Suppose a system has quartic degrees of freedom, instead of quadratic degrees of freedom: E(x) = cx4 where c is a constant and x is a continuous variable. Find the average energy of this system, in the same way that we derived the equipartition theorem.

Explanation:

The average energy of the system with quartic degrees of freedom. The quartic degrees of freedom is same as biquadratic since it means 4. Systems having quartic degrees of freedom are usually have their energies represented in terms of some variable raised to the power of 4.

The given system with quartic degrees of freedom here has E(x) = cx4 . The standard result from the statistical mechanics will be helpful here in calculating internal energy of the system, which is also its average energy.

U = kT^2\d(lnq)}/dT

Now, to find out q(x) we will use the equation  q(x) = \int^{+\infty}_{-\infty} exp\bigg(\frac{-E(x)}{kT}\bigg)dx = \int^{+\infty}_{-\infty} exp\bigg(\frac{-cx^4}{kT}\bigg)dx

For a quadratic system, you would get a Gaussian integral which has a standard result.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