Question

A certain substance has a heat of vaporization of 61.65 kJ/mol. At what Kelvin temperature will the vapor pressure be 3.50 times higher than it was at 301 K?

Answer #1

**Answer :- T2 = 317.13 K**

You must use the **Clausius-Clapeyron
equation****.**

Therefore,

**ln( P1 / P2 ) = ( ΔHvap / R )( 1 / T2 – 1 / T1
),**

where,

P1 - the vapor pressure measured at T1

P2 - the vapor pressure measured at P2

ΔHvap - the enthalpy of vaporization

R - the gas constant - expressed in *Joules per mol K*

You have everything you need to solve for T2. Since the pressure
measured at this new temperature will be **3.50
times** bigger than P1, you can write it as P2=3.50P1 and
use it in this form in the equation.

So, plug all in and you'll get

ln(P1 / 3.50P1 ) = [ ( 61650 J/mol ) / ( 8.314 J/mol k ) ] * ( 1 / T2−1 / 301 K )

ln( 1 / 3.50 ) = ( 7415.20 )( 1 / T2 ) – ( 7415.20 )( 1 / 301 )

−1.2528 = 7415.20 T2 – 24.6352

23.3824 = 7415.20 T2

T2 = 7415.20 / 23.3824

T2 = 317.13K

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T=_____K

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