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A certain liquid has a vapor pressure of 92 torr at 23 c and 300 torr at 45 c. calculate the delt hvap . and find it normal boiling point and give the answers, please

User Pmpm
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Final answer:

To find the ΔHvap, use the Clausius-Clapeyron equation. The ΔHvap is -38,160 J/mol. The normal boiling point is approximately 96 °C.

Step-by-step explanation:

Solution:

To find the ΔHvap, we can use the Clausius-Clapeyron equation:

ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)

We can substitute the given values: P₁ = 92 torr, T₁ = 23 °C = 296 K, P₂ = 300 torr, and T₂ = 45 °C = 318 K. R is the ideal gas constant (0.0821 L·atm/mol·K).

Solving for ΔHvap, we get:

ΔHvap = (-8.314 J/mol·K)(ln(300/92))/(1/318 - 1/296))

ΔHvap = -38,160 J/mol

The normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level. Since the atmospheric pressure is about 760 torr, we can use this pressure to find the normal boiling point.

Substituting the given values into the Clausius-Clapeyron equation and solving for T, we get:

T = (-8.314 J/mol·K)(ln(760/92))/(1/318 - 1/296))

T ~ 369 K ~ 96 °C

User Mark Van Proctor
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