Given : Planet’s atmospheric pressure = 380 mmHg
Clausius- Clapeyron Equation
R = Ideal Gas Constant approx≈ 8.314 kPa * L/mol*K or J / mol * k
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Solve for L:
ln( 52.3/22.1) = - L /(8.314 \frac{J}{ mol * k}) * (\frac{1}{380K} - \frac{1}{328K}) ln(52.322.1)=−L8.314Jmol⋅k⋅(1380K−1328K)
ln(2.366515837…) * (8.314 \frac{J}{ mol * k}) / (\frac{1}{380K} - \frac{1}{328K}) = -Lln(2.366515837…)⋅8.314Jmol⋅k1380K−1328K=−L
0.8614187625 * (8.314\frac{ J}{mol * k}) / (\frac{1}{380K} - \frac{1}{328K}) = -L 0.8614187625⋅8.314Jmol⋅k1380K−1328K=−L
0.8614187625 * (8.314\frac{ J}{mol * k})/(-4.1720154*10^-4K)0.8614187625⋅8.314Jmol⋅k−4.1720154⋅10−4K
L \approx 17166 \frac{J}{mol} L≈17166Jmol
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We know that a substance boils when it’s vapor pressure is greater than or equal to atmospheric pressure thus, we need to solve for the temperature at which the vapor pressure is greater than or equal to 380mmHg:
Solve for T:
ln(380/52.3) = (-17166 \frac{J}{mol}) / (8.314\frac{ J}{mol * k}) * (1/T - frac{1}{380K})ln(38052.3)=−17166Jmol8.314Jmol⋅k⋅(1T−1380K)
ln(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) = (1/T - 1/380K) ln(38052.3)⋅8.314Jmol⋅k−17166Jmol=(1T−1380K)
[ln(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) ] + (1/380) = (1/T) [ln(38052.3)⋅8.314Jmol⋅k−17166Jmol]+(1380)=(1T)
T = 1/ [ [ln(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) ] + (1/380)] T=1[ln(38052.3)⋅8.314Jmol⋅k−17166Jmol]+(1380)
T approx 598.4193813 K approx 598 K T≈598.4193813K≈598K
Thus the boiling point is approx 598 K ≈598K
