The standard boiling point of A(l) is 400 K . The enthalpy of vaporization of A(l) at 400 K and 1 ba r is 20 k c a l / m o l . Which of the following information is/are correct regarding the process:

Question

The standard boiling point of A(l) is 400 K . The enthalpy of vaporization of A(l) at 400 K and 1 ba r is 20 k c a l / m o l . Which of the following information is/are correct regarding the process: A ( l , 400 K , 1 ba r ) → A ( g , 400 K , 0.5 ba r ) Given: Molar heat capacities ( C p , m ​ ) values (in c a l / K ⋅ m o l ): A(g) = 5 ; A(l) = 10 ; ln 2 = 0.7 , ln 5 = 1.6 , R = 2 c a l / K ⋅ m o l . Assume ideal behavior for A(g).

TestHub · Accepted Answer

Question Explanation: Determine which of the following statements are correct.

Concept: For a change of state, the following thermodynamic equations apply:

$Δ H_{t o t a l} = Δ H_{v a p} + Δ H_{g a s}$

$Δ S_{t o t a l} = \frac{Δ H _{v a p}}{T _{b}} + R ln \frac{P _{1}}{P _{2}}$

$Δ G = Δ H - T Δ S = - RT ln \frac{P _{2}}{P _{1}}$

Solution: At $T = 400 K$ , for an ideal gas, $Δ H_{g a s} = 0$ .

Therefore, $Δ H = Δ H_{v a p} = 20 k c a l / m o l$ .

The change in entropy is:

$Δ S = Δ S_{v a p} + Δ S_{g a s} = \frac{20000}{400} + 2 ln \frac{1}{0.5} = 50 + 2 (0.7) = 51.4 c a l / (K \cdot m o l)$

The Gibbs free energy change is:

$Δ G = Δ H - T Δ S = 20000 - 400 (51.4) = 20000 - 20560 = - 560 c a l / m o l$

Since $Δ G < 0$ , the process is spontaneous under these conditions.

Final answer: Options (A, B & C)

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Options:(select one or more)

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