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.
Use the Clausius-Clapeyron equation.
You know p = 1 atm when T = 351K. Solve for p2 when T = 1.35 atm.
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.
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.