52671 + Interview Questions in Chemistry in Class 11 Page 1005 InterviewSolution

50201.	A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25^(@)C produces 6.11 litres of CO_(2). Find out the amount of heat evolvedon burning one litre of the mixture. The heatsof combustion of ethylene and methane are - 1423 and - 891kJ mol^(-1) at 25^(@)C
Answer» Solution :Combustion REACTION of ETHYLENE and methane are `C_(2)H_(4) + 3O_(2)rarr 2CO_(2) + 2H_(2)O , Delta H = - 1423kJ` `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2)O , Delta H = - 891 kJ` Suppose volume of `C_(2)H_(4)` in the mixture of `C_(2)H_(4)` in the mixture`=X` litres. Then volume of `CH_(4)` in the mixture `= ( 3.67 -x) ` litres From the abovereaction `:` 1 litres of `C_(2) H_(4)` gives `CO_(2)= 2 `litres `:. ` x litre of `C_(2)H_(4)` gives ` CO_(2) = 2 x `litres 1 litre of `CH_(4)` gives `CO_(2) = 1 ` litres `:.( 3.67 - x)` litres of `CH_(4)` gives `CO_(2) = ( 3.67 - x)` litres Total `CO_(2)` produced `= 2 x + ( 3.67 - x ) = ( 3.67 + xx) ` litres `:. 3.67+ x = 6.11` or ` x= 2.44` `:. ` 1 litre of the mixture will contain `C_(2)H_(4) = ( 2.44)/( 3.67 )= 0.66 `litre and ` CH_(4) = 1 - 0.66 = 0.34 ` litre Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `0^(@)C= 22.4L` Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `25^(@)C = ( 22.4 xx 298 ) /( 273)= 24.45 L` 24.45 litres of `C_(2)H_(4)`at `25^(@)C` give heat `= 1423kJ` `:. 0.66` litre of `C_(2) H_(4)` will give heat`= ( 1423)/( 24.45) xx 0.66 kJ = 3841kJ` `24.45` litre of`CH_(4)` give heat`= 891 kJ` `:. ` 0.34litre of` CH_(4)` will give heat`= (891)/(24.45) xx0.34 kJ = 12.39 kJ` `:. `Total heatproduced `= 38.41 +12.39 =50.38 kJ`

Discussion

50202.	A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25^@C and at 1 atm pressure produce 6.11 lit of carbon dioxide. Find out the amount of heat evolved in kJ, during this combustion. (DeltaH_C(CH_4)=-890 "kJ mol"^(-1)) and (DeltaH_C(C_2H_4)=-1423 "kJ mol"^(-1)) .
Answer» SOLUTION :Given : `DeltaH_C(CH_4)=-890 "kJ mol"^(-1)` `DeltaH_C(C_2H_4)=-1423 "kJ mol"^(-1)` Let the mixture containx lit of `CH_4` and (3.67-X) lit of ethylene . `underset" x lit"(CH_4)+2O_2 to underset"x lit"(CO_2) + 2H_2O` `underset"(3.67-x)lit"(C_2H_4 + 3O_2) to underset"2(3.67-x)lit"(2CO_2 + 2H_2O)` Volume of CARBON dioxide formed =x +2 (3.67 -x) =6.11 lit x+7.34 -2x =6.11 x=1.23 lit Given mixture CONTAINS 1.23 lit of methane and 2.44 lit of ethylene , hence `DeltaH_C=[(DeltaH_C(CH_4))/"22.4 lit"xx(x)"lit"]+[(DeltaH_C(C_2H_4))/"22.4 lit"xx(3.67-x)"lit"]` `DeltaH_C=[(-890"kJ mol"^(-1))/"22.4 lit"xx"1.23 lit"]+[(-1423)/"22.4 lit"xx(3.67-1.23)"lit"]` `DeltaH_C=[-48.87 "kJ mol"^(-1) ]+[-155 kJ "mol"^(1)]` `DeltaH_C=-203.87 "kJ mol"^(-1)`

Discussion

50203.	A gas mixture contains equal number of molecules of N_(2) and SF_(6) Some of it is passed through gaseous effussion apparatus Calculate how many molcules of N_(2) are pressent in the product gas for every 100 molecules of SF_(6) ? (At wt of F=20) .
Answer» ANSWER :233

