What is the root mean squared velocity of a helium atom in one mole of an ideal gas at a pressure of 2.00 atmospheres and a temperature of 30°C? (He = 4 u)

To solve this problem, we need to use the formula for root mean square velocity, which is given by:

v_rms = sqrt(3kT/m)

where:

• k is the Boltzmann constant (1.38 x 10^-23 J/K)
• T is the temperature in Kelvin
• m is the mass of one molecule

First, we need to convert the temperature from Celsius to Kelvin. The conversion formula is T(K) = T(C) + 273.15. So, T = 30 + 273.15 = 303.15 K.

Next, we need to find the mass of one helium atom. We know that one mole of helium has a mass of 4 grams. Since 1 mole = 6.022 x 10^23 atoms (Avogadro's number), the mass of one helium atom is 4/6.022 x 10^23 = 6.644 x 10^-24 g. But we need the mass in kilograms, so we convert it to 6.644 x 10^-27 kg.

Now we can substitute these values into the formula:

v_rms = sqrt(3 * 1.38 x 10^-23 * 303.15 / 6.644 x 10^-27) = 1367 m/s.

So, the root mean square velocity of a helium atom in one mole of an ideal gas at a pressure of 2.00 atmospheres and a temperature of 30°C is approximately 1367 m/s.