Question

The normal boiling point of liquid ethanol is 351 K. Assuming that its molar heat of vaporization is constant at 37.5 kJ/mol, the boiling point of C2H5OH when the external pressure is 1.35 atm is K.

Answer 1

Use the Clausius-Clapeyron equation.

You know p = 1 atm when T = 351K. Solve for p2 when T = 1.35 atm.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the boiling point of ethanol when the external pressure is 1.35 atm, we can use the Clausius-Clapeyron equation:

ln(P1/P2) = ΔHvap / R * (1/T2 - 1/T1)

Where P1 is the initial pressure, P2 is the final pressure, ΔHvap is the molar heat of vaporization, R is the ideal gas constant, T1 is the initial temperature, and T2 is the final temperature.

Given:
P1 = 1 atm
P2 = 1.35 atm
ΔHvap = 37.5 kJ/mol
R = 8.314 J/(mol·K)
T1 = 351 K (normal boiling point of ethanol)

Substituting these values into the equation, we can solve for T2:

ln(1.35/1) = (37.5 × 10^3)/(8.314) * (1/T2 - 1/351)

ln(1.35) = 4510.74 * (1/T2 - 1/351)

Taking anti-logarithm of both sides:

1.35 = (1/T2 - 1/351)^4510.74

Taking reciprocals of both sides:

1/1.35 = (1/T2 - 1/351)^(-4510.74)

1 = (1/T2 - 1/351)^(-4510.74) * 1.35

(1/T2 - 1/351) = (1/1.35)^(1/(-4510.74))

(1/T2 - 1/351) = 0.8924

Now, rearranging the equation to solve for T2:

1/T2 = 0.8924 + 1/351

1/T2 = 0.8924 + 0.0028

1/T2 = 0.8952

T2 = 1/0.8952

T2 ≈ 1.1164 K

Therefore, the boiling point of ethanol when the external pressure is 1.35 atm is approximately 1.1164 K.

Answer 3

[AI-generated answer — Explain Bot]

To find the boiling point of ethanol (C2H5OH) when the external pressure is 1.35 atm, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

Where:
P1 is the initial pressure (normal boiling point in this case, 1 atm)
P2 is the final pressure (1.35 atm)
ΔHvap is the molar heat of vaporization (37.5 kJ/mol)
R is the ideal gas constant (8.314 J/mol·K)
T1 is the initial temperature (normal boiling point, 351 K)
T2 is the final temperature (what we want to find)

First, let's convert the heat of vaporization from kJ/mol to J/mol:
ΔHvap = 37.5 kJ/mol = 37.5 * 1000 J/mol = 37500 J/mol

Now let's rearrange the equation to solve for T2:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

ln(P2/P1) = (37500 J/mol / (8.314 J/mol·K)) * (1/351 K - 1/T2)

ln(1.35/1) = (37500 J/mol / (8.314 J/mol·K)) * (1/351 K - 1/T2)

Taking the natural logarithm of both sides:

ln(1.35) = (37500 J/mol / (8.314 J/mol·K)) * (1/351 K - 1/T2)

Now we can rearrange the equation to solve for T2:

(1/351 K - 1/T2) = (8.314 J/mol·K/37500 J/mol) * ln(1.35)

Simplifying:

(1/351 K - 1/T2) = 0.0002216 K⁻¹ * ln(1.35)

Now we can substitute the natural logarithm value of 1.35 into the equation:

(1/351 K - 1/T2) = 0.0002216 K⁻¹ * 0.30010

(1/351 K - 1/T2) = 0.0000667

Now let's solve for T2:

1/351 K - 0.0000667 = 1/T2

1/T2 = 0.0000667 + 1/351 K

T2 = 1 / (0.0000667 + 1/351 K)

Calculating T2:

T2 ≈ 331.5 K

Therefore, the boiling point of ethanol (C2H5OH) when the external pressure is 1.35 atm is approximately 331.5 K.