Question

Enter your answer in the provided box. The vapor pressure of ethanol is 1.00 x 10² mmHg at 34.90°C. What is its vapor pressure at 54.81°C? ( AH yap for ethanol is 39.3 kJ/mol.) mmHg

Answer 1

2,54x10² mmHg

Step-by-step explanation:

To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:

`Ln((P_(2))/(P_(1)) ) =( deltaH_(vap))/(R) ((1)/(T_(1))-(1)/(T_(2)) )`

Where:

P1 is 1,00x10² mmHg

ΔHvap is 39,3 kJ/mol

R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹

T1 is 34,90°C + 273,15 = 308,05 K

T2 is 54,81°C + 273,15 = 327,96 K

Thus:

`Ln((P_(2))/(1,0x10^(2)mmHg)) =(39,3kJ/mol)/(8,314x10^(-3)kJ/molK) ((1)/(308,05K)-(1)/(327,96K) )`

Thus, P2 is 2,54x10² mmHg

I hope it helps!