Ideal Gas Equation and Gas Constant

The Equation of State or Ideal Gas Equation(different figure numbering from book)

Let us consider a mole of gas contains in a cylinder fitted with a frictionless piston. Suppose the gas changes from initial state P₁, V₁, T₁ to the final state P₂, V₂, T₂ where P, V and T denote pressure, volume and temperature of a gas. To derive an ideal gas equation. Let us assume the change takes place in two steps:

From Boyls law ,we have .

$$p_1V_1=p_2V$$

$$or,\;V=\frac{p_1V_1}{P_2}\dots \dots(i)$$

Now the pressure p₂is kept constant and the temperature is increased up to the desired value T₂ so that volume also increased up to V_{2 .} From Charlie's law ,we have

$$\frac{V}{T_1}=\frac{V_2}{T_2}$$

$$or\;,V= \frac{V_2\;T_1}{T_2} \dots \dots(ii)$$

EquatingEqs.(i) and (ii),we get

$$\frac {p_1V_1}{p_2}=\frac{V_2T_1}{T_2}$$

$$or\;,\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$$

in general

$$\therefore \frac {pV}{T} = constant $$

This constant is same for all gasses for 1 mole of gas. This constant is called universal gas constant or molar constant. It is donated by ‘R’.

For 1 mole gas

$$\therefore \frac {pV}{T} = R$$

$$ PV = RT $$

The equation is an ideal gas equation for 1-mole gas.

For ‘n’ mole of gas, the ideal gas equation is

$$PV = nRT\dots(i)$$

If ‘m’ is total mass and ‘M’ is molar mass of the gas then no of moles

$$N=\frac{m}{M}$$

Now equation (i) becomes

$$PV = RT$$

$$\text{or,} PV = \frac{m}{M}RT $$

$$\text{or,}PV = mrT\dots(ii) $$

Where r = is an ordinary gas constant

For unit mass

$$PV = rT$$

Gas Constant

The gas constant is not only of simple constant, but it is the amount of work done per mole per unit Kelvin. For n moles of a gas.

$$PV = nRT$$

\(R = \frac{pV}{nT} = \frac{nm^{-2}m^2}{\text{No. of moles} \times K}\) = Jmole^-1 K^-1

Hence, unit of R is Jmole^-1 K^-1.

Numerical value of R

At standard temperature and pressure (s.t.p). pressure of gas, p = 760 mm Hg = 760 × 10^-3 m Hg.

Density of mercury at s.t.p is 13600 kg/m³. So,

$$ P = \rho gh = 13600×760×10^{-2}×9.8 N/m{-2}= 1.013×10^5 N/m^2$$

Standard temperature, T = 273 K

One mole gas at s.t.p occupies a volume of 22.4 liter or 22.4×10^-3m³. So

\(R = \frac{1.013×10^5 \times 22.4 \times10^{-3} }{273}= 8.32\) Jmole^-1 K^-1

In CGS-system,

$$1 cal = 4.2 J$$

$$ R =\frac{8.3}{4.2} cal mole^{-1}C^{-1} = 1.98 cal mole^{-1} C^{-1} $$

Gas Constant for Unit Mass

The gas constant for unit mass, r = where M is a molar mass. The equation of state in terms of r is

$$PV = mrT $$

The value of r depends upon the molecular mass of the gas whose value is different for different gases.

For hydrogen gas,

$$ r = \frac {R}{M} = r= \frac {8.31}{2 \times 10^{-3}} = 4155 Jkg^{-1}K^{-1}$$

Absolute Zero

Let be the volume of the gas at 0and be its volume at constant pressure. Then

$$V_\theta = V_o(1 + \gamma_p \times \theta)$$

$$V_\theta = V_o(1 + \frac { \theta}{273}) [\therefore \gamma_p = \frac{\theta}{273}]$$

$$ At \theta =-273^oC, V_\theta = 0 $$

Also, if be pressure of the gas at 0and be pressure at constant volume, then

$$p_\theta = p_o(1 + \gamma_v \times \theta)$$

$$p_\theta =p_o(1 + \frac { \theta}{273}) [\therefore \gamma_v = \frac{\theta}{273}]$$

$$ At \theta =-273^oC, p_\theta = 0 $$

Therefore, the temperature, at which volume or pressure of any gas reduces to 0, is referred to as absolute zero temperature. Its value is equal to -273^oCto 0 K. at absolute zero temperature all particles remain at rest.