States of Matter: Gateway to JEE (Main), NEET, ICAR, OUAT and All Similar Entrance Examinations

States of Matter: Gateway to JEE (Main), NEET, ICAR, OUAT and All Similar Entrance Examinations

    You need to be friendly with concepts and problems, before appearing any entrance examinations. A systematic flow of concepts and corresponding problems will build your confidence to crack entrance examinations. This gateway will help you rise step wise to reach and crack the entrance examinations.

Pratap Sir's Core Concept of States of matter (PSCC: 1): Ideal Gas Equation

This equation though applicable to hypothetical ideal gasses, it can be applied to real atmospheric gasses like oxygen, nitrogen, methane etc with little to no error at particular conditions like high temperature and low pressure.

    This equation contains all gaseous variables like pressure, temperature, volume and number of moles, given by the formula: 

PV = nRT = (w/m) RT ........ eq. 1

where n =  number of moles = given mass (w) / molecular mass (m) . 

R = Universal gas constant, the value of which depends on the units of all other parameters in the formula. For example when the pressure is taken in atmosphere and volume in litre, the value of R is 0.082 litre atm per kelvin per mole.

R= 0.0821 litre atm per kelvin per mole 

= 0.083 litre bar per kelvin per mole

= 82.1 ml atm per kelvin per mole 

= 8.314 joule per kelvin per mole  or N m per kelvin per mole

= 5.189 X 10^19  eV per kelvin per mole

= 1.987 or 2.0 cal per kelvin per mole

= 62.3 mm per kelvin per mole

    From the above relationship, problems may be asked to calculate the pressure, temperature, volume and number of moles while all other parameters except one are given.

Problem - 1: Calculate the temperature at which 1 mole of nitrogen (N2 gas) will ocuupy a volume of 10 litres at 2.46 atm.

Solution: PV = nRT Thus T = PV / nR = 2.46 X 10 / 1 X 0.082 = 300 K

Problem - 2: 9.4 gram of an oxide of nitrogen applies a pressure of 5 atm inside an one litre vessel at 25 degree centigrade. Predict the molecular formula of the oxide of nitrogen.

Solution: PV = nRT = (w/m) RT 

Thus m = wrt / pv = (9.4 X 0.0821 X 298) / (5 X 1) = 45.996 which approaches 46.

Thus formula of the oxide should be NO2 since its molecular mass is 46.

Problem - 3: What mass of ammonia will exert double the pressure as aplied by 18 g of H2S in the same container under similar temperature?

Solution: Under similar temperature and volume, pressure deppends on given mass and molecular mass of a gas. i.e., p ∝ w/m. 

Thus p1 = w1 /m1 and p2 = w2 / m2   (Here we may refer 1 to NH3 and 2 to H2S)

p1 /p2 = w1 . m2 / m1 . w2

Ammonia exerts double the pressure of H2S, hence p1 = 2p2

2p2 / p2 = w1 . 34 / 17 . 18

w1 = mass of ammonia = 2 X 17 X 18 / 34 = 18 g

Problem - 4: 10 gram of a gaseous mixture of Helium and Argon contained in a 10 litre vessel at 27 degree centigrade exert a pressure of 2 atm. Determine masss percentage of helium in the mixture?

Solution: Let the mass of He in the mixture is w g and that of Ar is (10 - w) g 

PV = nRT = (number of moles of He + number of moles of Ar) R T

(number of moles of He + number of moles of Ar) = PV / RT = 0.813

(w / 4) + {(10 - w) / 40} = 0.813

Solving the equation we get, w = 2.5 g

Mass percent of helium = (2.5 / 10) X 100 = 25 %

Problem - 5:  12 gram of a gaseous mixture of methane and helium was taken in a vessel to which 8 gram of oxygen was added at the same temperaure. The pressure increased by a factor 7/6. what was the weight percentage of methane in the 12 gram of original mixture?

Solution: Under similar temperature and volume, pressure depends on the number of moles.

Consider that 12 gram of original mixture contains w gram of methane and in total 'n' moles of methane and helium. 

Thus n ∝ p  ....... eq. 1              (where p is the pressure of original mixture)

When we add 8 g (or 8 / 32 moles ) of O2 to the original mixture, the total moles of the resulting mixture becomes n + (8/32) for which the pressure increases by factor 7/6 w.r.t. initial pressure.

Thus { n + (8/32) } ∝ 7/6 . p

Or (n + 0.25) ∝ 7/6 . p   ......... eq. 2

Taking ratio of eq. 1 and eq. 2 we get,

n / (n +  0.25) = 6/7 solving which we get, n = 1.5

Thus [w/16 + {(12 - w)/4}] = 1.5 solving which we get, w = 8

Mass percent of methane = (8/12) X 100 = 66.67 %

Do It Yourself (DIY) - 1: Find the total pressure in a mixture of 4g of O2 and 2g H2 confined in a vessel of 1L at 0 degree centigrade.

Note: In case composition of the gasses of the atmosphere is given in percent by weight or percent by volume, average molar mass of the air can be determined using the formula, M (air) = {m1.x1 + m2.x2 + m3.x3 + ... } / 100. For Example if air is 79% N2 and 21% O2 by volume, then M (air) = {28 x 79 + 32 x 21} / 100 = 28.84.

We have from eq. 1, PV = nRT = (w/m) RT, 

This equation can further be developed as follows:

Pm = (w/v) RT = dRT (Since density = given mass / volume = w/m)

The density (d) in this case is called absolute density whose unit is g/ltr and depends on the pressure and temperature. 

Note: The density of a gas at STP = molar mass / 22.4. 

For example, d (O2) at STP = 32 / 22.4 = 1.43 gram per litre

Problem - 6: Find the density of CO2 at 100 degree centigrgade and 800 mm pressure.

Solution: We know, Pm = dRT , temperature must be converted into kelvin and the pressure must be converted into atm to solve the numerical smoothly.

Thus d = Pm / RT =   {(800 / 760) . 44} / 0.0821 . 373 = 1.5124 g per litre

 

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