Determining final temperature, pressure and work done

  • Thread starter nombusoz
  • Start date
  • #1
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hi guys im 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
 

Answers and Replies

  • #2
993
13


P2 is ok.

Check the value of P1V1 in the calculation for the work.
 
  • #3
2
0


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

thus W using pa is equal to -1609.438 KJ
 
  • #4
993
13


P1= 1000kpa = 1000*10^3 pa =1000000000 pa
P1= 1000kpa = 1000*10^3 pa =1000000 pa
 
  • #5
TSny
Homework Helper
Gold Member
12,955
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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.
 

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