ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
where:
P1 = 1 atm
P2 = 2.5 atm
ΔHvap = 40.7 kJ/mol (latent heat of vaporization of water)
R = 8.314 J/mol*K (gas constant)
T1 = 373 K (boiling point of water at 1 atm)
T2 = ? (boiling point of water at 2.5 atm)
Rearranging the equation, we get:
T2 = T1 / (1 + ΔHvap/R * (ln(P2/P1)/T1))
Substituting the values, we get:
T2 = 373 K / (1 + 40.7 kJ/mol / (8.314 J/mol*K) * (ln(2.5/1)/373 K))
T2 ≈ 384.1 K
Therefore, the boiling point of water at 2.5 atm is approximately 384.1 K or 111 C.
