Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

89551.

A gas on subjecting to adiabatic expansion gets cooled due to:

Answer»

FALL in temperature
Loss of KINETIC energy
Decrease in velocity
Energy USED in doing work

Answer :D
Discussion
89552.

A gas occupies a volume of 250 ml at 700 mm Hg pressure and 25^(@)C . What additional pressure is required to reduces the gas volume to its 4//5 th value at the same temperature ?

Answer»

<P>225 mm Hg
175 mm Hg
150 mm Hg
265 mm Hg

Solution :`V_(2) = 250 ml` , `P_(1) `= 700mm Hg
`V_(2) = 250 xx ( 4)/( 5)`
`= 200 ml , P_(2) = ?`
According to Boyle.s LAW,
`P_(1) V_(1) = P_(2) V_(2)`
or `P_(2) = ( 250 xx 700)/( 200) = 875 mm Hg`
Additional pressure required
`= 875 - 700`
= 175 mm Hg
Discussion
89553.

A gas occupies 300ml at 27^(@)C and 730 mm pressure. What would be its volume at STP -

Answer»

162.2 ml
262.2 ml
362.2ml
462.2ml

Solution :Given at `T_(1) = 300 K,T_(2) = 273 K` (STP )
`4V_(1) = 300 ml = (( 300)/( 1000) ) `LITRE , `P_(1) =((730)/( 760))` ATM. `P_(2) = 1` atm, `V_(2)=?`
`(P_(1)V_(1))/( T_(1))=(P_(2)V_(2))/( T_(2))`,
`( 730 xx 300)/( 760 xx 1000 xx 300 ) =( 1 xx V_(2))/( 273) `
`:. V_(2) = 0.2622 `litre =262.2 ml
Discussion
89554.

A gas obey P(V-b)=RT which of the following are correct about this gas ?

Answer»

<P>Isochoric Curves have slope `=(R )/((V-B))`
Isobaric Curves have slope R/P and intercept b
For the gas compressibility FACTOR `=1+(Pb)/(RT)`
The attraction forces are overcome by repulsive forces

Solution :`P(V-b)RT`
`therefore P=(R )/((V-b))T`
Slope `=(R )/((V-b))`
`PV-Pb=RT`
`rArr V =(RT)/(P)+b`
Slope `=(R )/(P)`, Intercept =b
`Z=(PV)/(RT)=1+(Pb)/(RT), Z gt 1`.
Hence repulsive forces predominates.
Hence, (A), (B), (C) and (D) are the correct ANSWER.
Discussion
89555.

A gas mixture of CH_(4) and C_(3)H_(6) undergo complete cracking into C_(s) and H_(2)(g). The total mass of H_(2) (g) produced is 42 gm. If the total volume of the initial gas mixture at 1 atm and 0 .^(@)C is 224 litre then mole % of CH_(4) in original mixture is -

Answer»

`10%`
`20%`
`90%`
`80%`

SOLUTION :MOLE of mixture `= (224)/(22.4) = 10`
`CH_(4) RARR C (s) + 2H_(2) (g)`
`{:(CH_(4),rarr,C (s),+ 2H_(2) (g),),(x "mole",,,2x "mole",),(C_(3)H_(6),rarr,3C (s),+ 3H_(2) (g),),((10 - x) "mole",,,3 (10 - x) "mole",):}`
`[2x + 3 (10 - x )] = 21`
`2x + 30 - 3X = 21`
`x = 9` mole
`%` mole of `CH_(4) = (9)/(10) xx 100 = 90%`
Discussion
89556.

