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

𝑅 and Cv==

5

2

𝑅.

Calculate the following:

a. T2,P3, V1 and V3

b. Q,W,ΔU and ΔH

Answer 1

$V3 = 2.91×10^{-3}(\frac{10}{650})=4.48×10^{-4}$