To estimate the boiling point of water under an external pressure of 0.80 atm, we can make use of an empirical relationship known as the Antoine equation, which relates the vapor pressure of a substance to its temperature. The Antoine equation is typically expressed as:
[tex]\[ \log_{10}(P) = A - \frac{B}{T + C} \][/tex]
Where:
- [tex]\( P \)[/tex] is the vapor pressure,
- [tex]\(T\)[/tex] is the temperature in degrees Celsius,
- [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are substance-specific constants.
For water, the Antoine constants are:
- [tex]\( A = 8.07131 \)[/tex],
- [tex]\( B = 1730.63 \)[/tex],
- [tex]\( C = 233.426 \)[/tex].
Given the external pressure [tex]\( P \)[/tex] is 0.80 atm, we need to solve for the temperature [tex]\( T \)[/tex] in Celsius where this relationship holds true.
To do this, we rearrange the Antoine equation for [tex]\( T \)[/tex]:
[tex]\[ \log_{10}(P) = A - \frac{B}{T + C} \][/tex]
Rearranging to solve for [tex]\( T \)[/tex]:
[tex]\[ T + C = \frac{B}{A - \log_{10}(P)} \][/tex]
[tex]\[ T = \frac{B}{A - \log_{10}(P)} - C \][/tex]
Plugging in the values:
[tex]\[ \log_{10}(0.80) \approx -0.09691 \][/tex]
[tex]\[ T = \frac{1730.63}{8.07131 - (-0.09691)} - 233.426 \][/tex]
Let's calculate this step-by-step:
- Calculate the denominator:
[tex]\( 8.07131 - (-0.09691) = 8.07131 + 0.09691 = 8.16822 \)[/tex]
- Then calculate the fraction:
[tex]\( \frac{1730.63}{8.16822} \approx 211.88 \)[/tex]
- Finally, subtract the constant [tex]\( C \)[/tex]:
[tex]\( 211.88 - 233.426 = -21.546 \approx -21.55 \)[/tex]
The estimated boiling point of water under an external pressure of 0.80 atm is therefore approximately [tex]\( -21.55^\circ C \)[/tex].
To convert this to Kelvin, we add 273.15:
[tex]\[ -21.55 + 273.15 = 251.60 \, \text{K} \][/tex]
Hence, the boiling point of water under an external pressure of 0.80 atm is approximately [tex]\( -21.55^\circ C \)[/tex] or [tex]\( 251.60 \, \text{K} \)[/tex].
