Question

The normal boiling point of water is $373K$. Vaunt pressure of water at temperature $T$ is ${\text{19mmHg}}$. Its enthalpy of vaporization is ${\text{40}}{\text{.67 kg/mole}}$ there temperature $T$ would be.
(use:\;\log $$$2 = 03.\;R:\;8.3\;J{K^{ - 1}}\;MO{l^{ - 1}})$$ A. 250K$B.$291.4K$C.$230K$D.$290K$

Answer 1

The above problem is solved by clauses -clapeyron equation. First we know about the equation, then we solve the problem.

Complete step by step solution:
Normal boiling point is the temperature at which the presser equals to atmospheric answer which is $1$ atm on ${\text{760mmHg}}{\text{.}}$
the one asked to calculate the temperature at which the capon presser is 19mm hg.
This type of problem and solved by using clausius clapeyron equation:
In $\left( {\dfrac{{P2}}{{P1}}} \right) = \dfrac{{\vartriangle Havn}}{R}(\dfrac{1}{{T1}} - \dfrac{1}{{T2}})$
${P_1} = 760mmHg$
${T_1} = 373K$
${P_2} = 19mmHg,\;{T_2} = t$
$\Delta {H_{vap.}} = 40.67KJ/mol$
$R$ is universal gas constant and its value in terms of $KJ$ is ${\text{0}}{\text{.008134 KJ/male}}{\text{.k}}$
Lets plug in the values in the equation and solve it:
$ln(\dfrac{{19}}{{760}}) = \dfrac{{40.67}}{{0.008314}}\;(\dfrac{1}{{373}} - \dfrac{1}{t})$
Or, $- 3.69 = 4891.75(0.00268 - \dfrac{1}{t})$
Or, $\dfrac{{ - 3.69}}{{4891.75}} = (0.00268 - \dfrac{1}{t})$
Or, $- 0.000754 = (0.00268 - \dfrac{1}{t})$
Let’s keep the similar terms on same side:
$\dfrac{1}{t} = 0.00268 + 0.000754$
Or, $\dfrac{1}{t} = 0.003434$
Or, $t = 291.20$
$\therefore \;t = 291.20k$
So, the value of the temperature $\dfrac{T}{{(t)}}$would be
$291.20k,$
In the options which is B.291.4k
so, The normal boiling point of water is 373k.
Vapor pressure of water at temperature T is $19mmHg.$
If enthalpy of vaporization is ${\text{40}}{\text{.67 KJ/mole}}$, then temperature $T$ would be $B.291.4K$

The connection option is B.

Additional information:
Celsius -clopeynon equation pen trains the relationship between the pressure and temperature for conditions of equilibrium between two phases.
The two phares could be vapor and solid form sublimation of solid and liquid for melting.
The Celsius statement: it is imposingly to a constant $R$ device which operates $R$ cycle and produces no other form a cooler body to a hotter body. Equivalence of the clausius and kelvin Planck statements.

Note: we must eane of that unit should be same of the gas constant $R$ and $\Delta {H_{vap.}}$
The Celsius inequality applies to any heat engine cycle and implies a negative change in entropy on the cycle. That is, the entropy given to the environment dewing the cycle is large then the entropy talons transferred to the engine by heat from the hot hesenvoin.