Final answer:
Doubling the average molecular speed of an ideal gas at constant volume would quadruple the kinetic energy, which implies a fourfold increase in temperature, subsequently increasing the pressure due to more forceful molecular collisions with the container.
Step-by-step explanation:
If the temperature of an ideal gas is increased such that the average molecular speed is doubled, several changes occur at the microscopic level involving the kinetic energy and pressure of the gas. According to the kinetic theory of gases, the average kinetic energy (KEavg) of a gas molecule is directly proportional to the temperature of the gas. If the temperature doubles, and the volume is kept constant, the kinetic energy of the gas molecules also doubles since KEavg = (3/2)kT, where k is the Boltzmann constant and T is the temperature. Doubling the average speed (v0 to 2v0) also means an increase to four times the initial kinetic energy (since KE = 1/2 mv2, KE after increase will be 1/2 m(2v0)2 = 1/2 m4v02 = 4 x (1/2)mv02), suggesting that the temperature needed to double the average speed of gas molecules must be increased by a factor of four (since the relationship between kinetic energy and temperature is linear). Therefore, the gas's pressure would increase due to more forceful collisions with the container walls.