The normal boiling point of ammonia is -33.33 C. The boiling point of ammonia un

Question

The normal boiling point of ammonia is -33.33 C. The boiling point of ammonia under 23.74 psi is -23.33 C. Suppose the pressure on the liquid ammonia at -23.33 C is suddenly dropped from 23.74 psi to 14.71 psi (1 atmosphere). What will happen? Calculate the entropy change of the universe per mole of ammonia for this event. The heat vaporization of liquid ammonia at -33.33 C is 327.4 cal/g while that at -23.33 C is 320.2 cal/g. The heat capacity of gaseous ammonia is 35.06 J/K mol while that of liquid ammonia is 80.80 J/K.

Explanation / Answer

1 psi = 0.068 atm so, 23.74psi = 1.613 atm and 14.74psi=1 atm Initial temperature = 273-23.33 =249.67 K Since n R and V are constant so, P1/T1 =P2/T2 or, 1.613/249.67 = 1/T2 or T2= 154.75 K =-118.21 degree C So all ammonia will be converted to gas(as Boiling pt of ammonia at 1atm =-33.33degree C) or if temperature is constant volume of NH3 will increase. If this happens for 1 mole of ammonia 1 mole NH3 =17g of NH3 Vapourization will take place at -33.33 degree C= 239.67 K SO entropy change will be due to temperature change as well as vapourization So change in entropy (1 cal = 4.18J) =80.80(249.67-239.67) + (327.4*4.18*17) + 35.06(239.67-154.75) Heat for temperature Heat for Heat for temperature change for liq NH3 Vapourization change for gaseous NH3 from 249.67K to at -33.33degreeC from 239.67K to 154.75K 239.67K =(808 +23265.044+2977.29 ) J =27050.34 J

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