The normal boiling point of water is 373 K(at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporisation is 40.656 kJ the boiling point of water at 23 mm atmospheric pressure will be

Question

The normal boiling point of water is 373 K(at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporisation is 40.656 kJ the boiling point of water at 23 mm atmospheric pressure will be (A) 250 K (B) 298 K (C) 51.6 K (D) 12.5 K

Tardigrade Admin · Accepted Answer

Applying clausius clapeytron equation log ⋅ (P2/P1)=(Δ HV/2.303 R)[(T2-T1/T1 × T2)] log ⋅ (760/23)=(40656/2.303 × 8.314)[(373-T1/373 T1)] This gives T1=294.4 K.

Q. The normal boiling point of water is 373 K(at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporisation is 40.656 kJ the boiling point of water at 23 mm atmospheric pressure will be

250 K

298 K

51.6 K

12.5 K

Applying clausius clapeytron equation
$lo g \cdot \frac{P _{2}}{P _{1}} = \frac{Δ H _{V}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} \times T _{2}}]$
$lo g \cdot \frac{760}{23} = \frac{40656}{2.303 \times 8.314} [\frac{373 - T _{1}}{373 T _{1}}]$
This gives $T_{1} = 294.4 K$ .

Q. The normal boiling point of water is 373 K(at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vaporisation is 40.656 kJ the boiling point of water at 23 mm atmospheric pressure will be

250 K

298 K

51.6 K

12.5 K

Applying clausius clapeytron equation log⋅P1​P2​​=2.303RΔHV​​[T1​×T2​T2​−T1​​] log⋅23760​=2.303×8.31440656​[373T1​373−T1​​] This gives T1​=294.4K.

Applying clausius clapeytron equation
$lo g \cdot \frac{P _{2}}{P _{1}} = \frac{Δ H _{V}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} \times T _{2}}]$
$lo g \cdot \frac{760}{23} = \frac{40656}{2.303 \times 8.314} [\frac{373 - T _{1}}{373 T _{1}}]$
This gives $T_{1} = 294.4 K$ .