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The vapor pressure of liquid lead is 400 mm Hg at 1.90×10^3 K. Assuming that its...

The vapor pressure of liquid lead is 400 mm Hg at 1.90×10^3 K. Assuming that its molar heat of vaporization is constant at 181 kJ/mol, the vapor pressure of liquid Pb is________ mm Hg at a temperature of 1.92×10^3 K

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Answer #1

P1 = 400mmHg

T1 = 1.9*10^3 K = 1900K

P2 =

T2 = 1.92*10^3K = 1920K

H   = 181KJ/mole = 181000J/mole

logP2/P1     = H/2.303R [ 1/T1 -1/T2]

logP2/400   = 181000/2.303*8.314 [1/1900- 1/1920]

logP2/400    = 9453(0.000526-0.00052)

logP2/400    = 0.056718

P2/400     = 10^0.056718

P2/400     = 1.1395

P2          = 400*1.1395   = 455.8mmHg >>>>answer

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