Question

Certain experimental results indicate that the propellant gases of a liquid oxygen- gasoline reaction have a mean molecular mass of 23.2 kg/kg-mol and a specific heat ratio of 1.22. Compute the specific heat at constant pressure and at constant volume, assuming a perfect gas_____

Answer 1

To calculate specific heats for a perfect gas, use the mean molecular mass and the specific heat ratio with the universal gas constant to find Cp = 1.9853 kg m²/s² K and Cv = 1.6273 kg m²/s² K.

Step-by-step explanation:

To compute the specific heat at constant pressure (Cp) and at constant volume (Cv) for a perfect gas where the mean molecular mass (M) and specific heat ratio (γ) are given, we use the following equations:

The relation between Cp and Cv for an ideal gas is given by:

• Cp - Cv = R

Where R is the universal gas constant, which has a value of 8.314 J/(mol K) or 8.314 kg m2/s2 K mol.

The mean molecular mass M is given in kg/kg-mol, but we need R in terms of kg/kg-mol to be consistent with units:

• R = 8.314 kg m2/s2 K mol * 1 kg-mol/23.2 kg = 0.358 kg m2/s2 K kg-mol

The specific heat ratio γ is given as 1.22, which is also equal to Cp/Cv.

Thus, we can set up the following system of equations to solve for Cp and Cv:

1. γ = Cp / Cv
2. Cp - Cv = R

Solving these equations simultaneously, we get:

• Cv = R / (γ - 1)
• Cp = γ * Cv

Now, using the given values:

• Cv = 0.358 / (1.22 - 1) = 0.358 / 0.22 = 1.6273 kg m2/s2 K
• Cp = 1.22 * 1.6273 = 1.9853 kg m2/s2 K

The specific heat at constant volume is 1.6273 kg m2/s2 K and the specific heat at constant pressure is 1.9853 kg m2/s2 K.