Question

Unknown gas a vapor pressure of 52.3mmHg at 380K and 22.1mmHg at 328K on a planet where atmospheric pressure is 50% of Earths. What is the boiling point of unknown gas?

please help with steps

Answer 1

The boiling point is 598 K

Explanation:

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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Solve for L:

`ln( 52.3/22.1) = - L /(8.314 \frac{J}{ mol * k}) * (\frac{1}{380K} - \frac{1}{328K})`

`ln(2.366515837…) * (8.314 \frac{J}{ mol * k}) / (\frac{1}{380K} - \frac{1}{328K}) = -L`

`0.8614187625 * (8.314\frac{ J}{mol * k}) / (\frac{1}{380K} - \frac{1}{328K}) = -L`

`0.8614187625 * (8.314\frac{ J}{mol * k})/(-4.1720154*10^-4K)`

`L \approx 17166 \frac{J}{mol}`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) = (1/T - 1/380K)`

`[ln(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) ] + (1/380) = (1/T)`

`T = 1/ [ [ln(380/52.3) * (8.314\frac{ J}{mol * k}) / (-17166 \frac{J}{mol}) ] + (1/380)]`

`T approx 598.4193813 K approx 598 K`

Thus the boiling point is `approx 598 K`