Three moles of helium gas (molar mass MM = 4.00 g/molg/mol) are in a rigid container that keeps the volume of the gas constant. Initially the rms speed of the gas atoms is 850 m/sm/s. What is the rms speed of the gas atoms after 3600 J of heat energy is added to the gas?

The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Explanation:

Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg

The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)

where u = initial rms speed of the gas = 850 m/s

Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J

Then, 3600 J is added to the gas,

New kinetic energy of the gas = 4335 + 3600 = 7935 J

New kinetic energy of Helium atoms = (1/2)(m)(v²)

where v = final rms speed of the gas = ?

7935 = (1/2)(0.012)(v²)

v² = (7935×2)/0.012

v² = 1,322,500

v = 1150 m/s

Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Hope this Helps!!!

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.