You may only use anything written by you, or handed out in class. The following

Question

You may only use anything written by you, or handed out in class. The following information may or not be of use: R = 0.0821 l-atm/mole-^degree K = 8.314 J/mole-^degree K, 1.00 l-atm = 101.3 J and 1.00 cal = 4.184 J The heat capacity of NH_3 may be written: C_p = 28.0 + 26.3 times 10^-3 T (J/K-mole) One mole of NH_3, initially at 300 K, is plunged into a large temperature bath at 400K and allowed to come to equilibrium. 1.03 kJ of heat are absorbed by the NH_3 during this process. Find delta S_univ for this process.

Explanation / Answer

The heat absorbed the system ammonia is 1.03 kJ

The heat gained by the ammonia from the bath is =ncp*delta T

n is the moles of ammonia.

Basis: 1 mol of ammonia

Entropy of the surroundings = integral[n*cp*delta T]/T

Integrate between the limits 400 K and the 300 K.

Delta S (surroundings) =28 ln [400/300]+(26.3*10^-3)*(400-300)=8.05 J+2.63 J=10.68 J/K

Delta S of the system =[1.03*1000]J/400 K =-2.575 J/K

Thus the entropy of the universe =10.68-2.575 =8.105 J/K

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