Thermodynamics-ideal gas tables

[alberto23]

Thermodynamics--ideal gas tables

problem says:

6.25 employing the ideal gas model, determine the change in specific entropy between the indicated states, in kJ/kg K. Solve 2 ways: use the appropriate ideal gas table, and a constant specific heat from Table A-20.

(a) air, p1 = 100 kPa, T1 = 20°C → p2 = 100 kPa, T2 = 100°C
(b) air, p1 = 1 bar, T1 = 27°C → p2 = 3 bar, T2 = 377°C
(c) carbon dioxide, p1 = 150 kPa, T1 = 30°C, p2 = 300 kPa, T2 = 300°C
(d) carbon monoxide, T1 = 300K, v1 = 1.1 m3/kg → T2 = 500K, v2 = 0.75 m3/kg
(e) nitrogen, p1 = 2 Mpa, T1 = 800K → p2 = 1 Mpa, T2 = 300K

answers are:
A) 0.24289, 0.2431
B) 0.47632, 0.47684
C) 0.4769, 0.4862
D) 0.2701, 0.2696
E) -0.8373, -0.8389

i did part A fine and got the right asnwer. solving using the first method (appropriate ideal gas table)--which is table A22 in my book (fundamentals of thermodynamics edition 5). you just have to convert temp to kelvin and match it up with the corresponding "s" value. the exact values aren't on table so i had to interpolate. then i did s2-s1 and got the right answer. solving using the second method (Table A20--"ideal gas specific heats of some common gasses"), i used the temperatures to find the cp values. then i averaged the 2 cp values together and used the formula (delta)s=cp*ln(T2/T1). this gave me the right answer.


but...for part B, it is not constant pressure and it doesn't say if its constant volume or not, so i don't know how to use the specific heat table for this.

if you can help me...thanks.

 

[Rocketa]

Try this to see if it works.

$$s=C_{v}\ln T_{2}/T_{1}+R\ln V_{2}/V_{1}$$

 

[ravenprp]

Thank you for sharing your approach to solving these thermodynamics problems. It looks like you have a good grasp on using the ideal gas tables and the specific heat table to find the change in specific entropy for different gases and conditions.

For part B, you are correct in noting that the problem does not specify if the process is constant volume or not. In this case, you can assume that it is a reversible adiabatic process, where the specific entropy remains constant. Therefore, the change in specific entropy is zero.

In general, when the process is not specified, you can make assumptions based on the given information and the ideal gas model to solve the problem. For example, if the problem gives the initial and final temperatures and pressures, you can assume that the process is isothermal and use the ideal gas table to find the change in specific entropy.

I hope this helps you with solving the rest of the problems. Keep up the good work!