Discussion

50204.	A gas mixture contains accetylene and carbondioxide. 20 litres of this mixture required 20 litres of oxygen for complete combustion. If all gases are measured under similar conditions of temperature and pressure, the percentage of acetylene in the mixture is
Answer» `50%` `40%` `60%` `75%` SOLUTION :`C_(2)H_(2) = (20)/(50) XX 100` 1 lit `-(5)/(2)` lit 20 lit - ? 50 lit `%C_(2)H_(2) = (20)/(50) xx 100`

Discussion

50205.	A gas mixture consisting of 1mol of N_(2) and 3 mol of O_(2) has a pressure of 2 atmm at 0^(@)C. Keeping the volume and the temperature of the mixture constant, some amount of O_(2) was removed from the mixture. As a result, the total pressure of the mixture and the partial pressure of N_(2) in the mixture became 1.5 atm and 0.5 atm respectively. the amount of oxygen gas removed was-
Answer» 8 g 16 g 32 g 64 g Answer :C

Discussion

50206.	A gas mixture at 27^(@)C and 1 atm contains equal masses of He,H_(2),CO_(2) and CH_(4). How do their molecular velocities compare?
Answer» `He=H_(2)=CO_(2)=CH^(4)` `HeltH_(2)ltCO_(2)ltCH_(4)` `H_(2)HeltCH_(4)ltCO_(2)` `CO_(2)ltCH_(4)ltHeltH_(2)` Answer :d

Discussion

50207.	A gas is initially at 1 atm pressure. To compress it to 1/4 th of its initial volume, pressure to be applied is:1 atm, 2 atm, 3 atm, 1/4 atm
Answer» 1 ATM 2 atm 4 atm 1/4 atm. Solution :4 atm

Discussion

50208.	A gas is filled into a bulb connected to anopen limb manometer. The level of mercury in the open arm is 2.1 cm lower than that in the other arm of the manoneter. The atmospheric pressure is 740 mm. What is the pressure of the gas ?
Answer» Solution :As level in the open arm of the manometer is lower that that in the arm connected to the gas BULB, this means that atmospheric PRESSURE is greater than pressure of the gas by difference of the two LEVELS. HENCE, pressure of gas=740-21=719 MM.

Discussion

50209.	A gas is filled in a container whose mass is 916.4 g. The mass of the gas plus container is found to be 917.64g. If the container can hold 1107 cm^3 of water, find the density of the gas.
Answer» Answer :`1.1 XX 10^(-3) g CM^(-1)`

Discussion

50210.	a gas is enclosed in a room. The pressure, density, temperature and number of moles are p atm, g cm^(-3), t^(@)C andn moles repectively. What will be the pressure, temperature, density and number of moles, in each compartment. If room is partitioned into four equal compartments ?
Answer» Solution :Pressure in each compatment is same. P. TEMPERATURE will remain same `t^(@)C`. Density will remain same. `d GCM^(-3)` The number of MOLES in each COMPONENT will be, `(N)/(4)` moles.

Discussion

50211.	A gas is enclosed in a room. The temperature, pressure, density and the number of moles respectively are t^@C, p " atm, d g " cm^(-3)and n moles. What will be the pressure, temperature, density and number of moles in each compartment if the room is partitioned into four equal compartments ?
Answer» Solution :When the wall between compartments 1 and 2 is removed : Pressure = p atm. This is because the number of moles and volume of gas will INCREASE to the same EXTENT, i.e., will get doubled. DENSITY = dg `CM^(-3)` . Density of the gas remains unchanged because both mass and volume get doubled. Number of moles = `n/4+n/4=n/2`.

Discussion

50212.	A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy DeltaU of the gas in joules will be
Answer» `-500 J` `-505 J` `+505 J` `1136.25 J` SOLUTION :`DELTAU = q + W` For adiabatic process q = 0 `DeltaU = -w = -pDeltaV` ` = -2.5 atm (4.5 - 2.5)L` ` =-5 LATM = -5 xx 101.3 J = -506.5J`

Discussion

50213.	A gas is allowe to expandin a well insulated container againsta constant external pressure of2.5 atmfrom an initial volume of2.50 Lto a final volume of 4.50L. The change in internal energy DeltaU of thegas in Joules will be
Answer» <P>1136.25J` `-500J` `-505J` `+505J` SOLUTION :`DeltaU =q+w` For well insulated CONTAINER, process is adiabatic , `q=0` `:. DeltaU =w= - P DeltaV` `= - 2.5 atm ( 4.5 - 2.5) L` `=- 5L atm` `= 5xx 101.3 J= - 506.5 J ` `= -505J`

