Question

Boiling point of a liquid is 50K at 1 atm and deltaH =460.6 cal .What will be its boiling point at 10 atm

Answer 1

Answer:

approx=100k

Explanation:

from

Clausius-Clapeyron equation

l n(P2/P1) = (ΔHv/R).(1/T1 - 1/T2)

T1 and T2 are the boiling points in Kelvins at pressures P1 and P2 respectively

ΔHv is the enthalpy of vapourisation and R the gas constant, 8.314 J/mol.K

converting the units into SI system= 460.6*4.185 =1925.1 J

ln(10/1) = [1925.1/8.314].[(1/50) - (1/ T2)]

[(1/50) - (1/ T2)] = 2.303/231.9 = 9.931 x 10^(-3)

when we solve it we get

T2= 99.3K

which approximately equals to 100K