A gas mixture of4 litres of ethylene and methane on complete combustion at 25^(@)C produces 6 litres of CO_(2). Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are-1464 and -976 KJ mol^(-1) at 25^(@)C

Answer»

Solution :gt `X "moles""" (4-x)"moles" ""6"moles"`
Applying `POAC` for `C` atoms,
`2xxx+1xx(4-x)=1xx6, x=2 "lit"`
Thus, the volume of ` C_(2)H_(4)=2"lit"`, and volume of `CH_(2)=2"lit"`.
`:.` volume of `C_(2)H_(4)` in a `1` LITRE mixture `=2//4=0.5 "lit"`.
and volume of `CH_(4)` in a `1` litre mixture `=1-0.5=0.5"lit"`
Now, THERMOCHEMICAL reactions for `C_(2)H_(4)` and `CH_(4)` are
`{:(C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O, Delta H= -1464KJ,,,,),(CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O, Delta H= -976 KJ,,,,):}`
As `Delta H` values given at `25^(@)C`. let US first calculate the volume occupied by one mole of any gas at `25^(@)C` (supposiong pressures as `1 atm`)
Volume per mole at `25^(@)C=(298)/(273)xx22.4=24.4"lit"`
Thus, heat evolved in the combustion of `0.5` lit of `C_(2)H_(4)=(1464)/(24.4)xx0.5= -30KJ`
and heat evolved in the combustion of `0.5` lit of `CH_(4)=(976)/(24.4)xx0.5=-20KJ`.
`:.` total heat evolved in the combustion of `1` litre of the mixture `= -30+(-20)= -50 KJ`
Discussion
89557.

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 evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are -1423 and -891 "kJ mol"^(-1) at 25^(@)C.

Answer»

Solution :
`or x "moles (3.67-x)moles6.11moles"`
Applying POAC for C atoms,
`2xxx+1xx(3.67-x)=1xx6.11,x=2.44 "lit"`
Thus, the volume of `C_(2)H_(4)=2.44` lit, and volume of `CH_(4)=1.23` lit.
`:.` volume of `C_(2)H_(4)` in a 1-LITRE mixture `=(2.44)/(3.67)=0.665` lit
and volume of `CH_(4)` in a 1-litre mixture `=1-0.665=0.335` lit.
Now, thermochemical reactions for `C_(2)H_(4)` and `CH_(4)` are
`C_(2)H_(4)+3O_(2) to 2CO_(2)+2H_(2)O,DeltaH=-1423 "kJ"`
`CH_(4)+2O_(2) to CO_(2)+2H_(2)O, DeltaH=-891 "kJ"`
As DeltaH values given are at `25^(@)C` (supposing pressure as 1 atm)
Volume per mole at `25^(@)C=(298)/(273)xx22.4=24.45 "lit"`
Thus, heat evolved in the combustion of 0.665 lit. of `C_(2)H_(4)`
`=-(1423)/(24.45)xx0.665=-38.70"kJ"`
and heat evolved in the combustion of 0.335 "lit". of `CH_(4)`
`=-(891)/(24.45)xx0.335=-12.20"kJ"`
`:.` TOTAL heat evolved in the combustion of 1 litre of the mixture
`=-38.70+(-12.20)`
`=-50.90 kJ`
Discussion
89558.

A gas mixture of 3.0 L of propaane and butane on complete combustion at 25^(@)C produced 10 L of CO_(2). Find out the composition of the gas mixture.

Answer»

Solution :`C_(3)H_(8)+5O_(2)rarr 3CO_(2)+4H_(2)O`
`C_(4)H_(10)+6(1)/(2)O_(2)rarr 4CO_(2)+5H_(2)O`
Suppose propane = xL. Then butane = (3-x) L
`1 L C_(3)H_(8)` PRODUCES 3 L of `CO_(2)` and 1 L of `C_(4)H_(10)` produced 4 L of `CO_(2)`
Hence, `CO_(2)` produced `=3x +4(3-x)=12-x=10 L`
or x = 2 L
`therefore` Propane = 2 L and butane = 3 - 2 = 1 L.
Discussion
89559.

A gas mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture.

Answer»

`19.97%`
`20.05%`
`50%`
`80.03%`

Solution :Solution CONTAIN `He +CH_(4)`
Their mol. `wt = 4+16 = 20`
`%`wt of `CH_(4) = (" wt of " CH_(4))/(" TOTAL wt") xx 100 = 16/20 xx 100 = 80.0%`
Discussion
89560.

A gas mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture?

