temperature at which most probable speed of ?? ?? becomes equal to root mean square speed of ?? ?? is [Given: ?? ?? at ?????? ° ?? ]: (A) ?????? ?? (B) ???????? ?? (C) ?????? ?? (D) ?????? ?? Ans: (B) v 2?? ?? 0 ?? O 2 = v 3???? ?? N 2 = 2×?? 0 32 = 3×700 K 28 ? T 0 = 1200 K Q7: A rigid container containing ???????? gas at some pressure and temperature. The gas has been allowed to escape (do not consider any effusion or diffusion) from the container due to which pressure of the gas becomes half of its initial pressure and temperature become ( ?? /?? rd of its initial. The mass of gas (in gms) escaped is: (A) 7.5 (B) 1.5 (C) 2.5 (D) 3.5 Ans: (C) Final pressure, temperature = ?? ?? , ?? ?? Initial pressure, temperature = P 0 'T i Mole ratio = Mass ratio ? Escaped gas = 10 - 10 × 3 4 = 2.5gms Q8 : if the absolute temperature and pressure of a gas are doubled, its volume would become: (A) Doubled (B) Halve (C) Increases four times (D) Remains the same Ans: (D) P 0 V 0 = ???? T 0 When pressure and temperature doubled ?? ?? ?? ?? = ???? ?? ?? 2?? 0 ?? ?? = ???? ( 2?? 0 ) ?? ?? = ???? ?? 0 ?? 0 = ?? 0 from (i) ? Volume remains same Q9: Two gases contained separately in flasks of equal pressures of 1 atmosphere each. What will be the resultant pressure if the flasks are connected? (A) 2 atmosphere (B) 1/2 atmosphere (C) 1 atmosphere (D) None of these Ans: (C) ?? 0 ?? 0 = ???? ?? 0 ?? ?? ?? ?? = ?? ?? ?? ?? ?? Now ?? ?? = ?? 0 , ?? ?? = 2?? 0 , ?? ?? = 2?? 0 ? ?? ?? = ?? 0 ?? ?? 0 ?? 0 = ?? 0 ?? ?? = ?? 0 = 1 atm Page 5 JEE Solved Example on Gaseous State JEE Mains Q1: Gas ?? ( ?? ?????? ) dissociates in a closed rigid container of volume 0.16 lit. as per following reaction ?? ?? ( ?? )? ?? ?? ( ?? )+ ???? ( ?? ) If degree of dissociation of ?? is 0.4 and remains constant in entire range of temperature, then the correct ?? vs ?? graph is [Given ?? = ?? . ???? lit-atm mol /?? ] Ans: ?? = 0.16l AIM ? to find slope of P -T curve ???? = ?????? ?? ?? = ???? ?? 2Ag( g)? 3 B( g)+ 2C ( g) 1 - 0.4 3 2 ( 0.4) 2 2 × 0.4 0.6 0.6 0.4 Total moles ? 0.6 + 0.6 + 0.4 ? 1.6 Slope tan ?? = ?? ?? = ???? ?? = 1.6 × 0.08 0.16 tan ?? = 0.8 ?? = tan -1 0.8 Q2: Calculate the ratio of rate of effusion of ?? ?? and ?? ?? from a container containing ???????? ?? ?? and ?????? ?? ?? : (A) ?? : ?? (B) ?? : ?? (C) ?? : ?? (D) 4: 1 Ans: (A) ?? O 2 ? ?? O 2 v M O 2 ? ?? O 2 v ?? O 2 ?? H 2 ? ?? H 2 v ?? H 2 ? ?? H 2 v ?? H 2 ?? O 2 ?? H 2 = v ?? H 2 ?? O 2 ?? O 2 ?? H 2 = v 2 32 ( 16 /32 2 /3 ) = 1 8 Q3: Kinetic energy of one mole of ???? ?? at ' ?? ' ?? is: (A) 3 RT (B) ?? /?? RT (C) ?? /?? RT (D) None of these Ans: (C) Kinetic energy of CO 2 at T K Total K.E. of energy ? 3 2 RT Q4: The mean kinetic energy of 1 mole of ???? ?? at ?????? ?? is ?? . The average kinetic energy at ???????? ?? will be: (A) ?? /?? (B) ???? (C) ?? /???? (D) ?? Ans: (B) K.E. = 3 2 ???? So at T = 300 K.E. = ?? = 3 2 ?? ( 300 ) At ?? = 1500 ?? . ?? . = 3 2 ?? ( 1500 )= ( 3 2 ?? ( 300 ) )( 5)= ?? 