1 mol of NH3 gas at 27C is expanded under adiabatic condition to make volume 8 times ( = 1.33). Final temperature and work done, respectively, are (a) 150 K, 900 cal (b) 150 K, 400 cal (c) 250 K, 1000 cal (d) 200 K, 800 cal

Gamma is given = 1.33
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I also calculated the temperature after some work. Could you help me with the Work done?
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Hlo...

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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 Now, using γ = Cp / Cv 1.33 = 4R / 3R = Cp /Cv So, by compairing both sides we can say that : Cv = 3R Using : ΔU= q +w As, q=0 ; ΔU= w w = nCvΔT w= 1 (3R) ( T2 -T1 ) w= 3R ( 150-300 ) w= -150R
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The correct answer is t2 is 150k & work done 900 cal

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