The vapour pressure of a certain liquid is given by the equation:
$l o g_{10} P = 3.54595 - \frac{313.7}{T} + 1.40655 l o g_{10} T$

where $P$ is the vapour pressure in mm and T = Kelvin temperature. Determine the molar latent heat of vaporisation at $T = 80 K$ .

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Solution

$\frac{d l n P}{d T} = \frac{Δ H}{R T^{2}} . . . (i)$
$l o g P = 3.54595 - \frac{313.7}{T} + 1.40655 l o g T$
$l n P = 3.54595 \times 2.303 - \frac{313.7}{T} \times 2.303 + 1.40655 l n T$
$\frac{d l n P}{d T} = \frac{313.7 \times 2.303}{T^{2}} + \frac{1.40655}{T} . . . (i i)$
Compairing equation (i) & (ii)
$Δ H = R [313.7 \times 2.303 + 1.40655 T]$
at $T = 80 K$
$Δ H = 1669.9 c a l$ .

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The vapour pressure of a certain liquid is given by the equation:log10P=3.54595−313.7T+1.40655log10T where P is the vapour pressure in mm and T = Kelvin temperature. Determine the molar latent heat of vaporisation at T=80K.

dlnPdT=ΔHRT2...(i)logP=3.54595−313.7T+1.40655logTlnP=3.54595×2.303−313.7T×2.303+1.40655lnTdlnPdT=313.7×2.303T2+1.40655T...(ii)Compairing equation (i) & (ii)ΔH=R [313.7×2.303+1.40655T]at T=80KΔH=1669.9cal.

The vapour pressure of a certain liquid is given by the equation:
$l o g_{10} P = 3.54595 - \frac{313.7}{T} + 1.40655 l o g_{10} T$

where $P$ is the vapour pressure in mm and T = Kelvin temperature. Determine the molar latent heat of vaporisation at $T = 80 K$ .