Answer to Question #300687 in General Chemistry for Kier

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 m^-3. Then, it went to isothermal process to decrease its pressure. Finally, the gas returns to its initial state. Take CP = 7/2 and CV = 5/2R.

a. T2, P3, V1and V3

b.Q,W,ΔU and ΔH

Answer 1

T1=650K

P1= 10bar

V2=2.91x10-3

Constant P and T

V1=?

Pdv= nRTPdv=nRT

10(V2-V1) = 8.314×650

2.91x10^-2 -10v1 =5404.1

V1=5.4×10³