Answer»

`19.97%`
`20.05%`
`50%`
`80.03%`

Solution :Equal volumes contain equal NUMBER of moles.
Hence, MOLAR ratio of `He:CH_(4)=1:1`
`therefore" Ratio by weight "=4:16=1:4`
`therefore CH_(4)` present by weight`=(4)/(5)xx100=80%`
Discussion
89561.

A gas mixture contains 50% helium and 50% methane by volume. What is the percentage by mass of methane in the mixture?

Answer»

0.1997
0.2005
0.5
`80.03%`

Solution :Molar and volume ratio will be same. i.e., 1:1
`therefore` Mass of 1 MOLE `CH_(4)` and He will be 16 and 4 g respectively.
Percentage by mass of `CH_(4)=("Mass of"CH_(4))/("Total mass")XX100`
`=(16)/(20)xx100=80%`
Discussion
89562.

A gas is said to behave like an ideal gas when the relation PV/T = constant. When does a real gas not behave like an ideal gas?

Answer»

When the TEMPERATURE is low
When both the temperature and PRESSURE are low
When both the temperature and pressure are low
When the temperature is HIGH and pressure is low

Answer :A::B::C
Discussion
89563.

A gas is taken isochorically from state A to state C as shown in the graph. Choose the correct statement:

Answer»

MOLES of gas FIRST REMAINS CONSTANT and then increases
Moles of gas first increase and then remains constant
Moles of gas first remains constant and then decreases
Moles of gas first decreases and then remains constant.

Answer :C
Discussion
89564.

A gas is said to behave like an ideal gas when the relation (PV)/T = constant. When do you expect a real gas to behave like an ideal gas ?

Answer»

when the TEMPERATURE is ow
when the temperature and PRESSURE are low
when both the temperature and pressure is HIGH
when TE temperature is high and pressure is low

Answer :D
Discussion
89565.

A gas is present of 2 atm. What should be in the increase so that the volume of the gas can be decreased to 1//4^(th) of the initial value if the temperature is maintained constant.

Answer»

Solution :PV = constant for a given mass of gas at constant pressure.
`rArr P_(1) V_(1) = P_(2) P_(2)``P_(1) = `2 atm `V_(1) =V`
`V_(2) = V//4` `P_(2) = ?`
Now, `2 xxV = P_(2) xx ( V)/( 4)rArrP_(2) = 8` atm
`:.` Pressue should be increased from 2 to 8 atm
`:.` total INCREASE = 8- 2 == 6 atm
Discussion
89566.

A gas is heated from0^@ C to 100^@ C at 1.0 atm pressure. If the initial volume of the gas is 10 litre, its final volume would be :

Answer»

7.32 litre
10.0 litre
13.66 litre
20.2 litre

Answer :C
Discussion
89567.

A gas is liquefied

Answer»

Above CRITICAL TEMPERATURE and below cricical pressure
Below critical temperature and above critical pressure
Below criticaltemperature and pressure
Above critical temperature and pressure

SOLUTION :The temperature above which the GAS cannot be liquefied is called the critical temperatureand the minimum pressure REQUIRED to liquefy the gas at the critical temperature is called the critical pressure.
Discussion
89568.

A gas is evolved when a piece of zinc metal placed in dilute sulphuric acid (H_(2)SO_(4)) . What is the gas ?

Answer»

Hydrogen
OXYGEN
Water vapour
Sulphur dioxide

Solution :Hydrogen gas is evolved when a piece of zinc METALS is PLACED in dilute sulphuric acid `(H_(2)SO_(4))` . The chemical reaction is
`Zn + H_(2)SO_(4) to ZnSO_(4) + H_(2) uarr`
Discussion
89569.

A gas is enclosed in a cylinder with a piston. Weights are added to the piston, giving a total mass of 2.20 kg. As a result the gas is compressed and the weights are lowered 0.25 m. At the same time, 1.50 J of heat evolved from the system. What is the change in internal energy of the system ?

Answer»


ANSWER :(3.89 J)
Discussion
89570.

