Question

Determine the unknown quantity, assuming air behaves as an ideal gas. R¯ = 8.314 kJ kmol−1 K −1 . MWair = 28.97 kg kmol−1 . If needed, average the specific heats between the beginning and ending temperatures.1. At I bar and 22 C, determine o and r. 2. At 10 bar and v 0.25 m3 kg 1, determine T and u. 3. At 600 K, determine cp, Cu, k, and v

Answer 1

1) o=1.18 kg/m^3; r=286.987 J/(kg K); 2) T=871.12 K, U=375 kJ; 3) cp=438.793 J/(kg K), cv=430.480 J/(kg K), v=1.0.193, k=52.8135

Step-by-step explanation:

1)

This questions is fully based on the Ideal Gas Law:

`pV=nRT`

As we need to find initial density, we should change temperature to Kelvins:

`T=22+273=295 K`

In addition, we should re-calculate amount of substance n, using mass m and MWair:

`n=m/MWair`

As it is known, density can be calculated, as:

`o=m/V`

Now, we can use all equations in the ideal gas law:

`pV=(m)/(MWair) RT\\pMWair=(m)/(V) RT\\o=(m)/(V) =(pMWair)/(RT)=100*10^3*28.97*10^-3/(8.314*295)=1.18 kg/m^(3)`

To calculate specific gas constant for the given scenario (r), we should do the following:

`r=R/MWair=8.314/(28.97*10^(-3))=286.987 J/(kg*K)`

Note, that we converted MWair to kg/mol using (10^(-3)).

2)

To solve this question, we should use again ideal gas law, from where we can directly find temperature of the gas:

`pV=nRT\\pV=(m)/(MWair)RT\\ T=(pVMWair)/(mR)=(10*10^(5)*28.97*10^(-3)*0.25)/(8.314*1)=871.12 K`

Note, that all values were converted to SI units for the calculations (1 bar= 10^5 Pa).

To find internal energy, we can use the following equation:

`U=(3)/(2) nRT=(3)/(2) (m)/(MWair) RT=(3)/(2)(1)/(28.97*-10^(-3))*8.314*871.12=375000 J=375 kJ`

3)

To find specific heat, we can use the following ideal gas laws:

`Cv=U/T=(3/2 nRT)/(T)=3/2nR=(3)/(2)(m)/(M) R\\ cv=Cv/m=(3)/(2)(R)/(M) =430.480 J/(kg K)`

Then, the cp, can be calculated using cv and R:

`cp=cv+R=438.793 J/(kg K)`

The thermal coefficient v represents the ratio between cp and cv:

v=cp/cv=1.0193

Finally, we can find k:

`k=(v)/(v-1)= 52.813`