Question

An ideal monatomic gas at an initial temperature of 500 K is expanded from 5.0 atm to a final pressure of 1.0 atm. Calculate w, q, DU, and (where applicable) DH and DT when the expansion is performed (a) reversibly and isothermally, and (b) reversibly and adiabatically.

Help Please!!!

Answer 1

Given,

T=500 K

P₁=5atm

P₂=1 atm

a) For reversible isothermal process:

                w= -RT ln (P₂/P₁)

                =-8.314*500*ln(1/5)=6690.4 J

                q=-w=-6690.4 J

                dU=0

                dH=dU-w=0+6690.4=-6690.4 J

                dT= 0 for isothermal process

b) For reversible adiabatic process:

                w=-RT₁((P₂/P_1­)^(n-1)/n -1)

                =-8.314*500*((1/5)^0.67/1.67-1)

                =-1973 J

                q=0 for adiabatic process

                dU=-w=1973 J

                dH=0

                dT= w/C_V

                        =1973/20.8=94.8 K