A gas is composed of 30.4% N and 69.6% O. Its density is 11.1 g/L at -20^@Cand 2.5 atm. What are the empirical and molecular formula of the gas

Answer»

SOLUTION :`NO_2, N_2O_4`
Discussion
89571.

A gas is allowed to expand at constant pressure from a volume of 1.0 litre to 10.0 litre against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w and DeltaE (Given 1 L atm = 101 J)

Answer»

`{:("Q","W",DELTA"E"),("250J","-455J","-205J"):}`
`{:("q","w",Delta"E"),("-250J","-455J","-710J"):}`
`{:("q","w",Delta"E"),("250J","455J","710J"):}`
`{:("q","w",Delta"E"),("-250J","455J","205J"):}`

ANSWER :A
Discussion
89572.

A gas has smell like rotten egg and turns lead acetate paper black. The gas is :

Answer»

`NO_(2)`
`H_(2)S`
`CO_(2)`
`SO_(2)`

ANSWER :B
Discussion
89573.

A gas has a volume 250 mL at 27^@C . The volume is doubled by heating the gas at constant pressure. What will be the new temperature ?

Answer»


ANSWER :`327^@ C`
Discussion
89574.

A gas has a vapour density 11.2. the volume occupied by 1 g of the gas at NTP is:

Answer»

1L
11.2L
22.4L
4L

Solution :Molar mass =`2xx11.2=22.4g`
Volume of 1 g compound at STP`=(22.4)/(22.4)=1L`
Discussion
89575.

A gas has a density of 1.2504 g/L at 0^(@)C and a pressure of 1 atm. Calculate the rms, average and the most probable speeds of its molecules at 0^(@)C.

Answer»

Solution :`p = 76 xx 13.6 xx 981 "dynes/cm"^(2)`
`d = 1.2504 g//L = 0.0012504 g//c c`
Now we have,
`C = sqrt((3P)/(d))`
`= sqrt((3 xx 76 xx 13.6 xx 981)/(0.0012504)) = 4.93 xx 10^(4) cm//s`.
`therefore` average SPEED `= 0.9211 xx 4.93 xx 10^(4)`
`= 4.59 xx 10^(4)` cm/s
and most probable speed `= 0.8165 xx 4.93 xx 10^(4)`
`= 4.03 xx 10^(4)` cm/s.
[RMS speed : AV. speed : m.p. speed = 1: 0.9211 : 0.8165]
Discussion
89576.

A gas formed by the action of alcoholic KOH on ethyl iodie, decolourises alkaline KMnO_(4). The gas is

Answer»

`C_(2)H_(6)`
`CH_(4)`
`C_(2)H_(4)`
`C_(2)H_(4)`

SOLUTION :Ethylene is formed by the action of alcoholic KOH on ethyl iodide
`C_(2)H_(5(I+KOH(alc.)toC_(2)H_(4)+KI+H_(2)O`
`C_(2)H_(4)` DECOLOURISES ALKALINE `KMnO_(4)` to form enthylene glycol.
`C_(2)H_(4)+(KMnO_(4)(alk.)) underset(H_(2)O+O) to underset("Ethylene glycol")({:(CH_(2)OH,),(|,),(CH_(2)OH,):}`
Discussion
89577.

A gas formed by the action of alcoholic KOH on ethyl iodide, decolourises alkaline KMnO_4 solution. The gas is

Answer»

`CH_4`
`C_2H_6`
`C_2H_4`
`C_2H_2`

Solution :`CH_3CH_2I+KOH(ALC) to CH_2 =CH_2 +KI + H_2O`
`CH_2=CH_2+underset"Alk PINK"(KMnO_4)overset(H_2O)to underset("COLOURLESS")(underset(OH)underset(|)CH_2-underset(OH)underset(|)CH_2+MnO_4)`
Discussion
89578.

A gas expands , isthemally and reversiby.the work done by the gasis

Answer»

ZERO
maximum
minimum
cannot be determined

Solution :WORK done during isothemal REVERSIBLE expansion of a GAS is maximum
Discussion
89579.

