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