The vapour pressure of water at `100^(@)C` is `760mm`. What is the vapour pressure at `90^(@)C` if `triangle_(vap)H` of water is `41.25 kJ mol^(-1)`?

Explanation

log `(P_(2))/(P_(1)) = (Delta_("vap") H)/(2.303 R) ((T_2 - T_1)/(T_1 T_2)) implies log "" (P_(2))/(760 mm) = (41.25 xx 10^(3) J mol^(-1))/(2.303 xx 8.314 J K^(-1) mol^(-1)) ((363 - 373 K)/((363 K) (373 K)))`
log `P_(2) "log" 760 = 0.1591 implies "log" P_(2) = 2.8808 - 0.1591 implies (P_(2)` = antilog 2.7217 ) = 529.6 mm