Answer to Question #145832 in Physical Chemistry for Dexter M. Reyes

Question

An ideal gas undergoes a mechanically reversible process (constant pressure, constant temperature and adiabatic process. The gas entering a T1=650K and P1=10bar decreases its temperature at constant pressure where V2=2.91x10- 3 m3. Then, it went to isothermal process to decrease its pressure. Finally, the gas returns to its initial state. Take

Cp=7

2

Answer 1

T1=650K

P1= 10bar

V2=2.91x10^-3

Constant P and T

V1=?

$Pdv= nRT$

10(V2-V1) = 8.314×650

2.91x10^-2 -10v1 =5404.1

V1=5.4×10^-3