Question

For mercury, the following relation exists between saturation pressure (bar) and saturation temperature (K) : 

log₁₀ p = 7.0323 – 3276.6/T– 0.652 log₁₀ T 

Calculate the specific volume v_g of saturation mercury vapour at 0.1 bar. 

Given that the latent heat of vapourisation at 0.1 bar is 294.54 kJ/kg. 

Neglect the specific volume of saturated mercury liquid.

Answer 1

Latent heat of vapourisation, 

h_fg = 294.54 kJ/kg (at 0.1 bar) ...(given) 

Using Clausius-Claperyon equation

Since v_f is neglected, therefore eqn. (i) becomes

\(\cfrac{dp}{dT}\) = \(\cfrac{h_{fg}}{v_gT}\)

Now, log₁₀ p = 7.0323 – \(\cfrac{32766}T\) - 0.652 log₁₀ T

Differentiating both sides, we get

From (i) and (ii), we have

Solving by hit and trial method, we get 

T = 523 K 

Substituting this value in eqn. (iii), we get