Gas Laws - 5

Equating (i) to zero

T =
T =
As v
    T T_boyle’s



Critical parameter pertaining to its liquefication.
(1) Critical pressure (2) Critical Temperature (3) Critical Volume.

Critical P_C: It is limiting value of pressure required for liquefication. If gas has pressure (P) less than P_C it cannot be liquefied.
For liquefication

P PC

Critical T_C: It is limiting value of temperature required for liquefication such that

T TC

Critical V_C: It is volume occupied by real gas at T_C & P_C.
There are calculated using Sudrew's critical

(v - b) = RT      (n = 1)


On we get.

Putting V_C = 3b in equation (ii)

Putting value of T_C & P_C in (i)


Easy Type:

Q.1. A balton balton filled with ideal gas is at sca surface & the taken depth of 100 m. What will its volume in terms of original volume ?
Ans: Pressure at surface = 76 cm of Hg = 76 x 13.6 cm of H₂O
                                  = 10.3 m of H₂O

Dumb Question: Why 76 cm of Hg = 76 x 13.6 cm of H₂O ?
Ans: Pressure = gh
height of Hg = 76 cm & density = 13.6 g/cm³
Height of H₂O = ? & density = 1 g/cm³
   76 x 13.6 x g = h x 1 x g
   h = 76 x 13.6 cm of H₂O
Pressure at 100 m depth = (100 + 10.3) m
   10.3 x v = 110.3 x v₂   => v₂ = 0.093 v = 9.3 % of v

Q.2. A given mass of gas occupies 919 ml in dry state at STP. Same mass when collected once water at 15⁰C & 750 mm pressure occupies 1 L. Calculate aq. tension (v - pressure) of water at 15⁰C ?
Ans: Initial (at STP)            Final
       v₁ = 919 ml                v₂ = 1000 ml
       P₁ = 760 mm              P₂ = ? (dry state)
       T₁ = 273 K                 T₂ = 273 + 15 = 288 K

By

V.P. of water = Pressure of moist gas - P(dry gas)
                  = 750 - 736.7 = 13.3 mm


Q.3. Two vessels of capacitis 1.5 L & 2 L containing H₂ at 750 mm & O₂ at 100 mm respectively are connected to each other through a valve. What will be final pressure of gaseous mixture ?
Ans: Final volume = 1.5 + 2 = 3.5 L
For partial pressure of H₂, P₁V₁ = P₂V₂
i.e. 750 x 1.5 = P₂ x 3.5


For O₂,   100 x 2 = Po₂ x 3.5   => Po₂ = 57.14 mm
P_mix = Po₂ + = 57.14 + 321.43
                           = 378.57 mm

Question: Calculate diameter of O₂ molecules.
Ans: b = 4v    => v = b/4 = 7.95 x 10^-3 L mol^-1
Volume occupied by 1 O₂ molecule =
ITr³ = 1.32 x 10^-23
on solving r = 2.932 A⁰

Q.5. 5g C₂H₆ are in a bulb of 1 L capacity. Bulb is burst if pressure exceeds 10 atm. At what temperature will be pressure of gas reach bursting value ?
Ans: Let bulb burst at TK⁰
PV = nRT   i.e.   PV = RT
10 x 1 = x 0.082 x T     (Mol. mass of C₂H₆ = 30)
T = 730.81 K

Q.6. O₂ is present in 1 L flask at a pressure of 7.6 x 10^-10 mm of Hg. Calculate no. of O₂ molecules at 0⁰C.
Ans: PV = nRT
     x 1 = n x 0.0821 x 273
    n = 4.46 x 10_-14 mole of O₂
No. of molecules of O₂ = 4.46 x 10^-14 x 6.023 x 10²³
                                   = 2.68 x 10¹⁰

Ans: P = atm, T = 273 + 100 = 373 K
Let density be d for CO₂
For CO₂ PV = RT
PM = RT    =>  PM = d RT
= 1.5124 g/L

Q.8. Pure O₂ diffuses through aperture in 224 seconds, whereas mix. of O₂ & another gas containing 80% O₂ diffuses from same in 234 sec. What is mol. wt. of gas ?
Ans: For gaseous mix. 80% O₂ (M_m) = ................ (i)

M_m = 34.92
Putting M_m 34.92
We get, Mol wt. of gas m = 46.6

Q.9. Under 3 atm, 12.5 L of certain gas meighs 15g. Calculate avg. speed of gaseous molecules.
Ans: For gas PV = RT
3 x 12.5 = 15/M x 0.0821 x T => T/M = 30.45
u_av = = 8.028 x 10⁴ cm/s