What is the average kinetic energy per particle in 1.00 mole of an ideal gas at a pressure of 1.50 atmospheres and a volume of 2.00 liters?

To solve this problem, we need to use the formula for the average kinetic energy of a gas, which is given by:

KE_avg = 3/2 kT

where k is the Boltzmann constant (1.38 x 10^-23 J/K) and T is the temperature in Kelvin.

However, we are not given the temperature directly. Instead, we are given the pressure and volume of the gas. We can use the ideal gas law to find the temperature. The ideal gas law is:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, we need to convert the pressure from atmospheres to Pascals, since the ideal gas constant R is usually given in terms of Pascals. 1 atmosphere is approximately 101325 Pascals, so:

P = 1.50 atm * 101325 Pa/atm = 151987.5 Pa

The volume is given in liters, but we need it in cubic meters. 1 liter is 0.001 cubic meters, so:

V = 2.00 L * 0.001 m^3/L = 0.002 m^3

We are given that we have 1.00 mole of gas, and the ideal gas constant R is 8.314 J/(mol*K), so we can now solve the ideal gas law for T:

T = PV / nR = (151987.5 Pa * 0.002 m^3) / (1.00 mol * 8.314 J/(mol*K)) = 36553.7 K

Now we can substitute this into the formula for the average kinetic energy:

KE_avg = 3/2 kT = 3/2 * 1.38 x 10^-23 J/K * 36553.7 K = 7.56 x 10^-21 J

So the average kinetic energy per particle is approximately 7.56 x 10^-21 Joules.