According to Boyle's law, for a given mass of a gas at constant temperature

According to Charles' law,

Combining (i) and (ii)

or

or

Now, if V_{1} is the volume of a gas at temperature T_{1} and pressure P_{1}_{.} V_{2} is the volume of same amount of gas at temperature T_{2} and pressure P_{2}, then

The above relationship is very useful for converting the volume of a gas from one set of conditions to another.

The numerical value of the constant of proportionality (K) depends upon the quantity of gas. The volume of a gas is directly proportional to the number of moles of gas at constant temperature and pressure (Avogadro's law). This means that 'K' is directly proportional to the number of moles, 'n', i.e.,K n

or K = nR,where 'R' is the universal gas constant of proportionality. The value of 'R' is same for all gases.

However, the numerical value of 'R' varies with the units in which pressure and volume are expressed. Therefore the Ideal gas Equation is derived as:##### For one mole (n = 1),

PV = RT (Ideal Gas Equation)

The Ideal Gas Equation is also known as the equation of state for gases as it expresses the quantitative relation ship between the four variables that describe the state of the gas. The word 'ideal' is used here because in reality no gas obeys the above condition and the gases, which deviate from ideality are called as real gases.*Universal gas constant (R)*

From the above ideal gas equation:

The universal gas constant is a measure of energy change (work done) per mole of the gas for one degree change in its temperature.

## Numerical Value of R

The magnitude and unit of 'R' depends upon the units in which pressure, volume and temperature are expressed.

### Conditions and problems

### I)When pressure is expressed in atmosphere and volume in litres

Under standard condition of temperature and pressure i.e., when

P = 1 atmT = 273.15 K

V = 22.414 L mol^{-}

^{1}

This gives,

### II

) When pressure is expressed in atmosphere, and volume in mL

Here,

P = 1 atmT = 273.15 K

V = 22414 mL mol^{-}

^{1}

This gives,

### III) R in energy units

When,

If 'R' is to be expressed in CGS units, the unit of pressure should be dyne cm^{-}^{2} and volume in cm^{3}mol^{-}^{1}. Then,

Normal pressure, P = (1 x 76 x 13.6 x 981) dyne cm^{-}^{2}

= 22414 cm^{3}mol^{-}^{1}

##### = 8.314 x 10

^{7}erg deg

^{-}

^{1}mol

^{-}

^{1}

Since, 4.182 x 10^{7}erg = 1calorie

### IV) In SI units

In the SI system of units, under NTP conditions,

P = 101325 N m^{-}

^{2}

T = 273.15 K

V = 22.414x10^{-}

^{3}m

^{3}

Therefore,

= 8.314 J mol^{-}^{1} K^{-}^{1}

### The value of Gas constant R in different units

### Problems

5. Calculate the number of moles of hydrogen (H_{2}) present in a 500 cm^{3} sample of hydrogen gas at a pressure of 760 mm of Hg and 27°C.

### Solution

According to ideal gas equation, PV = nRT

T = 27 + 273 = 300 K, R=82.1 cm^{3} atm K^{-}^{1} mol^{-}^{1}

n = 0.0203 mol = 2.03 x 10

^{-}

^{2}mol.

6. About 200 cm^{3}of a gas is confined in a vessel at 20°C and 740 mm Hg pressure. How much volume will it occupy at S.T.P.?

### Solution

We are given

P_{1}= 740 mm Hg P

_{2}= 760 mm Hg

T_{1} = 20 + 273 = 293K T_{2} = 273 K

_{1}= 200 cm

^{3}V

_{2}= ?

According to gas equation,

Substituting the values, we get

= 181.4 cm^{3}.

2.5 x 10^{5} Nm^{-}^{2} pressure and 300 K temperature.

### Solution

According to ideal gas equation,

n = 2 mol, T = 300 K, P = 2.50 x 10^{5} Nm^{-}^{2}

^{-}

^{1}mol

^{-}

^{1}

= 19.95 x 10

^{-}

^{3}m

^{3}=19.95dm

^{3}.

## 1 comment:

nice and helpful. keep helping

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