Discussion

50214.	A gas heated to 0^(@)C to 546^(@)C at 5 bar pressure at 5 bar pressure it its volume becomes the third then calculate final pressure.
Answer» ANSWER :45 BAR

Discussion

50215.	A gas has molecular formula (CH)_n.If vapour density of the gas is 39, what should be the formulaof the compound ?
Answer» `C_3H_3` `C_4H_4` `C_2H_2` `C_6H_6` Solution :Mol. Wt. `=2xxV.D. = 2xx39=78` `(CH)_n=(13)_n " or " 13xxn=78` N=78/13=6 `therefore " MOLECULAR formula " = C_6H_6`

Discussion

50216.	A gas has density 1.52gL^(-1)at 1.5atm. Calculate its RMS velocity.
Answer» Solution :`1.732 XX 10^(3) CMS^(-1)`

Discussion

50217.	A gas fired power plant burns methane, CH_(4)(g) for which standard enthalpy of combustion is -890 kJ mol^(-1). How many moles of CO_(2) gas are produced for every 1.0MJ ( mega joule) of energy produced by it ?
Answer» Solution :`1 MJ = 10^(6) J`

Discussion

50218.	A gas formed by the action of alcohlic KOH on ethyl iodide, decolourless alkaline KMnO_(4) solution. The gas is
Answer» `CH_(4)` `C_(2)H_(6)` `C_(2)H_(4)` `C_(2)H_(2)` Answer :C

Discussion

50219.	A gas filled in a bulb of capacity 25.1 mL at 27°C and 750 mm pressure weighs 0.072 g. If 1 litre of hydrogen at S.T.R weighs 0.09 g, calculate the molecular mass of the gas.
Answer» Solution :Let the volume of the gas at STP be `V_(2)` mL `P_(1) = 750 MM, V_(1) = 25.1 mL, T_(1) = 27^(@) C = 300 K` `P_(2) = 760 mm, V_(2) = ?, T_(2) = 273 K` According to the gas equation: `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` Substituting the values, we get `V_(2) = (750 xx 25.1 xx 273)/(300 xx 760)` = 22.5 mL Mass of 22.5 mL of `H_(2)` at S.T.P. `=0.09/1000 xx 22.5 = 2.025 xx 10^(-3) G` V.D. = `("Mass of 22.5 mL of gas at S.T.P")/("Mass of 22.5 mL of H_(2) at S.T.P")` `=0.072/(2.025 xx 10^(-3))` = 35.55 Hence, MOLECULAR mass `=2 xx V.D. = 2 xx 35.55 = 71.1`

Discussion

50220.	A gas filled freely callapsible ballon is pushed from the surface level of take to a depth of 100 meter Calculate what per cent of its original volume the balloon finally have ? Assume ideal gas nature .
Answer» ANSWER :`9.37%ofV_(i)`

Discussion

50221.	A gas expands isothermally and reversibly. The work done by the gas is
Answer» Zero Maximum Minimum Not known Answer :B

Discussion

50222.	A gas expands from a volume of 1m^(3) to a volume of2m^(3) against an external pressure of10^(5) Nm^(-2). The work done by the gas will be
Answer» `10^(5) kJ ` ` 10^(2) kJ` ` 10^(2) J` `10^(3)J` SOLUTION :`w = - PDeltaV =- 10^(5) NM^(-2) ( 2-1) m^(3) =- 10^(5) Nm = - 10^(5) J = -10^(2) kJ `

Discussion

50223.	A gas expands against a variable pressure given by P = (20)/(V) (where P in atm and V in L). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J. How much heat is absorbed by the gas during expansion?
Answer» 46 J 4660 J 5065.8 J 4260 J Solution :`intdw=int-P.DV` `"IMPLIES"w=-int20.(dV)/(V)=-20ln.(V_(2))/(V_(1))` `w=-46.06 " L-atm"=-4665.8 J` `DeltaU=q+wimplies400=q-4665.8` `q = 5065.8 J`

Discussion

50224.	A gas expands against a constant external pressure so thatthe work done is 607.8 J. The work done in litre atmosphere is
Answer» ANSWER :6

