The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 55.5°C?

Answer 1

The vapour pressure at 55.5 °C is 282 mmHg.

Chemists often use the Clausius-Clapeyron equation to estimate the vapour pressures of pure liquids:

#color(blue)(bar(ul(|color(white)(a/a) ln(P_2/P_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "#

where

#P_1# and #P_2# are the vapour pressures at temperatures #T_1# and #T_2#
#Δ_"vap"H# = the enthalpy of vaporization of the liquid
#R# = the Universal Gas Constant

In your problem,

#P_1 = "100 mmHg"#; #T_1 = "34.9 °C" = "308.05 K"#
#P_2 = "?"#; #color(white)(mmmmm)T_2 = "55.5 °C" = "328.65 K"#
#Δ_"vap"H = "42.3 kJ/mol"#
#R = "0.008 314 kJ·K"^"-1""mol"^"-1"#
#ln(P_2/("100 mmHg")) = (42.3 color(red)(cancel(color(black)("kJ·mol"^"-1"))))/("0.008 314" color(red)(cancel(color(black)("kJ·K"^"-1""mol"^"-1"))))(1/(308.05 color(red)(cancel(color(black)("K")))) - 1/(328.65 color(red)(cancel(color(black)("K")))))#
#ln(P_2/("100 mmHg")) = 5088 × 2.035 × 10^"-4" = 1.035#
#P_2/("100 mmHg") = e^1.035 = 2.816#
#P_2 = "2.816 × 100 mmHg = 282 mmHg"#

The on-line calculator here predicts a value of 286 mmHg. Close enough!

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Answer 2

To find the vapor pressure of ethanol at 55.5°C, you can use the Clausius-Clapeyron equation. However, since we only have one data point, we cannot directly apply this equation. Instead, we can use the Antoine equation, which is an empirical relationship that provides an approximation of vapor pressure over a range of temperatures. For ethanol, the Antoine equation is:

log10(P) = A - (B / (T + C))

where P is the vapor pressure in mmHg, T is the temperature in Celsius, and A, B, and C are constants specific to the substance. For ethanol, A = 8.20417, B = 1642.89, and C = 230.3.

Using this equation, you can calculate the vapor pressure of ethanol at 55.5°C by substituting T = 55.5°C into the equation and solving for P.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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