Question

Question #f4070

Answer 1

Given

• `n->"number of moles of the gas"=1mol`
• `P_i->"Initial pressure of the gas"=10atm`
• `P_f->"Final pressure of the gas"=1atm`

• `T->"Temperature of the gas"=27^@C=300K`

• `R->"Universal gas constant"`
  `" "=2calK^-1mol^-1xx4.184J/"cal"`
  `" "=8.368JK^-1mol^-1`

• `R->"Universal gas constant"=0.082LatmK^-1mol^-1`

The expansion may be done in two different paths
1) Isothermal reversible
2) Isothemal irrevresible

1) For Isothermal reversible expansion

work done `w_"rev"=-nRTln(P_i/P_f)`

`w_"rev"=-1xx8.368xx300ln(10/1)=-5.78xx10^3J`

2) For Isothermal Irreversible expansion

work done `w_"irrev"=-P_fxx(V_f-V_i)`

`=>w_"irrev"=-P_fxx((nRT)/P_f-(nRT)/P_i)`

`=>w_"irrev"=-nRT(1-P_f/P_i)`

`=>w_"irrev"=-1xx0.082xx300(1-1/10)=-22.14Latm`
`=-22.14Latmxx(101.3J)/(1Latm)~~-2.242xx10^3J`

PLEASE NOTE

The second part should be our answer as per given problem.
Since the expansion has been brought about under constant pressure of `1atm`. In the first case the pressure has been diminished slowly in reversible way from `10atm to 1atm`