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3 years ago

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.

Physics

1 answer:

Novosadov [1.4K]3 years ago

3 0

Answer:

389.78681 K

Explanation:

$P_1$ = Initial pressure = 55.1 mmHg

$P_2$ = Final pressure = 1 atm = 760 mmHg

$T_2$ = Boiling point

$T_1$ = Initial temperature = 35°C

$\Delta H_{vap}$ = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

$ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K$

The normal boiling point of the substance is 389.78681 K

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2 years ago

Resistor A has twice the resistance of resistor B. The two are connected in series and a potential difference is maintained acro

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Answer:

<em>The thermal energy dissipated in A would be twice that in B</em>

Explanation:

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