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The normal boiling point of cyclohexane (enthalpy of vaporization: 30.08kJmol1 ) is 80.75C . What will be its vapour pressure at 25C ?

Answer
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Hint: We can calculate the vapour pressure using the enthalpy of vapourization, temperature, and initial vapour pressure. We have to substitute the values of the enthalpy of vapourization, temperature, and initial vapour pressure in the expression below to get the vapour pressure at the given temperature.
Formula used: The formula to calculate the vapour pressure is,
logP2P1=ΔvapH2.303R(1T21T1)
Here P1 represents the initial vapour pressure
P2 represents the vapour pressure at the given temperature
T2 represents the final temperature
T1 represents the initial temperature
ΔHvap represents the enthalpy of vapourization

Complete answer:
The given data contains,
The value of T1 is 80.75C .
The value of T2 is 25C .
The value of P1 is 760mmHg .
The value of ΔHvap is 30.08kJmol1 .
Let us now convert the degree Celsius into Kelvin by adding 273.15 to the given temperature. The formula is written as,
K=C+273.15
Let us now substitute the given value of Celsius in the expression.
K=C+273.15
K=80.75C+273.15
On adding we get,
K=353.9K
The value of T1 in Kelvin is 353.9K .
K=C+273.15
K=25C+273.15
K=298.15K
The value of T2 in Kelvin is 298.15K.
Let us now use the expression to calculate the vapour pressure at the given temperature.
The formula to calculate the vapour pressure is,
logP2P1=ΔvapH2.303R(1T21T1)
Here P1 represents the initial vapour pressure
P2 represents the vapour pressure at the given temperature
T2 represents the final temperature
T1 represents the initial temperature
ΔHvap represents the enthalpy of vapourization
Let us now substitute the values of enthalpy of vapourization, temperatures, and pressure to get the vapour pressure at 25C.
The vapour pressure at 25C is calculated as,
logP2P1=ΔvapH2.303R(1T21T1)
On substituting the known values we get,
log(P2760mmHg)=30.08×103Jmol12.303(8.314JK1mol1)(1298.15K1353.9K)
We get,
log(P2mmHg)=log(760)30.08×103Jmol1(2.303)(8.314JK1mol1)[1298.15K1353.9K]
On simplification we get,
P2=112.4mmHg
We have calculated the vapour pressure at 25C is 112.4mmHg.

Note:
Based on the values and given variables we have to calculate the vapour pressure. We can also calculate the vapour pressure of the solution from mole fraction of the solvent and vapour pressure of the pure solvent. While calculating the vapour pressure from enthalpy of vapourization, we have to remember to convert the degree Celsius to Kelvin for temperatures failing which the exact vapour pressure could not be obtained.