Ideal Gas Law

An ideal gas is a theoretical gas composed of many randomly moving point particles that do not interact except when they collide elastically. The ideal gas law is the equation of state of an ideal gas. It relates the state variables of the gas: pressure \((P),\) volume \((V),\) and temperature \((T).\) Also included are the amount of the gas \((n)\) and the ideal gas constant \((R=8.314 \frac{\text{J}}{\text{K mol}}).\)

\[PV=nRT\]

To describe an ideal gas, a set of assumptions are made.

1) Gases consist of large numbers of tiny particles that are far apart relative to their size. This implies that the gas molecules have negligible volume compared to the volume of container in which they are placed.

2) Collisions between gas particles and between particles and container walls are elastic collisions (there is no net loss of total kinetic energy).

3) Gas particles are in continuous, rapid and random motion. They therefore possess kinetic energy, which is energy of motion.

4) There are no forces of interaction between gas particles. Thus they can move independent of each other. They only interact with each other through elastic collisions.

5) The average kinetic energy of a gas particle depends only on the temperature of the gas.

The last postulate of the kinetic molecular theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas. Thus, the average kinetic energy of the gas particles increases as the gas becomes warmer. Because the mass of these particles is constant, their kinetic energy can only increase if the average velocity of the particles increases. The faster these particles are moving when they hit the wall, the greater the force they exert on the wall. Since the force per collision becomes larger as the temperature increases, the pressure of the gas must increase as well. \[P \propto T\]
Here P is pressure and T is temperature in kelvin. In this, volume and number of moles of gas is taken constant. If temperature is represented in kelvin then the graph between pressure and temperature will be a straight line passing through origin.

If temperature is represented in celsius then the graph between pressure and temperature will be a straight line but will not pass through origin. On extrapolating, the graph will hit -273.15 degrees.

Gases can be compressed because most of the volume of a gas is empty space. If we compress a gas without changing its temperature, the average kinetic energy of the gas particles stays the same. There is no change in the speed with which the particles move, but the container is smaller. Thus, the particles travel from one end of the container to the other in a shorter period of time. This means that they hit the walls more often. Any increase in the frequency of collisions with the walls must lead to an increase in the pressure of the gas. Thus, the pressure of a gas becomes larger as the volume of the gas becomes smaller.

If temperature and amount of gas is fixed then pressure is inversely proportional to volume occupied by the gas.
\[P \propto 1/V\]
Here P is pressure and V is volume.

If temperature and number of moles of gas are fixed then the graph between pressure P and volume V will be a rectangular hyperbola. On increasing volume of gas pressure decrease and vice-versa. Such a process is also called as Isothermal Process

The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer. If they move faster, the particles will exert a greater force on the container each time they hit the walls, which leads to an increase in the pressure of the gas. If the walls of the container are flexible, it will expand until the pressure of the gas once more balances the pressure of the atmosphere. The volume of the gas therefore becomes larger as the temperature of the gas increases.

If pressure of an ideal gas is kept constant then volume of container is directly proportional to temperature (in kelvin) of the gas.
\[V \propto T\]
Here, V is volume of container and T is temperature of gas in Kelvin.

If the temperature is taken in degrees then the graph will be instead of passing through origin will hit temperature \(-273^\circ \)C

As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside. Thus, the volume of the gas is proportional to the number of gas particles.

If pressure and temperature of an ideal gas is kept constant then volume of container is directly proportional to the amount of gas (number of moles of gas) in the container.
\[V \propto N\]

Imagine what would happen, gases at different pressure but same temperature are added to a container. The total pressure would increase because there would be more collisions with the walls of the container. There is so much empty space in the container that each type of gas molecules hits the walls of the container as often in the mixture as it did when there was only one kind of gas. The total pressure will increase as more number of gas molecules hits the container walls but the pressure due to individual gas molecules remains same. The total number of collisions with the wall in this mixture is therefore equal to the sum of the collisions that would occur when each gas is present by itself. In other words,

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.
\[{P_t} = {P_1} + {P_2} + {P_3} + ...\]

It was first stated by Émile Clapeyron in 1834 as a combination of Boyle's law, Charles's law and Avogadro's Law. The ideal gas law is often written as:

\(PV=nRT \)

Where, \(R\) is the Gas Constant. Some values of \(R\) are given below.

\(R=8.314\) \(\text{J.mol}^{-1}\text K^{-1}\)

\(R=0.082\) \(\text{litre.atm.mol}^{-1}\text K^{-1}\)

\(R=8.2057\) \(\text m^{3}\text{.atm.mol}^{-1}\text K^{-1}\)

A gas at 27°C has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. Find the resulting temperature of the gas.

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From ideal gas equation, \[PV = nRT \] For a closed container, number of moles remains constant, therefore
\[\frac{{{T_2}}}{{{T_1}}} = \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\,\left( {\frac{{{V_2}}}{{{V_1}}}} \right) = \left( {\frac{{2{P_1}}}{{{P_1}}}} \right)\,\left( {\frac{{3{V_1}}}{{{V_1}}}} \right) = 6\]
\[{T_2} = 6{T_1} = 6 \times 300\]
\[{T_2} = 1800\,K = 1527^\circ C.\]

A balloon contains \(500\,{m^3}\) of helium at 27°C and 1 atmosphere pressure. The volume of the helium at – 3°C temperature and 0.5 atmosphere pressure will be

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From Ideal gas equation, \[PV = nRT \]

Since gas is trapped inside balloon therefore number of moles of gas remains unchanged. Let \({P_1}\), \({V_1}\) and \({T_1}\) are initial pressure, volume and temperature, and \({P_2}\), \({V_2}\) and \({T_2}\) are final pressure, volume and temperature
then,

\[\frac{{{V_2}}}{{{V_1}}} = \left( {\frac{{{T_2}}}{{{T_1}}}} \right)\,\left( {\frac{{{P_1}}}{{{P_2}}}} \right) = \left( {\frac{{270}}{{300}}} \right)\,\left( {\frac{1}{{0.5}}} \right) = \frac{9}{5}\]

\[{V_2} = 500 \times \frac{9}{5} = 900\,{m^3}\]