Question

Nitrogen gas is being withdrawn at the rate of 4.5 g/s from a 0.15-m3 cylinder, initially containing the gas at a pressure of 10 bar and 320 K. the cylinder does not conduct heat, nor does its temperature change during the emptying process. What will be the temperature and pressure of the gas in the cylinder after 5 minutes? What will be the rate of change of the gas temperature at this time? Nitrogen can be considered to be an ideal gas with Cp*=30 J/mol.K.

Answer 1

Answer:

Final temperature = 152.57K,

Pressure = 0.6907 bar.

dT/dt = - 1,151 K/s.

Explanation:

The first thing to do here is to write out the equation for mass balance as given below:

dN/dt = N -------------------------------------------------------------------------------------------(1).

N = P/T, then, substitute the values given in the question into:

d[p/T]/ dt = [- 4.5/28 × 8.314]/0.15 = - 8.9 × 10⁻⁵ bar/K.s.

Thus, there is the need to integrate, Integrate [p/T]f = 10/320 - 8.9 × 10⁻⁵ bar/K.s. ------------------------------------(2).

NB; fT = final temperature, fP = final pressure and iT = initial temperature.

Also, [ fT]³⁰/₈.₃₁₄/ [fP] = [iT]³⁰/₈.₃₁₄/ Pi] = [ 320]³⁰/₈.₃₁₄/ 10.

Therefore, [fT]³⁰/₈.₃₁₄ = 109.52 × 10⁶.

Final temperature=  [fP]³⁰/₈.₃₁₄ × 169.05.

Note that fP/ [fP]³⁰/₈.₃₁₄ × 169.05 = 10/320 - 8.9 × 10⁻⁵.

Therefore, [fP]¹ ⁻ ³⁰/₈.₃₁₄ = 0.7651.

Hence, Final temperature = 152.57K,

Pressure = 0.6907 bar

dT/ dt = N[RT]² / Cv . PV.

R = 30 - 8.314 = 21.86 J/mol K.

Then, the rate of change of the gas temperature at this time = dT/dt = - 1,151 K/s.