Answer to Question #304590 in Mechanical Engineering for Jecko

Question

A gas initially at P1= 827,400 Paa and v1=2.6 li undergoes a process and change state to a point where p2=275,800 Paa and v2=6.2li. If the internal energy decreases 123 KJ and Cp=295 J/kgm-K, determine (a) Cp, (b) delta U, and (c) R.

Answer 1

The values given in the question are as follow; P1= 827400 Pa

P2= 275800 Pa

V1= 2.6 litre

V2 = 6.2 litre

Internal energy decreases = 123 KJ

Cp = 295 J/kg-K

(a) Cp

the value of Cp = 295 J/kg-K (given in the question) Ans

(b) delta U

basically delta U is the change in the internal energy of the system, as mentioned in question 123KJ

delta U = -123 KJ Ans.

(c) R

from ideal gas law

P1 V1n = P2 V2n

827400* 2.6n = 275800 * 6.2n n= 1.26

We kno that , Cp = n*R/(n-1)

R = (n-1)*Cp /n

R = (1.26-1)*295 /1.26 R = 61.85 J/ kg-K Ans