Question

What is the vapor pressure of SiCl4 in mmHg at 32.5 ∘C? The vapor pressure of SiCl4 is 100 mmHg at 5.4∘C, and ΔHvap = 30.2 kJ/mol.

Answer 1

Given

Here we use Classius clayperon equation.

The equation

Ln(P2/P1) = - Delta H / R ( 1 / T2 - 1/ T1)

We are given vapor pressure of SiCl4 at 5.4 ⁰ C   ( 5.4 + 273.15 K ) = 100 mmHg and we have to find it at

32.5 ⁰C ( 32.5+273.15 K )

Delta H = 30.2 kJ/mol = 30200 J /mol

Lets P1 = 100 mmHg , T1 = ( 5.4 + 273.15 K)

P2 = unknown , T2 = ( 32.5+273.15 K )

Value of R (gas constant ) = 8.314 J / (K mol)

Lets plug given value and get the P2.

Ln (P2/ 100 mmHg)

= - 30200 J per mol/ ( 8.314 J / (K mol ) * [ 1/( 32.5+273.15 K ) – 1 / (5.4 + 273.15 K) ]

P2 = 323.80 mmHg

So the vapor pressure at 32.5 deg C = 323.80 mmHg