One mole of an ideal gas expands from 10 atm against a constant pressure of 1 atm at 27 degree Celsius . The magnitude of work done by the gas is? (R=2 Cal/mol/K)

#degree celcius #

#degree celcius #

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dk_ch Share
Jan 14, 2017

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#

drawn

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#

drawn

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 and reversible from #10atm to 1atm #

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