Can I get help with this.

Question

Calculate the vapor pressure of octane  (b.p. 126.0 ∘C126.0⁢ ∘C) at the gas chromatography column temperature of 83.0 ∘C83.0⁢ ∘C using the form of the Clausius–Clapeyron equation shown

ln(𝑃1𝑃2)=−(Δ𝐻vap𝑅)(1𝑇1−1𝑇2)ln⁡(P1P2)=−(Δ⁢HvapR)⁢(1T1−1T2)

where 𝑅R is the ideal gas constant, Δ𝐻vapΔ⁢Hvap is the enthalpy of vaporization, 𝑇1T1 and 𝑇2T2 are two different temperatures, and 𝑃1P1 and 𝑃2P2 are the vapor pressures at the respective temperatures. The enthalpy of vaporization can be estimated using Trouton's rule, Δ𝐻∘vap=(88 J·mol−1⋅K−1)×𝑇bp.

P=Torr?

Calculate the vapor pressure of acetone (b.p. 56.0 ∘C56.0⁢ ∘C) at the same column temperature.

P=Torr?

For compounds of the same chemical class, what is the relationship between vapor pressure of a compound and retention in gas chromatography?

As the vapor pressure increases, retention time increases.

As the vapor pressure decreases, retention time decreases.

There is no relationship between vapor pressure and retention.

As the vapor pressure decreases, retention time increases.

Answer 1

Just to be clear : The CC equation is ln(P₂/P₁) = -(ΔH_v/R)(1/T₂ - 1/T₁)

T_NBP = 126°C = 399.15 K where the vapor pressure is 760 mmHg by definition. You want P_VP (83.0°C = 356.15 K)

ΔH_v = (88 J/mole-K)(399.15 K) = 3513 J/mole near BPt., R = 8.314 J/mole-K.

Let BP = case 1 and 83° be case 2:

After rearranging to solve for P₂

P₂ = (760 mmHg)exp[-3513/8.314)(1/356.15 - 1/399.15) = 670 mmHg

Same calculation for acetone.

I imagine that the more volatile substance (higher P_VP will be detected and pass through faster (shorter

times)

Please consider a tutor. Take care.