Q 1 :

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is            [2013]

  • 1 : 4 

     

  • 1 : 2 

     

  • 6 : 9 

     

  • 8 : 9

     

(4)

P1M1=P1RT  and  P2M2=P2RT

  P1P2×M1M2=P1P2

43×23=P1P2

  P1P2=89



Q 2 :

A real gas behaves like an ideal gas if its          [2010]

  • pressure and temperature are both high

     

  • pressure and temperature are both low

     

  • pressure is high and temperature is low

     

  • pressure is low and temperature is high

     

(4)

A real gas behaves as an ideal gas at low pressure and high temperature.



Q 3 :

Which of the following graphs correctly represents the variation of β=-dV/dPV with P for an ideal gas at constant temperature?                    [2002]

  • [IMAGE 504]

     

  • [IMAGE 505]

     

  • [IMAGE 506]

     

  • [IMAGE 507]

     

(1)

According to Boyle's law, PV=constant.

  PdV+VdP=0

PdVdP=-V; β=-(1V)(dVdP)=(1P)

β×P=1

Hence graph between β and P will be a rectangular hyperbola.



Q 4 :

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. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be ________ (take the acceleration due to gravity = 10 ms-2 and the universal gas constant = 8.3 J mol-1K-1).             [2020]

[IMAGE 508]



(6)

[IMAGE 509]

Initially partition is held at a distance of 4m from the top.

Temperature, T=300 K

PV=PAh=nRT

P04A=0.1R×300     (For both)

Let piston shifts by x meter and final temperature is T, then

P2A(4-x)=0.1RT

  P2A=0.1RT4-x=RT10(4-x) and P1A=RT10(4+x)

Finally P2A-P1A=mg=83=RT10(14-x-14+x)

RT10(2x16-x2)=83

0.2Cv×300+mg·x=0.2CvT

32R×210×300+83x=210×32RT

(900R10+83x)=3(RT10)

(300R10+833x)=RT10(30R+83x3)(2x16-x2)=83

83(3+x3)(2x16-x2)=83(9+x)3(2x16-x2)=1

18x+2x2=48-3x25x2+18x-48=0

  x=1.782

Hence distance from top when reached equilibrium=4+2=6 m



Q 5 :

A cylindrical furnace has height (H) and diameter (D) both 1 m. It is maintained at temperature 360 K. The air gets heated inside the furnace at constant pressure Pa and its temperature becomes T = 360 K. The hot air with density ρ rises up a vertical chimney of diameter d = 0.1 m and height h = 9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density ρa=1.2 kg m-3, pressure Pa and temperature Ta=300 K enters the furnace. Assume air as an ideal gas, neglect the variations in ρ and T inside the chimney and the furnace. Also ignore the viscous effects.

[Given the acceleration due to gravity g=10 m s-2 and π=3.14]

[IMAGE 510]

Q.      Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is ________ gm s-1.  [2023]



(47.10)

Since, pressure P=constant ρaTa=ρT

1.2×300=ρ(360)

  ρ=1 kg/m3

Applying Bernoulli's theorem between upper and bottom point

Assuming velocity of hot air inside the furnace0

Pa+0+0=Pa-ρag(h)+ρg(h)+12ρV2

  V=2(ρa-ρ)g×9ρ=2(0.2)90=6

Therefore the steady mass flow rate of air existing the chimney

Q=ρπ(d24)V=1×3.14×(0.1)24×6

=0.0471 kg/s=47.10 gms-1



Q 6 :

A cylindrical furnace has height (H) and diameter (D) both 1 m. It is maintained at temperature 360 K. The air gets heated inside the furnace at constant pressure Pa and its temperature becomes T = 360 K. The hot air with density ρ rises up a vertical chimney of diameter d = 0.1 m and height h = 9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density ρa=1.2 kg m-3, pressure Pa and temperature Ta=300 K enters the furnace. Assume air as an ideal gas, neglect the variations in ρ and T inside the chimney and the furnace. Also ignore the viscous effects.

 [Given the acceleration due to gravity g=10 m s-2 and π=3.14]                

[IMAGE 511]

Q.   When the chimney is closed using a cap at the top, a pressure difference ΔP develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of ΔP is ________ Nm-2.  [2023]



(18)

[IMAGE 512]

Pressure Pa=Pin+ρg(h)

Pa=Pinside+ρg(9)

Pinside=Pa-ρg(9)

And, Poutside=Pa-ρag(9)

  ΔP=Pinside-Poutside

              =(ρa-ρ)g×9

=(1.2-1)×10×9=18 N/m2