5 = 5?? Answer = B Q5: At what temperature the RMS velocity of oxygen molecules is equal to that of ???? ?? at ?????? ?? ? (A) +?????? ° ?? (B) -?????? ° ?? (C) ???? ° ?? (D) ?????? ?? Ans: (B) RMS velocity = v 3???? ?? Let temperature be to RMS O 2 = RMS SO 2 ? v 3RT MO 2 = v 3RT( 300 ) M SO 2 ? T 0 = 300× 32 64 = 150 K = -123 ° C Q6: Temperature at which most probable speed of ?? ?? becomes equal to root mean square speed of ?? ?? is [Given: ?? ?? at ?????? ° ?? ]: (A) ?????? ?? (B) ???????? ?? (C) ?????? ?? (D) ?????? ?? Ans: (B) v 2?? ?? 0 ?? O 2 = v 3???? ?? N 2 = 2×?? 0 32 = 3×700 K 28 ? T 0 = 1200 K Q7: A rigid container containing ???????? gas at some pressure and temperature. The gas has been allowed to escape (do not consider any effusion or diffusion) from the container due to which pressure of the gas becomes half of its initial pressure and temperature become ( ?? /?? rd of its initial. The mass of gas (in gms) escaped is: (A) 7.5 (B) 1.5 (C) 2.5 (D) 3.5 Ans: (C) Final pressure, temperature = ?? ?? , ?? ?? Initial pressure, temperature = P 0 'T i Mole ratio = Mass ratio ? Escaped gas = 10 - 10 × 3 4 = 2.5gms Q8 : if the absolute temperature and pressure of a gas are doubled, its volume would become: (A) Doubled (B) Halve (C) Increases four times (D) Remains the same Ans: (D) P 0 V 0 = ???? T 0 When pressure and temperature doubled ?? ?? ?? ?? = ???? ?? ?? 2?? 0 ?? ?? = ???? ( 2?? 0 ) ?? ?? = ???? ?? 0 ?? 0 = ?? 0 from (i) ? Volume remains same Q9: Two gases contained separately in flasks of equal pressures of 1 atmosphere each. What will be the resultant pressure if the flasks are connected? (A) 2 atmosphere (B) 1/2 atmosphere (C) 1 atmosphere (D) None of these Ans: (C) ?? 0 ?? 0 = ???? ?? 0 ?? ?? ?? ?? = ?? ?? ?? ?? ?? Now ?? ?? = ?? 0 , ?? ?? = 2?? 0 , ?? ?? = 2?? 0 ? ?? ?? = ?? 0 ?? ?? 0 ?? 0 = ?? 0 ?? ?? = ?? 0 = 1 atm Q10: At very low pressure, the Vander Waal's equation for one mole is written is: (A) ???? - ???? = ???? (B) ???? + ?? = ???? (C) ???? = ???? (D) ( ?? + ?? /?? ?? ) ( ?? )= ???? Ans: (D) At very low pressure, attractive forces are low, volume occupied is higher compared to volume occupied by molecule ? V >> b ? Real gas equation ( P + a V 2 )V = RT Q11: A gas cylinder contains 0.3 mol of ?? ?? , ?? . ?? mol of ?? ?? and 0.5 mol of helium. If the total pressure is ?? atmospheres, what will be the partial pressure of nitrogen? (A) ?? /?? ?????? (B) ?? . ?? × ?? × ?????? ???? (C) ?? ?? .?? × ?????? ???? (D) ?? ?? .?? × ?????? Ans: (B) ?? 1 mole fraction of N 2 = 0.3 0.3+0.2+0.5 = 0.3 ? Partial pressure = ?? P P = 0.3Patm = 0.3P× 760 mm of Hg Q12: ???? ?? each of sulphur dioxide, phosphine and hydrogen are kept in three flasks. Decreasing order of number of atoms is: (A) Phosphine, sulphur dioxide, hydrogen (B) Hydrogen, phosphine, sulphur dioxide (C) Sulphur dioxide, phosphine, hydrogen (D) Hydrogen, sulphur dioxide, phosphine Ans: (B) 10 g each of SO 2 , PH 3 and H 2 are kept in 3 flasks We know, Number of atom ? Number of molecule And, for a given mass of gas Number of mole = Mass Molecule weight Which impliesRead More
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