# 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 P_{C} |

__Critical T__: It is limiting value of temperature required for liquefication such that

_{C} T T_{C} |

__Critical V__: It is volume occupied by real gas at T

_{C}_{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 H

_{2}O

= 10.3 m of H

_{2}O

__Dumb Question__: Why 76 cm of Hg = 76 x 13.6 cm of H

_{2}O ?

Ans: Pressure = gh

height of Hg = 76 cm & density = 13.6 g/cm

^{3}

Height of H

_{2}O = ? & density = 1 g/cm

^{3}

76 x 13.6 x

h = 76 x 13.6 cm of H

_{2}O

Pressure at 100 m depth = (100 + 10.3) m

10.3 x v = 110.3 x v

_{2}=> v

_{2}= 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 15

^{0}C & 750 mm pressure occupies 1 L. Calculate aq. tension (v - pressure) of water at 15

^{0}C ?

Ans: Initial (at STP) Final

v

_{1}= 919 ml v

_{2}= 1000 ml

P

_{1}= 760 mm P

_{2}= ? (dry state)

T

_{1}= 273 K T

_{2}= 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 H

_{2}at 750 mm & O

_{2}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 H

_{2}, P

_{1}V

_{1}= P

_{2}V

_{2}

i.e. 750 x 1.5 = P

_{2}x 3.5

For O

_{2}, 100 x 2 = Po

_{2}x 3.5 => Po

_{2}= 57.14 mm

P

_{mix}= Po

_{2}+ = 57.14 + 321.43

= 378.57 mm

__Question__: Calculate diameter of O

_{2}molecules.

Ans: b = 4v => v = b/4 = 7.95 x 10

^{-3}L mol

^{-1}

Volume occupied by 1 O

_{2}molecule =

ITr

^{3}= 1.32 x 10

^{-23}

on solving r = 2.932 A

^{0}

Q.5. 5g C

_{2}H

_{6}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 TK

^{0}

PV = nRT i.e. PV = RT

10 x 1 = x 0.082 x T (Mol. mass of C

_{2}H

_{6}= 30)

T = 730.81 K

Q.6. O

_{2}is present in 1 L flask at a pressure of 7.6 x 10

^{-10}mm of Hg. Calculate no. of O

_{2}molecules at 0

^{0}C.

Ans: PV = nRT

x 1 = n x 0.0821 x 273

n = 4.46 x 10

_{-14}mole of O

_{2}

No. of molecules of O

_{2}= 4.46 x 10

^{-14}x 6.023 x 10

^{23}

= 2.68 x 10

^{10}

Ans: P = atm, T = 273 + 100 = 373 K

Let density be d for CO

_{2}

For CO

_{2}PV = RT

PM = RT => PM = d RT

= 1.5124 g/L

Q.8. Pure O

_{2}diffuses through aperture in 224 seconds, whereas mix. of O

_{2}& another gas containing 80% O

_{2}diffuses from same in 234 sec. What is mol. wt. of gas ?

Ans: For gaseous mix. 80% O

_{2}(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 10

^{4}cm/s