The force a gas exerts on the walls of its container arises from collisions between the particles in — Physical Chemistry — Kinetics Chemistry Question
Ideal and not-so-ideal gases
The force a gas exerts on the walls of its container arises from collisions between the particles in the gas and the surface. In a single collision, the magnitude of the impulsive of the force exerted on the surface is equal to the change in the momentum normal to the surface, m∆v. The force on the surface is then the impulse, multiplied by the rate at which the particles collide with the surface. Since the motion of particles within a gas is random, the number of collisions occurring per unit time is a constant for a gas at constant temperature. The temperature of a gas reflects the distribution of particle velocities within the gas. For a given gas, the particle speeds will be higher, on average, at higher temperatures.
Given the above information, and assuming the gas is initially at room temperature and atmospheric pressure, consider how carrying out the following actions would be likely to affect the pressure. Would the pressure double, halve, increase slightly, decrease slightly, or remain unchanged?
i) Doubling the number of particles in the gas.
ii) Doubling the volume of the container in which the gas is confined.
iii) Doubling the mass of the particles in the gas (assume that the particle velocities remain constant).
iv) Increasing the temperature by 10°C.
Model Answer
i) pressure would double
ii) pressure would halve
iii) pressure would double
iv) pressure would increase slightly
The ideal gas model assumes that there are no interactions between gas particles. Particles in a real gas do interact through a range of forces such as dipole–dipole forces, dipole–induced–dipole forces, and van der Waals interactions (induced–dipole–induced–dipole forces). A typical curve showing the potential energy of interaction between two particles is shown right:
The force between two particles in a gas at a given separation r may be calculated from the gradient of the potential energy curve i.e. F = –dV / dr.
What is the force at the four points marked A, B, C and D on the figure?
(attractive / repulsive / approximately zero)
Model Answer
A – approximately zero
B – attractive
C – approximately zero
D – repulsive
The deviation from non-ideality in a gas is often quantified in terms of the compression ratio, Z.
Z = mV / m0V
where mV is the molar volume of the (real) gas, and m0V is the molar volume of an ideal gas under the same conditions of temperature, pressure etc.
Match the following values of Z with the dominant type of interaction in the gas.
[ Z = 1 ] [ Z < 1 ] [Z > 1 ]
Attractive forces dominate
Repulsive forces dominate
No intermolecular forces, ideal gas behaviour
Model Answer
Z = 1 - no intermolecular forces, ideal gas behaviour
Z < 1 - attractive forces dominate
Z > 1 - repulsive forces dominate
The compression ratio is pressure dependent. Consider the average separation between particles in a gas at different pressures (ranging from extremely low pressure to extremely high pressure), and the regions of the intermolecular potential that these separations correspond to. Sketch the way in which you think the compression ratio will vary with pressure on the set of axes below.
[Note: do not worry about the actual numerical values of Z; the general shape of the pressure dependence curve is all that is required.]
Model Answer
The solution graph shows the compression ratio Z on the y-axis and pressure p on the x-axis. The curve starts at Z=1 for p=0, initially dips downwards (indicating Z < 1), reaches a minimum, and then rises steadily and linearly to values greater than 1 as pressure continues to increase.