Discussion

50225.	A gas diffuses one-thrid as fast as O_(2) at 100^(@)C. This gas could be:
Answer» He(M=4) `C_(2)H_(5)(M=48)` `C_(7)H_(12)(M=96)` `C_(5)F_(12)(M=288)` ANSWER :d

Discussion

50226.	A gas diffuses four times as quickly as oxygen. The molecular weight of the gas is
Answer» 2 4 8 16 ANSWER :A

Discussion

50227.	A gas described by van der Waals equation
Answer» behaves similar to an ideal gas in the limit gas in the limit of large molar volumes behaves similar to an ideal gas in the limit of large pressures is characterised by van der Waals coefficients that are dependent on identity of the gas but are independent of the pressure has the pressure that is lower than the pressure EXERTED by the same behaving ideally Solution :van der Waals equation is `(P+(an^(2))/(V^(2)))(V-nb)=nRT`. (a) is correct because for large molar volumes, `an^(2)//V^(2)` is small and can be neglected in comparison to P and nb can be neglected in comparison to V. Eqn. becomes PV=nRT. (c ) is correct because van der Waals coefficients depend only on the NATURE of the gas. (d) is correct because due to inward pull on the MOLECULES striking the WALL, the pressure is lower.

Discussion

50228.	A gas decolourises alkaline KMnO_4 solution but does not give precipitate with AgNO_3, it is
Answer» `CH_4` `C_2H_4` `C_2H_2` `C_2H_6` SOLUTION :`C_2H_4 (CH_2=CH_2)` DECOLOURISES `KMnO_4` solution but does not GIVE PPT. with `AgNO_3` solution.

Discussion

50229.	A gas cylinder contains oxygen gas at 25^@C and 15.0 atm. If the temperature of the surroundings rises to 42^@C, whatwould be the pressure of the gas in the cylinder ?
Answer» Solution :In the present CASE, `P_1 = 15.0 " atm,"T_1 = 25^@C = 25 + 273 = 298 K, P_2 = ?, T_2 = 42^@C = 42 + 273 = 315 K` Using the pressure-temperature law `P_1/T_1 = P_2/T_2`and substituting the values, we get `P_2 = (P_1T_2)/(T_1) = (15.0 x 315)/(298) = 15.86` atm

Discussion

50230.	A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmospheres. The pressure gauge of the cylinder indicates 12 atmosphere at 27^(@)C. Due to sudden fire in the building, the temperature starts rising.At what temperature the cylinder will explode ?
Answer» SOLUTION :`P_(1)`=12 ATM, `T_(1)`=300 K, `P_(2)`=14.9 atm, `T_(2)`=? As volume of the CYLINDER is constant, `(P_(1))/(T_(1))=(P_(2))/(T_(2))`

Discussion

50231.	A gas cylinder can withstand a pressure of 15 bar. The pressure gauze indicates 12 bar at 27^(@)C. IF the building catches fire suddenly. At what temperature will the cylinder explode ?
Answer» Solution :`P_(1) = 12 " bar"P_(2) =`bar `T_(1) = 273 + 27 = 300 K ""T_(2) = ?` `(P_(1))/(T_(1)) = (P_(2))/(T_(2)) ""T_(2) = (P_(2) xx T_(1))/(P_(1)) = (15 xx 300)/(12)` TEMPERATURE at which cylinder explodes = 375 K = `102^(@)C`

Discussion

50232.	A gas cylinder withstands a pressure of 14.9 atm. Its pressure gauze indicates 12 atm at 27^@C. If the building catches fire suddenly, at what temperature the cylinder explodes?
Answer» ANSWER :above `99.5^@C`

Discussion

50233.	A gas cylinder can withstand a pressure of 14.9 bar at 27^@C. The pressure gauze indicates 12 bar. If the building catches fire suddenly, at what temperature the cylinder expodes?
Answer» SOLUTION :above `99.5^@ `C

Discussion

50234.	A gas mixture contains acetylene and carbondioxide. 20 lit of this mixture requires 20 lit of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is
Answer» 0.5 0.4 0.6 0.75 Answer :B

Discussion

50235.	A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure or 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.
Answer» Solution :Given DATA q = 400 J, `V_(1) = 5L , V_(2) = 10 L` `Delta U` = q - W (heat is given to the system (+q), work is done by the system (-w) `Delta u = q - Pd V ` = 400 J - 1 atm ( 10 - 5) L = 400 J - 5 atm L[`:'`1 L atm = 101 . 33 J ] = 400 J - ` 5 xx 101 . 33 ` J = 400 J - 506 65 J = - 106 . 65 J .

