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2 years ago

• How did your experimental absolute zero value compare to the accepted value?

Chemistry

1 answer:

Murrr4er [49]2 years ago

5 0

The experimental absolute zero value is less when compared to the accepted value of absolute zero.

<h3>What is absolute zero?</h3>

Absolute zero is defined as the temperature in which the lowest energy possible is attained in a thermodynamic system.

Absolute zero temperature has an accepted values of 0 Kelvin or -273.15 degrees Celsius.

At absolute zero, it is assumed that the volume of an ideal gas becomes zero. However, it has not been possible to cool any gas to absolute zero.

Based on the graph of temperature against volume of gases, the experimental absolute zero extrapolated from the graph where volume of the gases becomes zero is -285 degrees Celsius.

Therefore, the experimental absolute zero value is less when compared to the accepted value.

Learn more about absolute zero at: brainly.com/question/1191114

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Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propa

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Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of propane at $25.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.

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