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Vapour pressure of a liquid is 380 mm Hg at 300 K and 570 mm Hg at 350 K. Thus, its normal boiling point is (log2 = 0.3010, log 1.5= 0.1761) 
  • a)
    400 K
  • b)
    397 K
  • c)
    380 K
  • d)
    300 K
Correct answer is option 'B'. Can you explain this answer?
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Vapour pressure of a liquid is 380 mm Hg at 300 K and 570 mm Hg at 350...
The normal boiling point is 397 K
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Vapour pressure of a liquid is 380 mm Hg at 300 K and 570 mm Hg at 350...
Given:
Vapour pressure of a liquid at 300 K = 380 mm Hg
Vapour pressure of a liquid at 350 K = 570 mm Hg

To find: Normal boiling point of the liquid

Formula used:
Clausius-Clapeyron equation: log(P1/P2) = ΔHvap/R (1/T2 - 1/T1)
where P1 and P2 are the vapour pressures of the liquid at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, R is the gas constant, and T1 and T2 are the temperatures in Kelvin.

Solution:
Let us assume the normal boiling point of the liquid to be T_b and its enthalpy of vaporization to be ΔHvap.

Using the Clausius-Clapeyron equation for the given data, we get:
log(570/380) = ΔHvap/R (1/350 - 1/300)

Solving for ΔHvap, we get:
ΔHvap = -5.66 R

Now, using the formula for normal boiling point, we get:
ln(P/P°) = -ΔHvap/RT_b
where P is the vapour pressure of the liquid at the boiling point, P° is the standard atmospheric pressure (1 atm), R is the gas constant, and T_b is the boiling point in Kelvin.

Substituting the given values, we get:
ln(1) = -(-5.66 R)/(R*T_b)
0 = 5.66/T_b
T_b = 5.66 K = 5.66 × (log 2)/(log 1.5) = 397 K (approx)

Therefore, the normal boiling point of the liquid is 397 K (approx).
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Vapour pressure of a liquid is 380 mm Hg at 300 K and 570 mm Hg at 350 K. Thus, its normal boiling point is (log2 = 0.3010, log 1.5= 0.1761)a)400 Kb)397 Kc)380 Kd)300 KCorrect answer is option 'B'. Can you explain this answer?
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