# Gas Laws - 6

__Question__: & 0.0566 L mol. Calculate a, b & R

Ans: Given T_{C} = 374 + 273 = 647 K, P_{C} = 218 atm

V_{C} = 0.0566 L mol^{-1}

= 0.0189 L mol^{-1}

a = 3P_{C}V_{C}^{2} = 3 x 218 x (0.0566)^{2} = 2.095 L^{2} atm mol^{-2}

R = = 0.05086 L atm K^{-1}mol^{-1}

__Medium Type__:

Q1. A large flask filled with stop cock is evacuated & weight its mass is found to be 134.567 g. It is then filled to pressure 735 mm at 31^{0}C with a gas of unknown molecular mass & then reweighed; its mass is 137.456 g. Flask is then filled with H_{2}O & weighed again, its mass is now 1067.9 gm. Calculate molar mass of gas if gas is ideal.

Ans: Mass of meter filling flask = (1067.9 - 134.567)g

= 933.333 g

Volume of flask = Volume of meter = 933.333 ml

__Dumb Quetion__: Why volume of meter is 933.333 ml ?

Ans: Because density of meter is 1 g/cm^{3}

P = 735 mm, T = 273 + 31 = 304 K, V = 933.33

By PV = nRT

n = 0.036 mol

Mass of 0.036 mol of gas = (137.456 - 134.567) g

= 2.889 g

Mass of 1 mol of gas = = 80.25 g

Q2. Avg. speed at T_{1}K & Most probable speed at T_{2}K of CO_{2} gas is 9 x 10_{4} cm/s. Calculate value of T_{1} & T_{2}.

Ans: Avg. speed = Most probable speed =

Avg. speed at T_{1}K = MP speed at T_{2}K for CO_{2}

……………………………….. (i)

For CO_{2} U_{mP} = = 9 x 10^{4}

= 9 x 10^{4}

T_{2} = 2143.37 K

By eq. (i) T_{1} = 1684 K

Q3. Mass of molecule A is twice mass of molecule B. RMS speed of A is twice rms speed of B. If two samples of A & B contains same no. molecules. What will be ratio of P of two samples in separate containes of equal volume.

Ans: Given, m_{A} = 2m_{B}

Mol. wt. of A = 2 x Mol. wt. of B

Given U_{rms} of A = 2 U_{rms} of B

For gas A P_{A}V_{A} = M_{A} u^{2}_{rms A}

Fr gas B P_{B}V_{B} = M_{B} u^{2}_{rms B}

Given V_{A} = V_{B}

= 2 x (2)^{2} = 8

P_{A} = 8 P_{B} |

Q4. One way of writing eq. of state of writing eq. of state of real gases is PV = RT[1 + B/V + ……….] where B is constant. Derive expression for B in terms of Vander Waal’s constant a & b ?

Ans: According to Vander Waal’s Equation

[V - b] = RT

P =

Multiply by V

PV =

PV = RT

PV = RT

PV = RT

On comparing

Q5. A glass capillary tube scaled at both ends is 100 cm long. It lies horizontally with middle 10 cm containing Hg. Two ends are equal in length contain air at 27

^{0}C & pressure > 6 cm od Hg. Tube is kept in horizontal position, the one end is at 0

^{0}C & other end is at 127

^{0}C. Calculate length of air column & pressure which is at 0

^{0}C.

Ans:

2 L + 10 = 100

=> L = 45 cm

Let initial pressure be P

_{o}atm on each side. When one end is cooled & other heated, expansion of gas takes place at hoter end till pressure of two sides becomes same.

i.e. P

_{1}= P

_{2}= P

For end A. …………………….. (i)

a area of x - section of tube.

For end B, …………………………… (ii)

Equating. P

_{1}= P

_{2}

x = 8.49 cm

P

_{1}= = 82.25 cm of Hg

__Hard Question__:

Q. An LPG cylinder weighs 14.8 Kg when empty, when full, it weighs 29 Kg & shows pressure of 2.5 atm. During use at 27

^{0}C weight of full cylinder reduced to 23.2 Kg. Fid out volume of gas in m

^{3}used up at normal condition & final pressure inside cylinde. Assume LPG to be n-butane.

Ans: Wt: of butane in cylinder = 29 - 14.8 = 14.2 Kg = 14.2 x 10

^{3}

P = 2.5 atm, T = 300 K, Mol. wt. of butane = 58

PV = RT

2.5 x V = x 0.0821 x 300

V = 2.4 x 10

^{3}L = 2.4 m

^{3}

Now, wt. of as left after use = 23.2 - 14.8 = 8.4 x 1

^{3}g

Volume remains constant

PV = RT

P x 2.4 x 10

^{3}= x 0.0821 x 300

wt. of gas given out = 29 - 23.2 = 5.8 Kg

= 5.8 x 10

^{3}g

Volume of gas given out

1 x V = x 0.0821 x 300

V = 2.4 x 10

^{3}L = 2.4 m

^{3}