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`        1mole of NH3 gas(gamma=1.33) at 27°C is allowed to expand adiabatically reversibly so that the final volume becomes 8 times. What will be the work done?`
2 years ago

Chhavi Jain
92 Points
```							Dear student, Given : Rev. adiabatic expansion(q=0) , Vf =8 Vi   , gamma= 1.33 , n = 1 , T= 300K                                       using TVγ-1 = constant                                                                T1 / T2 = (V2 / V1 ) ^ (γ-1)                                   300/ T2 =  8 ^ ( 1.33-1)                                      T2 = 150 K ( approx ) Now, using γ = Cp / Cv                                1.33 = 4R / 3R = Cp /Cv So, by compairing both sides we can say that :    Cv = 3R Using : ΔE= q +w          As,   q=0    ;  ΔE= w                                w = nCvΔT                                w= 1 (3R) ( T2 -T1 )                                w= 3R ( 150-300 )                                w= -150R   Chemistry ConsultantAskiitians
```
2 years ago
Rishi Sharma
614 Points
```							Dear Student,Please find below the solution to your problem.Given : Rev. adiabatic expansion(q=0) , Vf=8 Vi , gamma= 1.33 , n = 1 , T= 300K            using TVγ-1= constant                 T1/ T2= (V2 / V1)^(γ-1)                 300/ T2= 8^ ( 1.33-1)                  T2= 150 K ( approx )Now, usingγ = Cp/ Cv              1.33 = 4R / 3R = Cp/CvSo, by compairing both sides we can say that :  Cv= 3RUsing : ΔE= q +w     As, q=0  ;ΔE= w               w = nCvΔT                w= 1 (3R) ( T2-T1)                w= 3R ( 150-300 )                w= -150RThanks and Regards
```
7 days ago
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