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Grade: 11
        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?
9 months ago

Answers : (1)

Chhavi Jain
92 Points
							
Dear student,
 
Given : Rev. adiabatic expansion(q=0) , Vf =8 V  , gamma= 1.33 , n = 1 , T= 300K 
              
                        using TVγ-1 = constant
                              
                                  T/ T= (V2 / V1 ) ^ (γ-1)
                                   300/ T2 =  8 ^ ( 1.33-1) 
 
                                    T= 150 K ( approx )
 
Now, using γ = Cp / Cv
              
                  1.33 = 4R / 3R = Cp /C
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) ( T-T1 )
                                w= 3R ( 150-300 )
                                w= -150R 
 
 
Chemistry Consultant
Askiitians 
9 months ago
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