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

Answer»

Zero
Minimum
Maximum
Equal to WORK done

Answer :C
Discussion
89580.

A gas expands in volume from 2L to 5L against a pressure of1 atm at constanttemperature. The work done by the gas will be

Answer»

3 J
`-303.9` J
`-303.9L*` ATM
`303.9 L*`atm

Answer :B::C
Discussion
89581.

A gas expands in an insulated container against a constant external pressure of 1.5 atm from an initial volume of 1.2 L to 4.2 L. The change in internal energy (DeltaU) of gas is nearly (1 L-atm = 101.3 J)

Answer»

`- 456 J`
`+ 456 J`
`- 402 J`
`+ 402 J`

ANSWER :A
Discussion
89582.

A gas expands from a volume of 3.0 dm^3 to 5.0 dm^3against a constant externalpressure of 3.0 atm. The work done during expansion is used to heat 10 moles of water at a temperature 290 K. Calculate the final temperature of water. Specific heat of water = 4.184 Jg^(-1) K^(-1)

Answer»

Solution :`W = -p.DeltaV, 1 atm = 101325 PA , 1dm^3 = 10^(-3) m^3`
290.807 K
Discussion
89583.

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 in J is absorbed by the gas during expansion ?

Answer»


Solution :`dW=-Pdv=(20)/(v)xx DV = 20 int_(1)^(10)(dv)/(v)=20 ln 10 = 10 =-46.06` LATM `= -4665.878 J, "" Delta U = 400 J`
Using first LAW of thermodynamics Q = 5065.878 J.
Discussion
89584.

A gas expands adiabatically at constant pressure such that: T prop(1)/(sqrt(V)) The value of gamma i.e., (C_(P)//C_(V)) of the gas will be:

Answer»

`1.30`
`1.50`
`1.70`
2

Answer :B
Discussion
89585.

A gas expands from a volume of 1 m^(3) to a volume of 2m^(3) against an external pressure of 10^(5) N m^(-2) . The work done by the gas will be

Answer»

`10^(5) KJ`
`10^(2) kJ`
`10^(2) J`
`10^(3) J`

Solution :Work DONE, `w = - p. Delta V`
`= - 10^(5) Nm^(-2) (2 - 1) m^(3)`
`= - 10^(5) N - m`
`= - 10^(5) J or - 10^(2) kJ`
Discussion
89586.

A gas evolved with effervescence on treating a salt with dil. HCl may be CO_(2) or SO_(3). How will you distinguish between them?

Answer»

Solution :a. `SO_(2)` has a PUNGENT burning sulphur SMELL, WHEREAS `CO_(2)` is ODOURLESS. ltbr. (b). `SO_(2)` durns potassium dichromate paper GREEN whereas `CO_(2)` does not.
Discussion
89587.

A gas diffuses at a rate which is twice that of another gas B. The ratio of molecular weights of A to B is

Answer»

1
0.75
0.5
0.25

Answer :D
Discussion
89588.

A gas diffuses 1//5 times as fas as hydrogen. Its molar mass is :

Answer»

25
50
`25 SQRT( 2)`
`50 sqrt( 2)`

SOLUTION :`( r ( GAS))/( r ( H_(2))) = sqrt(( 2)/( M ( gas )))`
`( 1//5)/( 1) = sqrt((2 )/( M ( gas)))`
or ` ( 1)/( 25) = ( 2)/( M ( gas))`
or M ( gas ) = ` 2 xx 25 = 50`
Discussion
89589.

A gas diffuse (1)/(5) times as fast as hydrogen at same pressure. Its molecular weight is

Answer»

50
25
`25 SQRT(2)`
`50 sqrt(2)`

Solution :`r_(g)=(1)/(5).r_(H_(2))`
`(M_(g))/(M_(H_(2)))=[(r_(H_(2)))/(r_(g))]^(2)=(5)^(2)=25`
`M_(g)=2 xx 25 = 50`
Discussion
89590.