Discussion

50236.	A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is
Answer» 30.38 JOULE 25 Joule 5 k/Joule 16 Joule ANSWER :A

Discussion

50237.	A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is _____
Answer» SOLUTION :`DELTAV`=expansion in VOLUME =100 to 250 `DeltaV=V_2-V_1`=250-100 `DeltaV`=150 ml =0.15litre Work done = ? Pressure =2 atm `w=-P DeltaV` `=-2xx0.15 "LITRE" xx 101.3 J L^(-1) "atm"^(-1)` =-30.39 J

Discussion

50238.	A gas can be liquefied if :
Answer» FORCES of attraction are LOW under ORDINARY conditions forces of attraction are high under ordinary conditions forces of attraction are zero under ordinary conditions forces of attraction either high or low under ordinary conditions Solution :As long IMF is PRESENT, gas can be liquified.

Discussion

50239.	A gas can be liquefied by
Answer» COOLING Compressing Both 1 and 2 None of these Solution :APPLYING P below `T_C`.

Discussion

50240.	A gas can be liquefied by pressure alone when its temperature is
Answer» Higher than its critical temperature LOWER than its critical temperature EQUAL to its critical temperature Equal to its Boyle's temperature SOLUTION :LIQUIFIED only below `T_C`.

Discussion

50241.	A gas bulb containing air is connected to an open limb manometer at 27^(@) C and at 750 mm Hg . Assuming that intially the level of Hg in the both limbs were same . The bulb was heated to 77^(@)C , what will be differences in the levels of Hg in two limbs ? (Assuming the volume difference of the gas produced is negligible at higher temperature) .
Answer» <P>7.5 cm Hg 8 cm Hg 6 cm Hg 12.5 cm Hg Solution :`because (P_(1))/(T_(1)) = (P_(2))/(T_(2)) , (750)/(300) = (P_(2))/(350)` `therefore P_2 = 875` mm Hg `therefore` So , difference in height= `875 - 750 = 125` mm Hg = 12.5 cm of Hg

Discussion

50242.	A gas bulb of 1 litre capacity contains 2.0 xx 10^(21) molecules of nitrogen exerting a pressure of 7.57 xx 10^3 Nm^(-2) . Calculate the root mean square speed and the temperature of gas molecules. If the ratio of most probable speed to the root mean square speed is 0.82, calculate the most probable speed for the molecules at this temperature.
Answer» Solution :Number of moles of NITROGEN = `("number of molecules")/(6.023 xx 10^(23))` `= (2.0 xx 10^(11))/(6.023 xx 10^(23)) = 3.32 xx 10^(-3)` According to the gas equation PV = NRT In the PRESENT case, `P = 7.57 xx 10^3 N m^(-2), V = 1` `L = 10^(-3) m^(3)`, `R = 8.314 J K^(-1) mol^(-1)` T = ? Substituting the values, we have `7.57 xx 10^3 xx 10^(-3) = 3.32 xx 10^(-3) xx 8.314 XXT` `:. "" T= 274.2 K` RMS speed is given by `u = sqrt((3RT)/M)= sqrt((3 xx 8.314 xx 274.2)/(28 xx 10^(-3)))= 494.2 m s^(-1)` (Molar mass of `N_2 " in kg " = 28 xx 10^(-3)`) `:. " Most probable speed " = 0.82 xx RMS` speed `:. " Most probable speed of " N_2` under given conditions `= 0.82 xx 494.2 = 405.2 m s^(-1)`

Discussion

50243.	A gas believed to be the cause of explosin in coal mine is
Answer» `CH_4` `C_(2)H_6` `C_(3)H_(8)` `C_(4)H_(10)` ANSWER :A

Discussion

50244.	A gas at a pressure of 5 atm is heated from 0^(@)C" to "546^(@)C and is simultaneously compressed to one third of its original volume. Find the final pressure of the gas.
Answer» SOLUTION :`45 ATM`

Discussion

50245.	A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option above the gas and its compressibility factor (Z) is :
Answer» `Z lt1` and attractive forces are dominant `Z lt 1` and REPULSIVE forces are dominant `ZGT1` and attractive forces are dominant `ZGT 1` and repulsive forces are dominant. Solution :`Z lt1` and attractive forces are dominant

