Chemistry - Thermodynamics - II Question with Solution | TestHub
Consider the following phase transitions:
(i) H2O (s, 1 atm, 273 K) → H2O (l, 1 atm, 273 K)
(ii) H2O (s, 1 atm, 300 K) → H2O (l, 1 atm, 300 K)
(iii) H2O (s, 1 atm, 200 K) → H2O (l, 1 atm, 200 K)
(iv) H2O (s, 0.5 atm, 273 K) → H2O (l, 0.5 atm, 273 K)
(v) H2O (s, 2 atm, 273 K) → H2O (l, 2 atm, 273 K)
(vi) C6H6 (l, 1 atm, 353 K) → C6H6 (g, 1 atm, 353 K)
(vii) C6H6 (l, 1 atm, 400 K) → C6H6 (g, 1 atm, 400 K)
(viii) C6H6 (l, 1 atm, 300 K) → C6H6 (g, 1 atm, 300 K)
(ix) C6H6 (l, 2 atm, 323 K) → C6H6 (g, 2 atm, 323 K)
The normal freezing point of water is 273 K and the normal boiling point of benzene is 353 K.
Now, a four-digit number 'abcd' is defined as:
Digit 'a': number of phase transitions for which Δ Stotal = 0
Digit 'b': number of phase transitions for which Δ Stotal > 0
Digit 'c': number of phase transitions for which Δ Stotal < 0
Digit 'd': number of phase transitions for which Δ H > 0
Determine the number 'abcd'.
Answer:
Solution:
All nine are endothermic (melting or vaporization), so d = 9. Equilibrium (Δ Stot=0): (i) and (vi), so a = 2. Spontaneous (>0): (ii) T above melting point; (v) at 2 atm the melting point of ice falls below 273 K (ice melts with Δ Vfus<0, so dTm/dP<0), so 273 K is above it; (vii) above the boiling point; so b = 3. Non-spontaneous (<0): (iii) below the melting point; (iv) at 0.5 atm the melting point of ice rises above 273 K, so ice is stable; (viii) below the boiling point; (ix) at 2 atm benzene boils above 353 K, so 323 K is well below it; so c = 4. abcd = 2349.
