At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

820 K

Explanation:

As per Boltzman equation, kinetic energy (KE) is in direct relation to the temperature, measured in absolute scale Kelvin.

KE α T.

Then, the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be 3 times such temperature.

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

Second, convert 0°C to kelvin:

T (K) = T (°C) + 273.15 K = 273.15 K

Then, 3 times gives you: 3 * 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.