A gas deviates from ideal behavious at a high pressure because its molecules

Answer»

Have KINETIC energy
Are bound by covalent bond
Attract one another
Show the Tyndall effect

Solution :Some attraction forces exist between the molecules of real gases. When a molecule approaches the wall of the container it experiences an inward pull as a result of ATTRACTIVE by the neighbouring molecules inside the vessel. Therefore the observed pressure is LESS than the IDEAL pressure and hence GAS deviates from ideal behaviour at high pressure.
Discussion
89591.

A gas deviates from ideal behaviour at a high pressure because its molecules :

Answer»

have KINETIC ENERGY
are bound by covalent bonds
attract one another
show the Tyndall EFFECT

Answer :C
Discussion
89592.

A gas described by van der Waal's equation (i) behaves similar to an ideal gas in the limit of large molar volume (ii) behaves similar to an ideal gas in the limit of large pressure (iii) is characterised by van der Waal's coefficients that are dependent on the identity of the gas but are independent of the temperature (iv) has the pressure that is lower than the pressure exerted by the same gas behaving ideally

Answer»

(i) and (ii)
(i) and (iii)
(i), (ii) and (iii)
(ii) and (IV)

Solution :(i) At very LARGE molar volume
`P+a/(V_(m)^(2))~~P` and `V_(m)-b=V_(m)`
(iii) According to van der WAALS EQUATION .a. and .b. are independent of temperature.
Discussion
89593.

A gas decolourizes alk. KMNo-4 solution (Baseyers reagent) but not give ppt. with AgNO_3. The gas is

Answer»

`CH_4`
`C_2H_4`
butyne-2
`C_2H_6`

ANSWER :B::C
Discussion
89594.

A gas on passing through ammonical solution of AgNO_(3) does not give any percipitate but decolourises alkaline KMnO_(4) solution. The gas may be

Answer»

`CH_(4)`
`C_(2)H_(6)`
`C_(2)H_(4)`
`C_(2)H_(2)`

ANSWER :C
Discussion
89595.

A gas decolourises KMnO_4 solution but gives no precipitate with ammoniacal cuprous chloride is

Answer»

Ethane
Methane
Ethene
Acetylene

Solution :
Discussion
89596.

A gas decolourised by KMnO_(4) solution but gives no precipitate with ammoniacal cuprous chloride

Answer»

ETHANE
ethene
methane
acetylene.

Solution :
Discussion
89597.

A gas cylinder contains 370 g of O_(2) at 30 atm and 25^(@)C. What mass of O_(2) would escape if first the cylinder were heated to 75^(@)C and then the valve were held open until the gas pressure was 1 atm, the temperature being maintained at 75^(@)C ?

Answer»


Solution :First calculate the VOLUME of the cylinder and then WT. of `O_(2)` present at 1 ATM and `75^(@)C` USING PV = n RT
Discussion
89598.

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

Answer»

Solution :Suppose the CYLINDER will BURST at `T_(2)K`.
When the pressure will increase from 12 ATM to 14.9 atm we have,
`(p_(1)V_(1))/(T_(1)) = (p_(2)V_(2))/(T_(2))""p_(1) = 12 " atm ", T_(1) = (27+273)`
`p_(2) = 14.9` atm, `T_(2) = ?`
Here `V_(1) = V_(2)` as the volume does not change
`THEREFORE T_(2) = (p_(2)T_(1))/(p_(1)) = (14.9 XX 300)/(12)`
= 372.5 K
Discussion
89599.

A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of cylinder indicates 12 atm at 27^(@)C. Due to sudden fire in building the temperature starts rising. The temperature at which the cylinder will explode is.

Answer»

`42.5^(@)C`
`67.8^(@)C`
`99.5^(@)C`
`25.7^(@)C`

ANSWER :C
Discussion
89600.

A gas (C_("v,m")=5/2R) behaving ideally was allowed to expand reversibly and adiabaticaly from 1 litre to 32 litre. It’s initial temperature was 327^(@)C. The molar enthalpy change (in J/mole) for the process is

Answer»

`-1125` R
`-575` R
`-1575` R
None of these

Answer :C
Discussion