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Ideal Gas Equation and Gas Constant

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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

Changing temperature, pr5essure and volume of a gasw.
Changing temperature, pressure and volume of a gas.

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 V2 . 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 -273oCto 0 K. at absolute zero temperature all particles remain at rest.



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.

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

The temperature, at which volume or pressure of any gas reduces to 0, is referred to as absolute zero temperature.

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