Determining final temperature, pressure and work done

[nombusoz]

hi guys I am swamp with work and have not been able to go through my assignments proper

1. one cube meter of an ideal gas at 600K and 1000Kpa expands to 5 times its initial volume: by mechanically reversible, isothermal process and by a mechanically reversible, adiabatic process


2.for each case calculate the final temperature, pressure, and the work done by the gas. take Cp= 21 J/mol.K



3. For isothermal process:
T2= 600
P1V1=P2V2
thus P2= (1*1000)/5 =200Kpa
W= -P1V1 ln V2/V1
= -(1000000000*1) ln (5/1)
= -1609.438 KJ

 

[grzz]

P2 is ok.

Check the value of P1V1 in the calculation for the work.

 

[nombusoz]

P1= 1000kpa = 1000*10^3 pa =1000000000 pa
V1= 1m^3

thus W using pa is equal to -1609.438 KJ

 

[grzz]

P1= 1000kpa = 1000*10^3 pa =1000000000 pa

P1= 1000kpa = 1000*10^3 pa =1000000 pa

 

[TSny]

Your work looks good except that when a gas expands reversibly, the gas does a positive amount of work.

It's always important to distinguish between the work done on a system and the work done by the system.