Question

Find the temperature at which ethanol boils on a day in the mountains when the barometric pressure is 547 mmHg. (Given: The heat of vaporization of ethanol is 39.3 kJ/mol; the normal boiling point of ethanol is 78.3°C; (R = 8.314J/ K • mol).

Answer 1

lenghty question. i have attached the answer! let me know if you have any question. 

Answer 2

The temperature at which ethanol boils on a day in the mountains when the barometric pressure is 547 mmHg is 342.92 K (69.92 °C).  

We can calculate the temperature at which ethanol boils with the Clausius-Clapeyron equation:

$ln(\frac{P_{1}}{P_{2}}) = \frac{\Delta H_{v}}{R}(\frac{1}{T_{2}} - \frac{1}{T_{1}})$  

Where:

P₁: is the pressure 1 = 1 atm = 760 mmHg (atmospheric pressure)

P₂: is the pressure 2 = 547 mmHg

T₁: is the normal boiling point at P₁ = 78.3 °C = 351.3 K

T₂: is the boiling point at P₂ =?

R: is the gas constant = 8.314 J/K*mol  

$\Delta H_{v}$: is the heat of vaporization of ethanol = 39.3 kJ/mol

Hence, the new boiling point of ethanol is:

$ln(\frac{760 mmHg}{547 mmHg}) = (\frac{39.3 \cdot 10^{3} J/mol}{8.314 J/K*mol})*(\frac{1}{T_{2}} - \frac{1}{351.3 K})$  

Solving the above equation for T₂ we have:

$T_{2} = 342.92 K = 69.92 ^{\circ}C$

Therefore, ethanol will boil at 342.92 K (69.92 °C) when the barometric pressure is 547 mmHg.

You can find more about the Clausius-Clapeyron equation here https://brainly.com/question/13738576?referrer=searchResults

I hope it helps you!