Answer to Question #268274 in General Chemistry for Wafretz

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=2.91x10-³m³

Then, it went to isothermal process to decrease its pressure. Finally, the gas returns to its initial state. Take Cp=7/2 and Cv==5/2 R. 

Calculate the following:

a. T2,P3, V1 and 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