Discussion

50246.	A gas at 0^@C and 1 atmospheric pressure occupies 2.5 litres. What change in temperature would be necessary if the pressure is to be adjusted to 1.5 atmospheres and the gas has been transferred to a 2.0 litre container?
Answer» SOLUTION :`54.6^@C`

Discussion

50247.	A gasabsorbs photonof 355nm andemits two wavelengthif oneof theemissionis at 680nm the otheris at .
Answer» 1035 NM 325 nm 743 nm 518 nm Solution :`E= hv = (hc)/(lambda)` `(hc)/(lambda) = (hc )/(lambda)+ (hc)/( lambda_(2))` `(1)/(355)=(1)/(680)+ (1)/(lambda)` `=(680)/(355) xx (355)/( 680)` `lambda_(2)= (355 xx 680)/( 325)= 742.769 = 743nm`

Discussion

50248.	A gas absorbs a photon of 355 nm and emit at two wavelengths. If one of the emission is at 680
Answer» 325 nm 743 nm 518 nm 1035 nm Solution :As energies are additive, `E = E_(1) + E_(2)` or `(HC)/(LAMDA) = (hc)/(lamda_(1)) + (hc)/(lamda_(2)) or (1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2))` But `lamda = 355 nm, lamda_(1) = 680NM,` `:. (1)/(355) = (1)/(680) + (1)/(lamda_(2))` or `(1)/(lamda_(2)) = (1)/(355) - (1)/(680) = (325)/(355 xx 680)` or `lamda_(2) = (355 xx 680)/(325) = 745nm`

Discussion

50249.	A gas absorbs 400 J of heat and expands by 2 xx 10^(-3)m^(3) against a constant pressure of 1 xx 10^5 Nm^(-2). The change in internal energy is (1 L atm = 100 J)
Answer» ZERO `200` `-600J` `-200J` ANSWER :B

Discussion

50250.	A gas absorbs 100 calories of heat energy and is compressed from 10L to 5L by applying an external pressure of 2 atm. Change in internal energy in calories will be nearly.
Answer» 312 342 426 562 Solution :`DELTA E = Q + W`

Discussion

Explore topic-wise InterviewSolutions in .

A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25^(@)C produces 6.11 litres of CO_(2). Find out the amount of heat evolvedon burning one litre of the mixture. The heatsof combustion of ethylene and methane are - 1423 and - 891kJ mol^(-1) at 25^(@)C

A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25^@C and at 1 atm pressure produce 6.11 lit of carbon dioxide. Find out the amount of heat evolved in kJ, during this combustion. (DeltaH_C(CH_4)=-890 "kJ mol"^(-1)) and (DeltaH_C(C_2H_4)=-1423 "kJ mol"^(-1)) .

A gas mixture contains equal number of molecules of N_(2) and SF_(6) Some of it is passed through gaseous effussion apparatus Calculate how many molcules of N_(2) are pressent in the product gas for every 100 molecules of SF_(6) ? (At wt of F=20) .

A gas mixture contains accetylene and carbondioxide. 20 litres of this mixture required 20 litres of oxygen for complete combustion. If all gases are measured under similar conditions of temperature and pressure, the percentage of acetylene in the mixture is

A gas mixture at 27^(@)C and 1 atm contains equal masses of He,H_(2),CO_(2) and CH_(4). How do their molecular velocities compare?

A gas is initially at 1 atm pressure. To compress it to 1/4 th of its initial volume, pressure to be applied is:1 atm, 2 atm, 3 atm, 1/4 atm

A gas is filled into a bulb connected to anopen limb manometer. The level of mercury in the open arm is 2.1 cm lower than that in the other arm of the manoneter. The atmospheric pressure is 740 mm. What is the pressure of the gas ?

A gas is filled in a container whose mass is 916.4 g. The mass of the gas plus container is found to be 917.64g. If the container can hold 1107 cm^3 of water, find the density of the gas.

a gas is enclosed in a room. The pressure, density, temperature and number of moles are p atm, g cm^(-3), t^(@)C andn moles repectively. What will be the pressure, temperature, density and number of moles, in each compartment. If room is partitioned into four equal compartments ?

A gas is enclosed in a room. The temperature, pressure, density and the number of moles respectively are t^@C, p " atm, d g " cm^(-3)and n moles. What will be the pressure, temperature, density and number of moles in each compartment if the room is partitioned into four equal compartments ?

