If the temperature of a gas is increased from 20 to 40 degrees celcius, by what factor does the speed of of the molecules increase?

average KE= constant*abaolute temp

average KE= (273+40)/(273+20)

but KE is proportional to v^2

so velocity has increased by sqrt above.

To calculate the factor by which the speed of gas molecules increases when the temperature is increased, we can use the root mean square (RMS) speed formula. According to the kinetic theory of gases, the RMS speed of gas molecules is directly proportional to the square root of the absolute temperature.

The formula for RMS speed is:
v = √(3kT / m)

Where:
v = RMS speed
k = Boltzmann constant (1.380649 x 10^-23 J/K)
T = Absolute temperature in Kelvin
m = Mass of gas molecules

In this case, we are given the change in temperature from 20 to 40 degrees Celsius. To convert this to Kelvin, we add 273.15:

T1 = 20 + 273.15 = 293.15 K
T2 = 40 + 273.15 = 313.15 K

Now we can calculate the ratio of the RMS speeds:

v2 / v1 = √(3kT2 / m) / √(3kT1 / m)
v2 / v1 = √((3kT2) / m) * √(m / (3kT1))
v2 / v1 = √(T2 / T1)

Plugging in the values:
v2 / v1 = √(313.15 / 293.15)
v2 / v1 ≈ 1.067

Therefore, the factor by which the speed of the gas molecules increases when the temperature is increased from 20 to 40 degrees Celsius is approximately 1.067.

To determine how the speed of gas molecules changes when the temperature increases, we can use the relationship between temperature and the average kinetic energy of gas molecules.

The average kinetic energy of a gas molecule is directly proportional to its temperature. According to the Kinetic Theory of Gases, the average kinetic energy of a gas molecule is given by the equation:

KE = (3/2) * k * T

where KE is the average kinetic energy, k is the Boltzmann constant (1.38 × 10^-23 J/K), and T is the temperature in Kelvin.

To convert the given temperatures from Celsius to Kelvin, we can use the formula:

T(K) = T(°C) + 273.15

For the initial temperature of 20 degrees Celsius:

T1 = 20 + 273.15 = 293.15 K

And for the final temperature of 40 degrees Celsius:

T2 = 40 + 273.15 = 313.15 K

Now, let's calculate the ratio of the average kinetic energies at the two temperatures:

KE2 / KE1 = [(3/2) * k * T2] / [(3/2) * k * T1]
= T2 / T1
= 313.15 K / 293.15 K
≈ 1.068

So, the ratio of the average kinetic energies, or equivalently the factor by which the speed of gas molecules increases, is approximately 1.068.

Therefore, the speed of the gas molecules increases by a factor of approximately 1.068 when the temperature increases from 20 to 40 degrees Celsius.