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy DeltaU of the gas in joules will be

A gas is allowe to expandin a well insulated container againsta constant external pressure of2.5 atmfrom an initial volume of2.50 Lto a final volume of 4.50L. The change in internal energy DeltaU of thegas in Joules will be

A gas heated to 0^(@)C to 546^(@)C at 5 bar pressure at 5 bar pressure it its volume becomes the third then calculate final pressure.

A gas has molecular formula (CH)_n.If vapour density of the gas is 39, what should be the formulaof the compound ?

A gas has density 1.52gL^(-1)at 1.5atm. Calculate its RMS velocity.

A gas fired power plant burns methane, CH_(4)(g) for which standard enthalpy of combustion is -890 kJ mol^(-1). How many moles of CO_(2) gas are produced for every 1.0MJ ( mega joule) of energy produced by it ?

A gas formed by the action of alcohlic KOH on ethyl iodide, decolourless alkaline KMnO_(4) solution. The gas is

A gas filled in a bulb of capacity 25.1 mL at 27°C and 750 mm pressure weighs 0.072 g. If 1 litre of hydrogen at S.T.R weighs 0.09 g, calculate the molecular mass of the gas.

A gas filled freely callapsible ballon is pushed from the surface level of take to a depth of 100 meter Calculate what per cent of its original volume the balloon finally have ? Assume ideal gas nature .

A gas expands isothermally and reversibly. The work done by the gas is

A gas expands from a volume of 1m^(3) to a volume of2m^(3) against an external pressure of10^(5) Nm^(-2). The work done by the gas will be

A gas expands against a variable pressure given by P = (20)/(V) (where P in atm and V in L). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J. How much heat is absorbed by the gas during expansion?

A gas expands against a constant external pressure so thatthe work done is 607.8 J. The work done in litre atmosphere is

A gas diffuses one-thrid as fast as O_(2) at 100^(@)C. This gas could be:

A gas diffuses four times as quickly as oxygen. The molecular weight of the gas is

A gas described by van der Waals equation

A gas decolourises alkaline KMnO_4 solution but does not give precipitate with AgNO_3, it is

A gas cylinder contains oxygen gas at 25^@C and 15.0 atm. If the temperature of the surroundings rises to 42^@C, whatwould be the pressure of the gas in the cylinder ?

A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmospheres. The pressure gauge of the cylinder indicates 12 atmosphere at 27^(@)C. Due to sudden fire in the building, the temperature starts rising.At what temperature the cylinder will explode ?

A gas cylinder can withstand a pressure of 15 bar. The pressure gauze indicates 12 bar at 27^(@)C. IF the building catches fire suddenly. At what temperature will the cylinder explode ?

A gas cylinder withstands a pressure of 14.9 atm. Its pressure gauze indicates 12 atm at 27^@C. If the building catches fire suddenly, at what temperature the cylinder explodes?

A gas cylinder can withstand a pressure of 14.9 bar at 27^@C. The pressure gauze indicates 12 bar. If the building catches fire suddenly, at what temperature the cylinder expodes?

A gas mixture contains acetylene and carbondioxide. 20 lit of this mixture requires 20 lit of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is

A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure or 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.

A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is

A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is _____

A gas can be liquefied if :

A gas can be liquefied by

A gas can be liquefied by pressure alone when its temperature is

A gas believed to be the cause of explosin in coal mine is

A gas at a pressure of 5 atm is heated from 0^(@)C" to "546^(@)C and is simultaneously compressed to one third of its original volume. Find the final pressure of the gas.

A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option above the gas and its compressibility factor (Z) is :

A gas at 0^@C and 1 atmospheric pressure occupies 2.5 litres. What change in temperature would be necessary if the pressure is to be adjusted to 1.5 atmospheres and the gas has been transferred to a 2.0 litre container?

A gasabsorbs photonof 355nm andemits two wavelengthif oneof theemissionis at 680nm the otheris at .

A gas absorbs a photon of 355 nm and emit at two wavelengths. If one of the emission is at 680

A gas absorbs 400 J of heat and expands by 2 xx 10^(-3)m^(3) against a constant pressure of 1 xx 10^5 Nm^(-2). The change in internal energy is (1 L atm = 100 J)

A gas absorbs 100 calories of heat energy and is compressed from 10L to 5L by applying an external pressure of 2 atm. Change in internal energy in calories will be nearly.